Mathematics IA Algebra Practice Questions Questions marked with a * are more challenging, or contain further applications which are not examinable. 1. Matrices and Linear Equations Matrices 1 1 8 4 2 5 −4 0 , B = 1. Let A = 2 , C = , D = 3 5 7 2 4 6 0 0 −1 −4 −2 and E = . −1 Evaluate each of the following matrices if they exist: (i) At (ii) AB (iii) BAC (iv) A + C (v) A + B t (vi) ED (vii) AEDB t C t (viii) C 3 1 2 −1 t Solution: (i) A = 1 0 −4 70 −56 (iii) BAC = 20 −16 10 8 8 8 4 (ii) AB = 16 −16 −20 −26 9 3 (iv) A + C is not defined. (v) A + B t = 6 4 1 2 −6 −10 −14 (vi) ED = −3 −5 −7 −54 0 125 −100 . (vii) AEDB t C t = −72 0. (viii) C 3 = 0 0 108 0 a b c . Find expressions for AAt and At A. 2. Let A = d e f a2 + b2 + c2 ad + be + cf Solution: = . ad + be + cf d2 + e2 + f 2 2 a + d2 ab + de ac + df At A = ab + de b2 + e2 bc + ef . ac + df bc + ef c2 + f 2 AAt 3. Let A, B, C and D be matrices which satisfy the following conditions: AB t C 2 Dt is defined and is a 3 × 3 matrix, A has 7 columns, and B has 5 rows. Find the sizes of each of the matrices A, B, C and D. 1 Solution: Note that for C 2 to be defined then C must be square, let its size be n × n. Then for C 2 Dt to be defined then Dt must be n × 3 (since the overall product is 3 × 3. Also, since B has 5 rows, then B t has 5 columns, so for B t C 2 to be defined m = 5. B t must have 7 rows since AB t is defined and A has 7 columns. Also, A must have 3 rows since the overall product is 3 × 3. Putting all of this together we have that A is 3 × 7, B is 5 × 7, C is 5 × 5 and D is 3 × 5. 4. Let A be an m × n matrix, and suppose there exist n × m matrices C and D such that CA = In and AD = Im . Prove that C = D. Solution: Consider the product CAD = C(AD) = CIm = C (by question 4), but also CAD = (CA)D = In D = D. Thus C = D. 5. A square matrix A is said to be symmetric if A = At . Let A be an m × n matrix. Prove that AAt and At A are symmetric matrices. Solution: Note that both AAt and At A are square matrices. To show that AAt is symmetric, we need to show that (AAt )t = AAt . We have (AAt )t = (At )t At = AAt , since (At )t = A. Similarly we have (At A)t = At (At )t = At A. 3 8 1 2 . Find all possible matrices C and B = 6. Let A = 1 −2 −1 −2 such that both AB = AC and BA = CA. Solution: Obviously are there any oth- C = B is one solution but x + 2z y + 2w x y 5 4 . AB = = AC = . ers? Let C = −x − 2z −y − 2w z w −5 −4 Equating the matrix entries gives a set of 4 equations, but note that it is actually two equivalent pairs, so if z = s and w = t then x = 5 − 2s and y = 4 − 2t. Doing the same thing with BA = CA gives two more equations x = −5 + y and z = 3 + w. The second says that we can replace s with 3 + t which gives x = −1 − 2t. Note that the equation x = −5 + y is satisfied in terms of t so this equation gives no further −1 − 2t 4 − 2t information, thus the matrix C can be of the form 3+t t for any choice of t ∈ R. 2 7.* Two square matrices A and B are said to commute if AB = BA. Find all 2 × 2 matrices A with commute with every 2 × 2 matrix. a b (Hint: let A = ; choose some very simple matrices B — for c d instance matrices with almost every entry equal to 0 — and investigate what conditions must be satisfied by a, b, c and d in order to have AB = BA.) a b Solution: Suppose that A = commutes with every 2 × 2 c d 1 0 matrix. In particular A commutes with . Therefore 0 0 a b 1 c d 0 a =⇒ c 0 1 = 0 0 0 a = 0 0 0 0 a b c d b 0 =⇒ b = c = 0 0 1 . Therefore Similarly A commutes with 1 0 a 0 0 0 d 1 0 =⇒ d 0 1 = 1 0 0 a = a 0 a 0 0 d d 0 1 0 =⇒ a = d a 0 . Note that such a matrix commutes with every 2 × 2 0 a matrix (because such an A is a scalar multiple of the identity matrix I2 ). Therefore A commutes with every 2 × 2 matrix precisely when A a 0 . is diagonal, i.e. A = 0 a So A = Linear Equations 8. Solve the following systems of linear equations: (a) 2x1 + 6x2 4x1 − 3x2 = 12 = 19 (b) x1 − 3x2 + 2x3 2x2 − x3 x1 + x3 = -10 =6 =0 Solution: These systems of equations may be solved by the matrix method, or just by substitution. (a) The first equation can be rewritten as x1 = 6 − 3x2 , substituting into the second then gives 24 − 15x2 = 19 which gives the solution (x1 , x2 ) = (5, 1/3). 3 (b) The second and third equations give x2 = (x3 +6)/2 and x1 = −x3 , so we can substitute into the first equation to get −x3 − 32 (x3 + 6) + 2x3 = −10 which gives the solution (x1 , x2 , x3 ) = (−2, 4, 2). 9. Solve the following systems of linear equations: (a) 2x1 + 6x2 x1 + 3x2 =3 =1 (b) x1 + 2x2 x2 − x3 =4 =2 Solution: (a) Multiplying the second equation by 2 gives 2x1 + 6x2 = 2 which clearly contradicts the first equation so there are no solutions. Alternatively, in matrix form we have the augmented matrix 2 6 | 12 and halving the first row, and then subtracting it from 1 3 | 1 1 3 | 6 the second row gives , so the last row says that 0 = −5, 0 0 | −5 so there are no solutions. (b) This system has more variables than equations, so there will not be a unique solution. Let x3 = t then x2 = 2 + t, and x1 = 4 − 2x2 = 4 − 2(2 + t) = −2t, so the solutions are (x1 , x2 , x3 ) = (−2t, 2 + t, t) for any t ∈ R. 10. Consider the system of linear equations, x1 − 2x2 = 1 3x1 + ax2 = 4 (a) If this system has a unique solution, (x1 , x2 ) = (5, 2), then what is a? (b) If this system has no solutions then what is a? Solution: (a) The second equation is 15 + 2a = 4, so a = −11/2. (b) The first equation says that x1 = 1 + 2x2 . Substituting into the second we get 3 + 6x2 + ax2 = 4 or (6 + a)x2 = 1. This will give a solution for x2 (and hence also for x1 ) unless 6 + a = 0,that is, a = −6, in which case there are no solutions. Gauss-Jordan Elimination 11. For each of the following matrices in reduced row echelon form, give the set of solutions to the corresponding system of equations (you can call the variables x1 , x2 , . . . ). 1 0 | 0 1 0 0 | 0 (a) 0 1 | 1 (b) 0 1 0 | 1 0 0 | 0 0 0 0 | 0 4 1 2 0 3 0 | 4 (c) 0 0 1 2 0 | 3 (d) 1 4 8 3 | −1 0 0 0 0 1 | 2 Solution: (a) (x1 , x2 ) = (0, 1) (b) Let x3 = t. Then (x1 , x2 , x3 ) = (0, 1, t) = (0, 1, 0) + t(0, 0, 1). (c) Let x2 = s and x4 = t. Then we have x1 + 2x2 + 3x4 = 4, so x1 = 4 − 2s − 3t. Also x3 + 2x4 = 3 so x3 = 3 − 2t and x5 = 2. Putting these together we get (x1 , x2 , x3 , x4 , x5 ) = (4−2s−3t, s, 3−2t, t, 2) = (4, 0, 3, 0, 2)+s(−2, 1, 0, 0, 0)+ t(−3, 0, −2, 1, 0). (d) Let x2 = r, x3 = s and x4 = t. Then (x1 , x2 , x3 , x4 ) = (−1−4r−8s−3t, r, s, t) = (−1, 0, 0, 0)+r(−4, 1, 0, 0)+ s(−8, 0, 1, 0) + t(−3, 0, 0, 1). 12. For each matrix below, use reduced row echelon form: 0 1 | (a) 0 0 | 1 0 | row operations to put the matrix into 0 1 0 1 | 1 1 (b) 0 2 0 | 2 0 1 1 0 | 1 (c) 1 2 3 | 4 for some α 6= 0 α 0 1 | 2 Solution: (a) Swap R1 and R3 , and then swap R2 and R3 to get 1 0 | 0 0 1 | 0. 0 0 | 1 1 0 1 | 1 (b) Do R3 → R3 − R1 and R2 → 1/2 R2 ; giving 0 1 0 | 1. 0 1 −1 | 0 1 0 1 | 1 Next do R3 → R3 − R2 to get 0 1 0 | 1 then R3 → −R3 : 0 0 −1 | −1 1 0 1 | 1 1 0 0 | 0 0 1 0 | 1 and R1 → R1 − R3 gives the rref 0 1 0 | 1 0 0 1 | 1 0 0 1 | 1 1 2 3 | 4 (c) R2 → R2 − αR1 gives and then 0 −2α 1 − 3α | 2 − 4α 1 2 3 | 4 R2 → − 1/2α R2 gives and finally 0 1 − 1/2α + 3/2 | − 1/α + 2 1/α 2/α 1 0 | R1 → R1 − 2R2 gives − 1 3 − 1 0 1 /2α + /2 | /α + 2 5 13. Find the solution space of the following systems of linear equations: (a) 4x + 2y + 5z = 8 (b) 4x + 2y + 4z = 8 2x + 4y + 4z = 1 x + 4y + 4z −2y − z = 2 Solution: =1 2x − 6y − 3z = 6 (a) The corresponding augmented matrix is 4 2 5 | 8 2 4 4 | 1 0 −2 −1 | 2 1 0 1 | 5/2 which row reduces to 0 1 1/2 | −1. 0 0 0 | 0 Let z = t. Then x = 5/2 − t and y = −1 − 1/2t so the solutions are (x, y, z) = ( 25 − t, −1 − 1/2t, t) = (5/2, −1, 0) + t(−1, −1/2, 1) for t ∈ R. 4 2 4 | 8 4 | 1 which (b) The corresponding augmented matrix is 1 4 2 −6 −3 | 6 15 1 0 0 | /7 row reduces to 0 1 0 | −2/7. Thus there is a unique solution, 0 0 1 | 0 15 2 (x, y, z) = ( /7, − /7, 0). 14. If a 5 × 5 augmented matrix for a system of 5 linear equations in 4 unknowns is row reduced: (a) How many pivots must there be for a unique solution? (b) How many zero rows must there be for a unique solution? Solution: (a) 4 (one for each unknown). (b) 1, there are 5 rows and 4 pivots so there must be exactly one row with no pivot. 15. For which values of c have (i) no solutions, (ii) a When solutions exist, c). (a) x + cy cy does each of the following system of equations unique solution, (iii) infinitely many solutions? write down what they are (possibly in terms of =0 (b) =1 x + cz = c x + cy + cz = 0 x + cz + z = 1 6 1 c | 0 Solution: (a) In matrix form this is . If c = 0 then 0 c | 1 1 0 | −1 there are no solutions. If c 6= 0 then we can row reduce it to , 0 1 | 1/c so there is a unique solution. Thus there are no values of c which give infinitely many solutions. 1 0 c | c c | 0. We can subtract the (b) In matrix form this is 1 c 1 0 c+1 | 1 1 0 c | c first row from each of the second and third rows to get 0 c 0 | −c . 0 0 1 | 1−c 1 0 0 | 0 If c = 0 then the reduced row echelon form is 0 0 1 | 1 which 0 0 0 | 0 has infinitely many solutions (x1 , x2 , x3 ) = (0, 0, 1) +t(0, 1, 0). 1 0 c | c If c 6= 0 then we can divide the second row by c to get 0 1 0 | −1 0 0 1 | 1−c and thensubtract c times row 3 from row 1 to get the reduced row eche 1 0 0 | c2 lon form 0 1 0 | −1 which gives a unique solution (x1 , x2 , x3 ) = 0 0 1 | 1−c (c2 , −1, 1 − c). 16. Suppose 3 foods are used to make a meal. The foods have the following units per gram of vitamin B and vitamin C respectively. Food 1 Food 2 Food 3 Vitamin B Vitamin C 12 18 8 10 4 2 By setting up a system of linear equations, determine if it is possible to combine these foods into a meal of 500 grams with precisely 2000 units of Vitamin B and 1000 units of Vitamin C. If this is possible, how many ways can it be done? Solution: Let x1 equal the number of grams of Food 1, x2 equal the number of grams of Food 2 and x3 equal the number of grams of Food 3 in the meal. Since the meal is of weight 500 grams, we must have x1 + x2 + x3 = 500. On the other hand, from the table, the meal will contain 12x1 + 8x2 + 4x3 units of Vitamin B, therefore 12x1 +8x2 +4x3 = 2000. Similarly 18x1 +10x2 +2x3 = 1000. Therefore x1 , x2 and x3 satisfy the linear system x1 + x2 + x3 = 500 12x1 + 8x2 + 4x3 = 2000 18x1 + 10x2 + 2x3 = 1000 7 The augmented matrix of the system is 1 1 1 | 500 12 8 4 | 2000 18 10 2 | 1000 Do the row operations R2 → R2 − 12R1 , R3 → R3 − 18R1 to obtain the augmented matrix 1 1 1 | 500 0 −4 −8 | −4000 0 −8 −16 | −8000 Now do the row operation R2 → −1/4R2 to obtain 1 1 1 | 500 0 1 2 | 1000 0 −8 −16 | −8000 The row operations R1 → R1 − R2 and R3 → R3 + 8R2 then put the augmented matrix into reduced row echelon form: 1 0 −1 | −500 0 1 2 | 1000 0 0 0 | 0 There are infinitely many solutions of this system: x1 = t − 500, x2 = 1000 − 2t, x3 = t. However, we seek a solution with x1 ≥ 0, x2 ≥ 0 and x3 ≥ 0 (you can’t eat negative amounts of food). Therefore we must have t − 500 ≥ 0 and 1000 − 2t ≥ 0. Hence t = 500 and so there is only one way to make the meal: just take a meal with 500 grams of Food 3. 17.* Let A be an m × n matrix. If the system of equations Ax = b has a solution for every b, show that there is an n × m matrix D so that AD = Im . 1 0 0 0 1 0 , . , . . . . ∈ Rm consecutively . Solution: Choose b = . . . 0 1 0 and let x1 , x2 , . . . xm be the corresponding solutions of Ax = b for these choices. Then the required matrix is D = [x1 |x2 | . . . |xm ]. 18.* Let A be an m × n matrix and b be a vector in Rm . Prove that there cannot be exactly two solutions to Ax = b. 8 Solution: Let x1 and x2 satisfy Ax1 = b and Ax2 = b, with x1 6= x2 . Then A(2x1 − x2 ) = 2Ax1 − Ax2 = 2b − b = b. Note also that 2x1 − x2 cannot be equal to either x1 or x2 as this would imply that x1 = x2 . More generally note that A(x2 − x1 ) = 0 and so At(x2 − x1 ) = 0 for any t ∈ R, thus we have A(x1 + t(x2 − x1 )) = b, so (1 − t)x1 + tx2 is a solution to Ax = b for any t ∈ R. Hence given two distinct solutions we can then find infinitely many solutions. Linear Combinations 19. Decide whether each of the following vectors is in the span of {(1, 0, 1, 2, 0), (2, 1, 1, 0, 1), (0, 2, 2, 1, 1)} and where possible write the vector as a linear combination of the vectors in the span: (a) (7, −2, 1, 4, 0) (b) (4, 3, 5, 5, 1) (c) (−1, −2, −1/2, 5/2, −3/2) (d) (−1/2, 13, 17/2, −7/2, 15/2). Solution: To write a vector v as a linear combination of the vectors in we need to looks for solutions to Ax = v where A = the span 1 2 0 0 1 2 1 1 2. We can do all of the calculations at once by row reducing 2 0 1 0 1 1 1 2 0 | 7 4 −1 − 1/2 0 1 2 | −2 3 −2 13 1 1 2 | 1 5 − 1/2 17/2 . 2 0 1 | 4 5 5/2 − 7/2 0 1 1 | 0 1 − 3/2 15/2 1 0 0 | 3 0 0 − 9/2 0 1 0 | 2 0 0 2 11/2 . This row reduces to 0 0 1 | −2 0 0 0 0 0 | 0 1 0 0 0 0 0 | 0 0 1 0 From this we can see that the vectors in (b) and (c) are not in the span as there are no solutions. For (a) we have (7, −2, 1, 4, 0) = 3(1, 0, 1, 2, 0) + 2(2, 1, 1, 0, 1) − 2(0, 2, 2, 1, 1) and for (d) we have ( − 1/2, 13, 17/2, − 7/2, 15/2) = − 9/2(1, 0, 1, 2, 0)+2(2, 1, 1, 0, 1)+11/2(0, 2, 2, 1, 1). 1 5 20. Let A = −1 9 1 3 of the columns of 2 5 , and let W be the set of all linear combinations 1 A. 9 (a) Show that (4, 10, 2)t is in W (you should be able to do this by inspection, that is, without any row operations). (b) Solve the homogeneous system Ax = 0. (c) Given that (1, 2, −1)t satisfies Ax = b, what is the vector b? (d) Write down the general solution of Ax = b, without directly solving this system. 4 Solution: (a) By inspection we see that 10 is given by the follow2 1 5 2 ing linear combination of the columns: 0 −1 + 0 9 + 2 5. 1 3 1 (b) Row reducing gives: 1 5 2 −1 9 5 1 3 1 1 5 2 0 14 7 0 −2 −1 1 5 2 1 0 1 2 0 0 0 1 0 − 12 1 0 1 2 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 2 Let x3 = t, then x1 = 12 t and x2 = − 12 t, so (x1 , x2 , x3 ) = t − 21 , for 1 any t ∈ R. 9 1 (c) Multiplying A 2 gives b = 12. 6 −1 (d) The general solution of Ax = b is given by adding together the general solution of Ax= 0 anda particular solution of Ax = b, in 1 1 2 this case we have x = 2 + t − 12 , t ∈ R. −1 1 1 3 −4 b1 2 −6 and b = b2 . 21. Let A = 3 −5 −1 8 b3 (a) Show that the equation Ax = b is not consistent for all b ∈ R3 . (b) Find an equation for the set of all b for which Ax = b is consistent. Describe the set geometrically. 10 (c)* Use your answer to part (b) to deduce that 3 −4 0 1 1 span 3 , 2 , −6 = span 0 , 1 . −5 −1 8 1 −2 Solution: We row reduce the corresponding augmented matrix: b1 1 3 −4 3 2 −6 b2 −5 −1 8 b3 1 3 −4 b1 0 −7 6 b2 − 3b1 0 14 −12 b3 + 5b1 1 3 −4 b1 0 1 −6/7 −(b2 − 3b1 )/7 0 0 0 b3 + 5b1 + 2(b2 − 3b1 ) (there is no need to reduce further if we are just checking for consistency). This system is not consistent for all b, as it is not always true that b3 + 5b1 + 2(b2 − 3b1 ) = −b1 + 2b2 + b3 = 0, for example let 1 b = 0, then the last row says that 0 = −1. 0 (b) For this system to be consistent we require −b1 + 2b2 + b3 = 0. This equation defines a plane in R3 . (c) Observe that span{(1, 3, −5), (3, 2, −1), (−4, −6, 8)} is precisely the set of vectors b such that the system Ax = b is consistent. In part (b) we observed that the set of vectors b = (b1 , b2 , b3 ) such that the system Ax = b is consistent is precisely the set {(b1 , b2 , b3 ) ∈ R3 | b3 = b1 − 2b2 }. In other words, the set of vectors b for which the system Ax = b is consistent is the set {(b1 , b2 , b1 − 2b2 )| b1 , b2 ∈ R}. Since (b1 , b2 , b1 − 2b2 ) = b1 (1, 0, 1) + b2 (0, 1, −2) we see that this last set is precisely span{(1, 0, 1), (0, 1, −2)}. 22.* Let u, v and w be vectors in Rn . Prove that w ∈ span{u, u − v, v − w}. (In other words, you need to prove that w is a linear combination of u − v, v − w and u.) Solution: One to do this is by trial and error. A more systematic way is as follows. We want to find scalars x, y and z such that xu + y(u − v) + z(v − w) = w. In other words, we want the vector equation (x + y)u + (z − y)v − zw = w 11 to be satisfied. This is clearly satisfied if we choose x, y and z so that x + y = 0, z − y = 0 and −z = 1. Therefore take z = −1, y = −1 and x = 1. As a check we see that indeed u − (u − v) − (v − w) = w. 23.* Let v1 , v2 , . . . , vk be a sequence of vectors in Rn . Suppose that vk+1 ∈ span{v1 , . . . , vk }. Prove that span{v1 , . . . , vk } = span{v1 , . . . , vk , vk+1 }. (In other words you need to show that every linear combination of the vectors v1 , . . . , vk is also a linear combination of the vectors v1 , . . . , vk , vk+1 and vice versa). Solution: Let w ∈ span{v1 , . . . , vk } so that w = x1 v1 + · · · + xk vk for some scalars x1 , . . . , xk . But then w is also a linear combination of v1 , . . . , vk , vk+1 since w = x1 v1 + · · · + xk vk + 0vk+1 . Conversely, suppose that w ∈ span{v1 , . . . , vk , vk+1 }, so that there exist scalars x1 , . . . , xk+1 such that w = x1 v1 + · · · + xk vk + xk+1 vk+1 . Since vk+1 ∈ span{v1 , . . . , vk } we may write vk+1 = y1 v1 + · · · + yk vk for some scalars y1 , . . . , yk . But then w = x1 v1 + · · · + xk vk + xk+1 vk+1 = x1 v1 + · · · + xk vk + xk+1 (y1 v1 + · · · yk vk ) = (x1 + xk+1 y1 )v1 + · · · + (xk + xk+1 yk )vk and hence w is a linear combination of v1 , . . . , vk . Homogeneous Equations 24. (a) Use Gauss-Jordan elimination to solve the following homogeneous system x1 − x2 − 4x3 = 0, x1 + x2 + 4x3 = 0, 2x2 + x3 = 0. (b) Find by inspection a particular solution of the system x1 − x2 − 4x3 = −4, x1 + x2 + 4x3 = 6, 2x2 + x3 = 3, and hence write down the general solution, using the results of (b). 12 Solution: (a) We row reduce the corresponding augmented matrix: 1 −1 −4 0 1 1 4 0 0 2 1 0 1 −1 −4 0 0 2 8 0 0 2 1 0 1 0 0 0 0 1 4 0 0 0 −7 0 1 0 0 0 0 1 0 0 0 0 1 0 x1 0 which shows that there is a unique solution x2 = 0. x3 0 (b) By inspection we see that substituting in x1= x2 =x3= 1 gives x1 0 1 the desired result, hence the general solution is x2 = 0 + 1 = 0 1 x3 1 1. 1 Elementary Matrices and the Inverse Matrix 25. Let A,B and C be invertible n × n matrices. Find expressions in terms of A, B, C and their inverses for the inverses of the following matrices: (i) ABC (ii) AB −1 A (iii) 3At BC 2 (iv) −BA−1 C t A Solution: (i) C −1 B −1 A−1 . (ii) A−1 BA−1 (iii) 31 (C −1 )2 B −1 (A−1 )t (iv) −A−1 (C −1 )t AB −1 . 26. Suppose that A and B are invertible matrices. Is (A + B)−1 = A−1 + B −1 ? Give reasons to justify your answer. Solution: Let A = B. Then A+B = 2A, so (A+B)−1 = 12 A−1 . On the other hand A−1 + B −1 = 2A−1 . Thus it is not true in general that (A + B)−1 = A−1 + B −1 . (There are various other counterexamples which could be used). 13 27. Which of the following are elementary matrices? 1 0 0 0 1 1 100 0 −1 1 0 (i) (ii) (iii) (iv) (v) 0 −1 0 1 0 0 1 1 0 0 0 Solution: (i) No, it is not square. (ii) Yes, it corresponds to swapping the two rows. (iii) Yes, it corresponds to adding 100 times row 2 to row 1. (iv) No, this would correspond to two separate row operations (a swap and multiplying a row by -1). (v) No, this corresponds to multiplying the second row by 0 which is not an elementary row operation. 0 28. Row reduce A = 2 1 down the elementary Solution: 1 0 2 2 to reduced row echelon form and write 0 2 matrix corresponding to each row operation. 1 0 2 First we swap rows 1 and 3 to get 2 2 2, so E1 = 0 1 0 0 0 1 0 1 0. 1 0 0 1 0 2 1 0 0 Next subtract 2R1 from R2 to get 0 2 −2 so E2 = −2 1 0. 0 1 0 0 0 1 1 0 0 1 0 2 Now multiply R2 by 1/2 to get, 0 1 −1, so E3 = 0 21 0. 0 0 1 0 1 0 1 0 2 1 0 0 Now subtract R2 from R3, 0 1 −1, with E4 = 0 1 0. 0 0 1 0 −1 1 Now do R1=R1-2R3 and R2=R2+R3 which reduces the matrix to I3 , 1 0 −2 and the last elementary matrices are E5 = 0 1 0 and E6 = 0 0 1 1 0 0 0 1 1 . 0 0 1 29. Give the inverse of each of the following elementary matrices: 1 0 0 0 0 1 1 0 0 (i) 0 34 0 (ii) 0 1 0 (iii) 0 1 0 0 0 1 1 0 0 0 −6 1 14 Solution: 0 (ii) 0 1 1 (iii) 0 0 0 1 0 0 1 6 1 0 0 (i) 0 43 0. 0 0 1 1 0. 0 0 0 1 30. Determine E a product of elementary matrices which when premultiplying A performs Gauss-Jordan pivoting on the (3, 3) entry of 1 0 −3 5 A = 0 1 3 4 . 0 0 2 6 Check E by computing EA. Solution: 1 The required row operations 1 + 3R3 and are R3 → /2R3 , R1 → R 1 0 0 1 0 3 R2 = R2 − 3R3 , so E1 = 0 1 0 , E2 = 0 1 0 and E3 = 0 0 1/2 0 0 1 1 0 0 0 1 −3. (Note that the order of E2 and E3 could be swapped.). 0 0 1 1 0 3/2 1 0 0 14 Then E = E3 E2 E1 = 0 1 −3/2. Then EA = 0 1 0 −5. 0 0 1/2 0 0 1 3 31. For each of the following matrices find the inverse by using elementary row operations. Write down the elementary matrix corresponding to each row operation that you use, and hence in each case express both the matrix and the inverse as products matrices. of elementary 1 0 0 0 0 1 1 3 (a) (b) 0 0 1 (c) 1 1 1. 4 3 1 2 3 0 1 2 Solution: (a) 1 3 4 3 1 3 0 −9 1 3 0 1 1 0 0 1 1 0 1 −4 1 4/9 − 1/3 4/9 0 1 0 1 0 E1 = E2 = − 1/9 1/3 − 1/9 15 E3 = 1 −4 1 0 1 0 0 1 0 − 1/9 −3 1 So A−1 = − 1/3 1/3 4/9 − 1/9 . 1 −3 1 Now E3 E2 E1 A = I, so = E3 E2 E1 = 0 1 0 1 0 1 0 1 3 and A = E1−1 E2−1 E3−1 = . 4 1 0 −9 0 1 A−1 0 1 0 , −4 1 − 1/9 (b) 1 0 1 1 0 0 1 0 0 1 0 0 1 0 0 0 0 2 0 0 2 0 2 0 0 1 0 0 1 0 0 1 3 0 1 3 0 3 1 0 3/2 1 0 0 1 1 0 0 1 0 −1 1 −1 0 1 − 1/2 0 1 − 1/2 0 0 1 0 0 1 0 0 0 1 0 0 1 0 − 3/2 1 0 0 1 0 0 1 0 1 0 0 1/2 0 0 1/2 0 E1 = E2 = E3 = E4 = 1 0 −1 1 0 0 1 0 0 1 0 0 0 so A−1 = − 1/2 − 3/2 1/2 . 0 1 0 1 1 0 0 1 0 0 A−1 = E4 E3 E2 E1 = 0 1 − 3/2 0 1/2 0 0 0 0 0 1 0 0 1 and 1 0 0 1 0 0 1 A = E1−1 E2−1 E3−1 E4−1 = 0 1 0 0 0 1 0 1 0 1 0 1 0 0 1 0 1 0 0 0 1 0 1/2 0 0 1 0 0 0 1 0 1 0 0 0 1 0 − 3/2 1 0 16 1 0 0 0 0 0 1 0 1 0 1 0 −1 0 1 0 0 1 0 0 2 0 0 1 3/2. 0 1 0 0 1 (c) 0 1 0 1 0 0 1 0 0 1 0 0 1 0 0 1 0 0 0 1 1 0 0 1 1 0 1 0 0 0 1 1 2 1 1 0 1 0 1 0 0 0 1 1 2 0 0 1 1 1 0 1 0 1 2 0 0 1 0 1 1 0 0 0 −1 0 1 −1 1 2 0 0 1 0 1 1 0 0 0 0 1 1 −1 1 2 0 0 1 1 0 0 0 1 0 0 1 1 −1 1 0 −2 0 1 0 1 1 0 0 1 1 −1 so A−1 = −2 0 1 . 1 0 0 Then we have E1 = E2 = E3 = E4 = E5 = 0 1 0 1 0 0 0 0 1 1 0 0 0 0 1 0 1 0 1 −1 0 0 1 0 0 0 1 1 0 1 0 1 0 0 0 1 1 0 0 0 1 −2 0 0 1 A−1 = E5 E4 E3 E2 E1 1 0 0 1 0 1 1 −1 0 1 0 0 0 1 0 = 0 1 −2 0 1 0 0 1 0 0 0 1 1 0 0 0 0 1 0 0 1 0 0 1 0 1 0 0 0 1 and A = E1−1 E2−1 E3−1 E4−1 E5−1 1 0 −1 1 0 0 1 1 0 0 1 0 1 0 0 = 1 0 0 0 0 1 0 1 0 0 1 0 0 1 2 . 0 0 1 0 0 1 0 0 1 0 0 1 0 1 0 32. Find all values of α for which the following matrix is not invertible. 1 4 −3 A = −2 −7 6 . −1 3 α Solution: 1 0 −3 0 . To further row reduce to the Row reducing A gives 0 1 0 0 α−3 identity we need α − 3 6= 0, so α 6= 3. Thus A is not invertible if α = 3. 17 33. Let A and B be matrices of the same size and suppose that B is invertible. (a) Prove that A is invertible if and only if AB is invertible. (b)* Hence prove that either A + B and I + AB −1 are both invertible or else both not invertible. Solution: (a) This is an ‘if and only if’ statement. Therefore, we need to prove two things (with the assumption given, that B is invertible): (i) if A is invertible then AB is invertible, and (ii) if AB is invertible, then A is invertible. The key property of invertible matrices that we will use is the following fact: if X and Y are invertible matrices of the same size then XY is also invertible (in fact we saw in class that (XY )−1 = Y −1 X −1 ). We prove (i). We are given that A is invertible and B is invertible. Therefore, AB is invertible by the key property above. We prove (ii). Now we are given that B is invertible and AB is invertible. Observe that (AB)B −1 = A(BB −1 ) = AI = A. But this shows that A is the product of two invertible matrices, namely AB and B −1 , and hence is invertible by the key property above (recall that we saw in lecture that B −1 is invertible with (B −1 )−1 = B). (b) Since B is invertible we may write A + B = (AB −1 + I)B. By part (a), A + B is invertible if and only if AB −1 + I is invertible, which is equivalent to the statement that A + B and AB −1 + I are either both invertible or both not invertible. 34. Let a,b, c be fixed non-zero constants. a 0 0 (a) Find the inverse of 0 b 0 in terms of a, b, c. 0 0 c a a 0 (b) Find the inverse of 0 a a. a 0 a Solution: 1/a 0 (a) 0 1/b 0 0 0 0 . 1/c (b) We can write this 1 1 so the inverse is a 0 1 (after using the usual 1 1 0 as a 0 1 1, 1 0 1 −1 1/2a − 1/2a 1 0 1 1/2a 1 1 /2a = − 1/2a 1/2a 0 1 method to find the inverse). 18 1/2a − 1/2a 1/2a 35. True or False? Examine each of the following statements carefully and decide whether they are true or false. Give a short reason for your decision in each case. (a) Let A and B be square matrices such that AB = O. If A is invertible then B = O. (b) Let A and B be invertible matrices of the same size. Then (A + B)−1 = A−1 + B −1 . (c) If A is an m × n matrix, then there is an invertible matrix E such that EA is in reduced row echelon form. (d) The matrix 1 −1 2 3 7 0 0 1 5 0 1 −2 2 0 4 −3 1 8 A= 1 −1 2 3 7 0 4 8 11 −21 0 −7 3 5 −6 2 1 4 is invertible. Solution: (a) True. Multiply both sides of AB = O by A−1 to get B = A−1 AB = A−1 O = O. (b) False. For instance A = I2 and B = −I2 are both invertible, but A + B = O is not even invertible. (c) True. If R is the reduced row echelon form of A, then there is a sequence of elementary matrices E1 , . . . , Ek such that E1 · · · Ek A = R. The matrix E = E1 · · · Ek is invertible since it is a product of invertible matrices. (d) False. Since Row 1 = Row 4, the reduced row echelon form of A has a zero row. Therefore A is not row equivalent to I6 , and hence A is not invertible. 36. Prove that if A and B are row equivalent matrices (i.e. you can get from one to the other by a series of elementary row operations) then the equations Ax = 0 and Bx = 0 have the same solutions. (Hint: the proof should use elementary matrices) Solution: If A and B are row equivalent then B = EA where E is a product of elementary matrices, and thus is invertible. Let x satisfy Bx = 0. Then EAx = 0 so Ax = E −1 0 = 0. On the other hand, suppose y satisfies Ay = 0. Then EAy = E0 = 0 so By = 0. 19 37. Let A be an n × n matrix with two identical columns. Explain why A is not invertible. Solution: Method 1: At has two identical rows, it was shown in lectures that this means that At is not invertible (since the rref must have a zero row), so A must not be invertible since if it was then At would have inverse (A−1 )t . Method 2: Suppose A is invertible. Then there is a pivot in each column of rref(A), which occurs in the corresponding row. Suppose the ith and jth columns are identical. Then carrying out the same row operations on the entries in these columns must give the same results, however the ith column should row reduce to give a pivot in the ith row, and the jth column should give a pivot in the jth row, hence we have a contradiction and A cannot be invertible. 38. Suppose that A is an invertible n×n matrix satisfying A3 −3A+2I = 0. Find an expression for A−1 in terms of A and I. Solution: A(A2 − 3I) = −2I so A 3I). − 1/2(A2 − 3I) = I thus A−1 = − 1/2(A2 − 2. Determinants Basic Definitions 39. Calculate the following determinants (answers should be formulas in terms of a and b): 1 2 a 2 0 a b a 1 1 1 0 a a a 0 . (c) (b) b (a) 1 a 0 0 0 0 1 −a −b −a a 1 0 1 a b b Solution: 1 a (a) Expanding along the 3rd column we have 1 = 1 − a2 . a 1 b 0 b a (b) Expanding along the first row we have 0−a +b = −a −a −a −b −a(−ab − 0) + b(−b2 + a2 ) = a2 b − b3 + a2 b = 2a2 b − b3 = b(2a2 − b2 ). 20 1 2 (c) Expanding along the 3rd row gives −1 1 0 1 a determinant along the 2nd row, giving (−1) −1 a a . We expand this b 2 a 1 2 −a = a b 1 a (2b − a2 ) + a(a − 2) = 2b − a2 + a2 − 2a = 2(b − a). a 5 3 b 0 6 0 0 a b (your answer should be = 4, evaluate 40. Given that 0 7 −1 0 c d c 8 4 d a number). Solution: a 3 b Expanding along the 2nd row we get 6 0 −1 0 , and then expanding c 4 d a b along the second row of this matrix we have 6 −1 = −24. c d 41. Let Jn be the n × n matrix all of whose entries example, 1 1 1 , J3 = 1 J1 = [1], J2 = 1 1 1 are equal to 1. For 1 1 1 1 . 1 1 Prove that if n > 1, then the matrix In − Jn is invertible with inverse (In − Jn )−1 = In − 1 Jn n−1 (here In is the n × n identity matrix). 1 Solution: We have to check that (In − Jn )(In − n−1 Jn ) = In = 1 (In − n−1 Jn )(In − Jn ). In fact, we proved a fact in lectures which allows us to cut out a bit of work. By Theorem 1.4, it is sufficient to prove that 1 (In − Jn )(In − Jn ) = In . n−1 We expand the left hand side: (In − Jn )(In − 1 1 1 Jn ) = In − Jn − Jn + J2 n−1 n−1 n−1 n n 1 = In − Jn + J 2. n−1 n−1 n 21 We need to think about the matrix power Jn2 . The (i, j) entry of this matrix is 1 1 [1 1 · · · 1] . = n .. 1 and so we see that Jn2 = nJn . Therefore (In − Jn )(In − 1 n 1 Jn ) = In − Jn + J2 n−1 n−1 n−1 n n n = In − Jn + Jn n−1 n−1 = In . 42. Show that all of the determinants D2 , D3 , D4 , . . . Dn , . . . are equal to 1: 1 1 1 1 1 1 1 1 1 1 2 2 2 D3 = 1 2 2 D2 = D4 = ... 1 2 1 2 3 3 1 2 3 1 2 3 4 1 1 1 1 ... 1 1 2 2 2 ... 2 1 2 3 3 ... 3 Dn = 1 2 3 4 ... 4 . . . . ... . 1 2 3 4 ... n Solution: 1 1 D2 = =2−1=1 1 2 1 1 1 1 1 0 1 2 2 2 D4 = = 1 2 3 3 0 0 1 2 3 4 In general,for n ≥ 2, Dn 1 1 1 1 1 1 1 1 D3 = 1 2 2 = 0 1 1 = =1 1 2 1 2 3 0 1 2 1 1 1 1 1 1 1 1 1 = 1 2 2 = 1. 1 2 2 1 2 3 1 2 3 = Dn−1 = . . . = D4 = D3 = D2 and D2 = 1. Row Operations On Determinants a b c 43. Given that d e f = 5, find g h i 22 d e f (i) g h i (ii) a b c a − 4g d g (iii) b − 4h e h . c − 4i f i d e f 2a + d 2b + e 2c + f g h i Solution: Given that a d g f i c b c e f = 5, h i d e g h = (−1)(−1) × 5 = 5 (R1 ↔ R2 , R2 ↔ R3 ) a b d e f (ii) 2a + d 2b + e 2c + f = (−1)(2) × 5 = −10 g h i a − 4g d g (iii) b − 4h e h = 5 c − 4i f i where the row operations are: (ii) R1 ↔ R2 ; R2 → 2R2 and R2 → R2 + R1 ; (iii) R1 → R1 − 4R3 ; then take the transpose which does not affect the determinant. (i) 2 a a 1 44. Let A = b2 b 1. c2 c 1 (a) Use row operations to show that det A = (b − a)(c − a)(b − c). (b) Calculate the adjoint and inverse of A when a, b, c are distinct. (c) Use (b) to find α, β, γ so that y = f (x) = αx2 + βx + γ has f (1) = 1, f (2) = 0, f (3) = 2. Solution: (a) a2 a 1 b2 b 1 c2 c 1 a2 a 1 − a2 b − a 0 = c2 − a2 c − a 0 b+a 1 = (b − a)(c − a) c+a 1 b2 23 = b2 − a2 b − a c2 − a2 c − a = (b − a)(c − a)(b − c) . b − c b2 − c2 bc(b − c) (b) [Aij ] = a − c a2 − c2 ac(a − c) ; a − b a2 − b2 ab(a − b) b−c c−a a−b a2 − c2 b2 − a2 . adj A = c2 − b2 bc(b − c) ac(c − a) ab(a − b) 1 adj A A−1 = (b−a)(c−a)(b−c) (c) y1 = f (a), y2 = f (b), y3 2 a b2 c2 = f (c) implies a 1 α y1 b 1 β = y2 c 1 γ y3 Using f (1) = 1, f (2) = 0, f (3) = 2 and (b), 2 −1 −1 2 −1 1 1 1 1 5 −8 3 22 2 1 = (2−1)(3−1)(2−3) 2 −6 6 −2 3 3 1 1/2 1 −1 /2 = − 5/2 4 − 3/2 3 −3 1 α = y1 /2 − y2 + y3 /2 = 3/2; β = −5y1 /2 + 4y2 − 3y3 /2 = − 11/2; γ = 3y1 − 3y2 + y3 = 5 y = 3x2 /2 − 11x/2 + 5. 45.* The numbers 22253, 62951, 54043, 70499 and 19635 are all divisible by 17. Use this fact to show that if 2 6 D= 5 7 1 2 2 4 0 9 2 9 0 4 6 5 5 4 9 3 3 1 3 9 5 then the number D is divisible by 17, without evaluating the determinant. Solution: Multiply the first column by 104 , the second column by 103 , the third column by 102 , and the fourth column by 10. Then 20000 60000 10 10 D = 50000 70000 10000 24 2000 2000 4000 0 9000 200 900 0 400 600 50 50 40 90 30 3 1 3 9 5 Now add column 2 to column 1, then column 3 to column 1, then column 4 to column 1 and finally column 5 to column 1. We get 22253 62951 10 10 D = 54043 70499 19635 2000 2000 4000 0 9000 200 900 0 400 600 50 50 40 90 30 3 1 3 9 5 Since each of the numbers in the first column are divisible by 17, we can take out a factor of 17 and deduce that the determinant is divisible by 17. But this means that 1010 D is divisible by 17 (clearly the determinant is an integer in this case). Since 17 is prime the only way that this can happen is if D is divisible by 17. General Properties of Determinants 46. Suppose A and B are n × n matrices with det A = 4 and det B = −3. Find each of the determinants below: (a) det(AB) (b) det(A2 ) (c) det(B −1 A) (d) det(2A) (e) det(3B t ) (f) det(AAt ) Solution: (a) det(A) det(B) = −12. (b) det(A) det(A) = 16. (c) (d) 1 det(B) det(A) = −4/3. 2n det(A) = 4.2n = 2n+2 . (e) 3n det(B) = −3n+1 . (f) det(A) det(At ) = det(A)2 = 16. 47. What can you say about det(A) if a square matric A satisfies (a) A2 = A (such a matrix is called idempotent). (b) Am = 0 for some m > 1 (such a matrix is called nilpotent). Solution: 25 (a) We have det(A)2 = det(A), so det(A)(det(A) − 1) = 0 and either det(A) = 0 or det(A) = 1. (b) det(A)m = 0 so we must have det(A) = 0. 48.* (a) Let A be a 2 × 3 matrix. Then AT A is a 3 × 3 matrix. Show that det(AT A) = 0. (b) Suppose again that A is a 2 × 3 matrix; is it necessarily true that det(AAT ) = 0? Solution: (a) Since A is a 2 × 3 matrix, there exists a vector x 6= 0 such that Ax = 0 (less equations than variables). Thus there exists x 6= 0 with AT (Ax) = 0 or (AT A)x = 0. This means that the square matrix AT A is not invertible; that is, det(AT A) = 0. (b) There are 2 × 3 matrices A with det(AAT ) 6= 0. For example, 1 1 1 3 2 1 1 1 , det(AAT ) 6= 0; AT A = 1 2 2 AAT = A= 2 2 0 1 1 1 2 2 T but det(A A) = 0. The Adjoint 1 2 3 49. Consider the matrix A = 4 5 6 . 7 8 10 (a) Find the matrix [Aij ] of minors of A. (b) Find the matrix [Cij ] of cofactors of A. (c) Find the adjoint, adj(A), of A. (d) Find det(A). (e) Hence write down the inverse of A. Solution: 2 −2 −3 (a) [Aij ] = −4 −11 −6 −3 −6 −3 2 2 −3 (b) [Cij ] = 4 −11 6 −3 6 −3 26 2 4 −3 (c) adj A = [Cij ]T = 2 −11 6 −3 6 −3 (d) |A| = 1A11 − 2A12 + 3A13 = −3 − 2/3 − 4/3 1 1 adj A = − 2/3 11/3 −2 (e) A−1 = |A| 1 −2 1 50.* Let A = [aij ] be an n × n matrix. If det(A) = 1 and each entry of A is an integer (i.e. aij ∈ Z for all i, j), prove that all of the entries A−1 are also integers. Solution: We have (det(A))A−1 = adj(A) and hence A−1 = adj(A). Therefore to show that every entry of A−1 is an integer, we need to show that every cofactor Cij of A is an integer, and hence that every (i, j) minor Aij of A is an integer. But Aij is the determinant of the (n − 1) × (n − 1) matrix obtained by removing the i-th row and j-th column from A; since every entry of A is an integer also every entry of this (n − 1) × (n − 1) matrix is an integer. Therefore all we need to do is to show that the determinant of a matrix all of whose entries are integers, is itself an integer. But the determinant of a n × n matrix A is a sum of terms of the form ±a1i1 a2i2 · · · anin ; since each entry aij is an integer, so must all of the terms in this sum be integers, therefore det(A) ∈ Z. 3. Optimisation and Convex Sets Convex Sets 51. (a) Write down the definition of a convex set. (b) Define a vertex of a convex set. Solution: (a) A convex set C is a set of points (or region) in Rn such that for any two points in C, the line segment joining the two points lies completely in C. −→ (b) A point P with position vector OP = v is a vertex of the convex set C if v can not be written in the form (1 − t)u + tw 0≤t≤1 for any (distinct) points u, w ∈ C except when v = u (t = 0) or v = w ( t = 1). 27 52. Sketch the following sets in R2 and determine the vertices. (a) {(x1 , x2 ) | x1 + 3x2 ≤ 6, 2x1 − x2 ≥ 4, x1 ≥ 0, x2 ≥ 0} (b) {(x1 , x2 ) | x1 + x2 ≥ 2, x1 − x2 ≤ 4, x1 ≥ 1, x2 ≥ 0} (c) {(x1 , x2 ) | −x1 + 4x2 ≤ 5, x1 − 4x2 ≤ 2, x1 + 2x2 ≥ 2} Solution: Graphs: (a) (b) (c) Vertices: (a): (2,0), (6,0), (18/7,8/7) (b) (1,1), (2,0), (4,0) (c) (2,0), (-1/3,7/6) 53. Sketch the following sets in R2 and hence determine whether they are bounded or unbounded, and whether they are convex or not. In the case where the set is not convex, give an example of a line segment where the definition of convexity breaks down. (a) {(x, y) | x + y ≥ 1, x − y ≤ 2, x ≥ 0} (b) {(x, y) | 1 ≤ |x| ≤ 2, |y − 3| ≤ 2} (c) {(x1 , x2 ) | x21 + x22 ≤ 4 and x1 + x2 ≤ 1} (d) {(x1 , x2 ) | x21 + x22 ≥ 1 and x1 + x2 ≤ 4} 28 Solution: (a) Unbounded, convex b) bounded, not convex. Consider the line segment from (-1,1) to (1,1). (c) Bounded, convex. 29 (d) unbounded, not convex. Consider the line segment from (-1,0) to (1,0). 54. Using the definition of convexity, show that the set C = {(x1 , x2 ) ∈ R2 | (5, −1) · (x1 , x2 ) ≤ 7} is convex. Solution: Let u and v be any two points in the set C, and consider any point x on the line segment between them, so x = (1 − t)u + tv for 0 ≤ t ≤ 1. We need to show that (x1 , x2 ) is in C, that is x · (5, −1) ≤ 7. Consider x · (5, −1) = [(1 − t)u + tv] · (5, −1) = (1 − t)u · (5, −1) + tv · (5, −1) where we have used the basic properties of the inner product. Since each of u and v is in C, u · (5, −1) ≤ 7 and v · (5, −1) ≤ 7, thus (1 − t)u · (5, −1) + tv · (5, −1) ≤ 7(1 − t) + 7t = 7 as required. Note that the final inequality depends on both t ≥ 0 and (1 − t) ≥ 0 which is true since 0 ≤ t ≤ 1. 55. Let u and v be the points (−2, 1, 0) and (1, −5, 6) in R3 . (a) Let x be the point (−1, −1, 2). Find t such that x = (1 − t)u + tv. Is x on the line segment joining u and v? (b) Is the point y = (−4, 5, −4) on the line segment joining u and v? Is y on the line through u and v? 30 (c) Is the point z = (2, −4, 4) on the line segment joining u and v? Is z on the line through u and v? Solution: (a) We need t such that (−1, −1, 2) = (1 − t)(−2, 1, 0) + t(1, −5, 6) or (−1, −1, 2) = (−2 + 3t, 1 − 6t, 6t), so t = 1/3. Since 0 ≤ t ≤ 1 then this means that x is on the line segment joining u and v. (b) We need to solve (−4, 5, −4) = (−2 + 3t, 1 − 6t, 6t). This is satisfied when t = −2/3 but since t < 0 we can say that y is on the line through u and v but it is not on the line segment joining u and v. (c) This time we need to solve (2, −4, 4) = (−2 + 3t, 1 − 6t, 6t). The third entry requires that t = 2/3, however this is not consistent with the other entries so there is no solution. Thus z is not on the line through u and v, which also means that it is not on the line segment joining the two points. 56.* Let C = {(x1 , x2 ) ∈ R2 | x1 + x2 ≤ 1, x1 ≥ 0, x2 ≥ 0}. Consider v = (1, 0). Suppose there exist u = (u1 , u2 ) and w = (w1 , w2 ) in C, and t ∈ (0, 1), such that v = (1 − t)u + tw. (a) Show that if u1 < 1 or w1 < 1, then (1 − t)u1 + tw1 < 1. This means u1 = w1 = 1. Why? (b) Hence or otherwise show that v is a vertex. Solution: (a) Note that the requirement that u and w are in C means that 0 ≤ u1 ≤ 1 and 0 ≤ w1 ≤ 1, so that (1 − t)u1 ≤ 1 − t and tw1 ≤ t. Thus (1−t)u1 +tw1 ≤ (1−t)+t = 1. Now if either u1 < 1 or w1 < 1 then the corresponding inequalities become strict inequalities (i.e. (1 − t)u1 < 1 − t or tw1 < t, thus (1 − t)u1 + tw1 < 1). Thus we must have u1 = w1 = 1 since (1, 0) = v = (1 − t)u + tw implies that we must have (1 − t)u1 + tw1 = 1. (b) We have shown that if 0 < t < 1 then there are no two distinct points u and w such that v = (1 − t)u + tw. Thus if 0 ≤ t ≤ 1 then the the only possible ways of writing v = (1 − t)u + tw would be t = 0 or t = 1, hence v is a vertex. 31 57.* Let S = {u1 , . . . , un } be a set of vectors in Rn . A vector w is said to be a convex combination of the vectors u1 , . . . , un if w is a linear combination w = x1 u1 + · · · + xn un where the scalars x1 , . . . , xn are non-negative real numbers and x1 + · · · + xn = 1. Let Conv(S) be the set of all convex combinations of the vectors in S. Use the definition of convexity to show that Conv(S) is a convex set. Solution: Let u and v be points of Conv(S). We have to prove that for any 0 ≤ t ≤ 1, the vector (1 − t)u + tv belongs to Conv(S). Since u and v belong to Conv(S), there exist non-negative real numbers x1 , . . . , xn and y1 , . . . , yn such that x1 + · · · + xn = 1, y1 + · · · + yn = 1, u = x1 u1 + · · · + xn un , and v = y1 u1 + · · · + yn un . Therefore (1 − t)u + tv =(1 − t)(x1 u1 + · · · + xn un ) + t(y1 u1 + · · · + yn un ) =((1 − t)x1 + ty1 )u1 + · · · + ((1 − t)xn + tyn )un Since ((1 − t)x1 + ty1 ) + · · · + ((1 − t)xn + tyn ) = (1 − t)(x1 + · · · + xn ) + t(y1 + · · · + yn ) = (1 − t) + t = 1 and each (1 − t)xi + tyi ≥ 0, we see that (1 − t)u + tv ∈ Conv(S), as required. Methods for Solving Optimisation Problems 58. We wish to find the minimum value of g(x, y) = 5x/2 + 5y subject to the constraints 3x + 2y ≥ 36 3x + 5y ≥ 45 x≥0 y≥0 Sketch the feasible region, identify its vertices, and find the vertex or vertices at which the smallest value of g(x, y) is taken. Solution: The feasible region is shown in the diagram below: 32 The vertices are (15,0), (0,18) and (10,3). Substituting each of these into g(x, y) gives the values 37 1/2, 90 and 40 respectively, so the minimum value is 37 1/2 at the point (15,0). 59. Consider the convex set given by 2x1 + x2 ≤ 8 2x1 + 3x2 ≥ 12 x1 ≥ 0 x2 ≥ 0 (a) Add slack variables to write this in standard form C = {x ∈ Rn | Ax = b, x ≥ 0}. (b) Find the basic solutions of Ax = b and indicate those which are feasible. (c) Check your answer to the previous point by sketching the feasible region. (d) Find the maximum value of f (x) = x1 − x2 over C and write down the corresponding optimal choice for x. Solution: (a) Multiplying the second inequality by -1 and adding slack variables x3 , x4 ≥ 0 gives 2x1 + x2 +x3 −2x1 − 3x2 for x1 , x2 , x3 , x4 ≥ 0. (b) The basic solutions are as follows: 33 =8 + x4 = − 12 Pivots Matrix Basic Solution x3 , x4 2 1 1 0 | 8 −2 −3 0 1 | −12 x2 , x4 x1 , x4 x1 , x2 x1 , x3 Feasible? Vertex (0, 0, 8, −12) no 2 1 1 0 | 8 4 0 3 1 | 12 (0, 8, 0, 12) yes 1 1/2 1/2 0 | 4 0 −2 1 1 | −4 (4, 0, 0, −4) no (3, 2, 0, 0) yes (6, 0, −4, 0) no 1 0 3/4 1/4 | 3 0 1 −1/2 −1/2 | 2 1 3/2 0 −1/2 | 6 0 −2 1 1 | −4 2/3 1 0 −1/3 | 4 (0, 4, 4, 0) x2 , x3 4/3 0 1 1/3 | 4 Therefore the vertices are (0, 8),(3, 2) and (0, 4). (0, 8) (3, 2) yes (c) The feasible region and its vertices are shown below: (d) Evaluating f at each vertex gives f (0, 8) = −8, f (0, 4) = −4 and f (3, 2) = 1, so the maximum is 1 at (3, 2). 34 (0, 4) 60. Consider the region given by x≤4 x+y ≥1 x−y−z ≤8 x, y, z ≥ 0 (a) Formulate this problem using slack variables s1 , s2 and s3 . Write down the basic solution obtained by choosing the slack variables as pivots and state whether it is feasible. (b) Find the basic solution corresponding to pivots z, s1 and s2 and state whether is is feasible. (c) Find the basic solution corresponding to pivots y,s1 and s3 and state whether it is feasible. Solution: (a) Adding slack variables gives the following system of equations: x + s1 −x − y =4 =−1 +s2 x − y −z + s3 =8 with x, y, z, s1 , s2 , s3 ≥ 0. In matrix form 1 0 0 1 0 0 −1 −1 0 0 1 0 1 −1 −1 0 0 1 we have | 4 | −1 | 8 The resulting basic solution is (0, 0, 0, 4, −1, 8) which is not feasible. (b) We carry out the operation Pivot(3,3) to 1 0 0 1 0 0 −1 −1 0 0 1 0 −1 1 1 0 0 −1 get | 4 | −1 | −8 The resulting basic solution is (0, 0, −8, 4, −1, 0) which is not feasible. (c) We carry out the operation Pivot(2,2) on the original matrix to get 1 0 0 1 0 0 | 4 1 1 0 0 −1 0 | 1 2 0 −1 0 −1 1 | 9 which gives the basic solution (0, 1, 0, 4, 0, 9) which is feasible. 35 Formulation of Optimisation Problems 61. A pet food manufacturer produces two types of food: regular and premium. A 20kg bag of regular food requires 2 hours to prepare and 3 hours to cook; a 20kg bag of premium food requires 2 hours to prepare and 5 hours to cook. The materials used to prepare the food are available 8 hours per day and the oven used to cook the food is available 15 hours per day. The profit on a 20kg bag of regular food is $42 and on a 20kg bag of premium food it is $50. Determine how many bags of each type of food should be made to maximise the profit (fractional bags can be produced – they just sell them in smaller bags). Solution: Let x1 = number of bags of regular food, x2 = number of bags of premium food. We wish to maximise profit, P = 42x1 + 50x2 such that 2x1 + 2x2 ≤ 8, 3x1 + 5x2 ≤ 15, x1 , x2 ≥ 0. The feasible region is 5 4 3 2 ( 52 , 32 ) 1 1 0 1 2 3 4 5 6 7 - 1 (0, 0) = 0, P (4, 0) = 168, P ( 5 , 3 ) = 180 P (0, 3) = 150, P 2 2 So maximum is $180 when (x1 , x2 ) = ( 25 , 32 ). This means we want bags of regular food, and 32 bags of premium food. 5 2 62. A trucking company owns two types of trucks. Type A has 20 cubic metres of refrigerated space and 30 cubic metres of non-refrigerated space. Type B has 20 cubic metres of refrigerated space and 10 cubic metres of non- refrigerated space. A customer wants to haul some produce a certain distance and will require 160 cubic metres of refrigerated space and 120 cubic metres of non-refrigerated space. The trucking company figures that it will take 300 litres of fuel for the type 36 8 A truck to make the trip and 200 litres of fuel for the type B truck. Find the number of trucks of each type that the company should allow for the job in order to minimise fuel consumption (a) graphically, and (b) algebraically (i.e. by using slack variables). Solution: Let x1 = number of trucks of type A, x2 = number of trucks of type B (x1 , x2 ≥ 0). We wish to minimise f = 300x1 + 200x2 subject to 20x1 + 20x2 ≥ 160, 30x1 + 10x2 ≥ 120. (a) The feasible region is 12.5 10 7.5 (2, 6) 5 2.5 - .5 2 0 2.5 5 7.5 10 12.5 15 Vertices: (0, 12), (2, 6), (8, 0) f (0, 12) = 2400, f (2, 6) = 1800, f (8, 0) = 2400 ∴ Min is 1800L for 2 type As and 6 type Bs. (b) Add slack variables x3 , x4 ≥ 0 : −20x1 − 20x2 +x3 = −160 −30x1 − 10x2 + x4 = −120 Number of possible basic solutions is 42 = 6. The first colum gives the choices of pivots. x1 x2 x3 x4 Basic soln Feasible? Vertex Value of f X X (0, 0, −160, −120) N − − X X (0, 8, 0, −40) N − − X X (8, 0, 0, 120) Y (8, 0) 2400 X X (4, 0, −80, 0) N − − X X (2, 6, 0, 0) Y (2, 6) 1800 X X (0, 12, 80, 0) Y (0, 12) 2400 So again min is 1800 at (2,6). 37 63. Two oil refineries produce three grades of gasoline, A, B and C. At each refinery the various grades of gasoline are produced in a single operation so that they are in fixed proportions. Assume that one operation at Refinery 1 produces 1 units of A, 1 unit of B, and 3 units of C. One operation at Refinery 2 produces 1 unit of A, 4 unit of B, and 2 units of C. Refinery 1 charges $400 for one operation, and Refinery 2 charges $300 for one operation. A consumer needs 170 units of A, 200 units of B, and 360 units of C. How should the orders be placed if the consumer’s needs are to be met most economically? (Use graphical means to solve). Solution: Let x be the number of operations at refinery 1, and y be the number of operations at refinery 2. Then we wish to maximise the cost, C(x, y) = 400x + 300y (in dollars), subject to the constraints: x + y ≥ 170 (units of A) x + 4y ≥ 200 (units of B) 3x + 2y ≥ 360 (units of C) x, y ≥ 0 (non-negativity). The feasible region is: The vertices are (0, 180), (200, 0), (160, 10), (20, 150). Evaluating the cost at each vertex gives C(0, 180) = 54000, C(200, 0) = 80000, C(160, 10) = 67000, and C(20, 150) = 53000. Therefore the cheapest option is to carry out 20 operations at refinery 1 and 150 operations at refinery 2. 38 64. Formulate as an optimisation problem, but do not solve: A manufacturer of a particular product is planning the production schedule for the months of October, November, and December. The demands for the product are for 70 units in October, 90 units in November, and 120 units in December. The industrial plant has the capacity to produce at most 100 units per month. It costs $150 to produce each unit in October, and $160 in each of November and December. As well, each unit held over in storage from one month to the next, attracts a storage fee (inventory cost). There is no storage cost in the month of production, but there is a cost of $30 in any following month. (Thus the storage cost for one unit is found by multiplying the number of items still in storage on the first day of the month by $30.) The manufacturer plans to have no items in storage at the beginning of October, and none in storage at the end of December. The manufacturer wishes to determine the number of units which should be produced in each month in order to minimise costs. (Fractional units can be produced.) Solution: Let xi be the number of items made in month i (i = 1 corresponds to October, etc). Then to satisfy demand, we need x1 ≥ 70. At the end of October, we will have x1 − 70 items left, which will incur a storage charge of $30 per item in November. However, we will then only need to ensure (x1 − 70) + x2 ≥ 90, the demand for November. That is, x1 +x2 ≥ 160. There will then be (x1 −70)+x2 −90 = x1 +x2 −160 items in storage at the end of November. Proceed similarly for December. The objective is then to minimise costs of 150x1 + 160(x2 + x3 ) + 30(x1 − 70) + 30(x1 + x2 − 160) = 210x1 + 190x2 + 160x3 − 6900 dollars. Thus the LP is: Minimise z = 210x1 + 190x2 + 160x3 − 6900 (cost in dollars) Subject to x1 x1 +x2 x1 +x2 +x3 x1 x2 x3 ≥ 70 (demand for October) ≥ 160 (demand for November) = 280 (demand for December; none left) ≤ 100 (monthly plant capacity) ≤ 100 (monthly plant capacity) ≤ 100 (monthly plant capacity) and x1 , x2 , x3 ≥ 0 (nonnegativity). 4. The Vector Space Rn Linear Dependence and Independence 65. Let u = (2, 1, 4, 3), v = (3, 4, 1, 2), w = (1, 2, −1, 0). 39 (a) Are the vectors {u, v, w} linearly independent? Justify your answer. (b) Are the vectors {v, w} linearly independent? Justify your answer. (c) Are the vectors {v, w, 0} linearly independent? Justify your answer. (d) Are the vectors {u, w, 5u − 3w} linearly independent? Justify your answer. Solution: (a) No, as 2u + (−3) v + 5w = 0. (b) Yes, as v is not a multiple of w. (c) No, as 0 is in the list. (d) No, as 5.u + (−3).w + (−1).(5u − 3w) = 0. 66. Which of the following lists of vectors are linearly independent: (a) {(3, 4), (4, 3)} (b) {(2, 1, −3, 6), (5, 3, 7, 8), (1, 1, 13, −4)} (c) {(2, 1, 3), (2, −2, −5), (7, 3, 9)} (d) {e1 , e1 + 2e2 , e1 + 2e2 + 3e3 , . . . , e1 + 2e2 + 3e3 + · · · + nen } where ei ∈ Rn is the vector which has 1 in the i-th place and 0 everywhere else. Solution: (a) Linearly independent: the 2 vectors are not multiples of each other. 2 5 1 1 0 −2 1 3 1 0 1 1 (b) Linearly dependent: as −3 7 13 row reduces to 0 0 0 6 8 −4 0 0 0 so 2u − v + w = 0. 2 2 7 1 0 0 (c) Linearly independent: as 1 −2 3 row reduces to 0 1 0. 3 −5 9 0 0 1 1 1 1 ... 1 0 2 2 . . . 2 0 0 3 · · · 3 (d) Linearly independent: as A = row reduces to .. .. .. . . . . . . . .. 0 0 0 ... n the identity n×n matrix. (In fact, the determinant of this matrix is 1 × 2 × . . . × n 6= 0 so it’s invertible which means that there is a unique solution to Ax = b for any b.) 40 67. For what value(s) of d are the following sets of vectors linearly dependent? Justify your answers. (a) {(1, −1, 4)), (3, −5, 5), (−1, 5, d)} (b) {(3, 7, −2), (−6, d, 4), (9, 1, −4)} Solution: (a) Linear dependent only if d = 10 as 1 3 −1R − 1 −5 5R4 5 d rowreduces to 1 0 5R0 1 −2R0 0 −10 + d . (b) Linear dependent only if d = −14 as 3 −6 9R7 d 1R − 2 4 −4 row reduces to 1 −2 0R0 d + 14 0R0 0 1 . 68. Determine, with reasons, if the following statements are (A) always true, (B) always false, or (C) might be true or false. (a) If v1 , ..., v5 ∈ R5 and v1 = v2 + v3 then the set {v1 , ..., v5 } is linearly dependent. (b) If v1 and v2 ∈ R2 lie on the same straight line through the origin, then the set {v1 , v2 } is linearly dependent. (c) If {v1 , v2 , v3 } is linearly dependent, then so is {v1 , v2 }. (d) If {v1 , v2 } is linearly independent, then so is {v1 + v2 , v1 − v2 }. (e) There is a set {v1 , . . . , v5 } of linearly independent vectors in R4 . Solution: (a) (A) Always true, since one vector is a linear combination of the others, or v1 − v2 − v3 + 0v4 + 0v5 = 0. (b) (A) Always true, since one vector is a multiple of the other. (c) (C) Neither e.g. result is false for v1 = (1, 0), v2 = (0, 1), v3 = (1, 1), but true for v1 = (1, 0), v2 = (2, 0), v3 = (1, 1). (d) (A) Always true. Suppose a(v1 + v2 ) + b(v1 − v2 ) = 0. Then (a + b)v1 + (a − b)v2 = 0. Since v1 , v2 are linearly independent, this gives a + b = 0, a − b = 0 giving a = b = 0 only. (e) (B) Always false. The maximum number of linearly independent vectors in R4 is four. 41 69. What is the definition of a subspace of Rn ? Solution: W ⊆ Rn is a subspace of Rn if it satisfies the following three conditions: • 0 ∈ W; • u ∈ W ⇒ c u ∈ W for any c ∈ R; • u, v ∈ W ⇒ u + v ∈ W . Subspaces 70. Which of the following are subspaces of Rn for some n? Give a reason for each answer. (a) W1 = {(x, y, z) ∈ R3 | 2x − y = 3z + x = 0} (b) W2 = {(x, y, z, w) ∈ R4 | x + y = zw} (c) W3 = {(α, β, γ) ∈ R3 | γ = 0} (d) W4 = {(x, y, z) ∈ R3 | x + y + z = 1} (e) W5 = {(x, y) ∈ R2 | x2 + y 2 = 1} ∪ {(0, 0)} (f) W6 = {(x, y) ∈ R2 | x2 + y 2 = 0}. Solution: (a) W1 is a subspace as it is the solution set for two homogeneous linear equations. (b) W2 is not a subspace as (1, 1, 1, 2) ∈ W2 , but 2(1, 1, 1, 2) = (2, 2, 2, 4) 6∈ W2 , so W2 is not closed under scalar multiplication. (c) W3 is a subspace of R3 since it is the solution set of a homogenous linear equation. (d) W4 is not a subspace, since (0, 0, 0) = 0 6∈ W4 . (e) W5 is not a subspace, since even though 0 ∈ W5 , it is not closed under addition or scalar multiplication, for example (1, 0) ∈ W5 since 12 + 02 = 1 but 2(1, 0) = (2, 0) ∈ / W5 since 22 + 02 6= 1 (f) W6 is a subspace, since W6 = {(0, 0)} ⊂ R2 . 71. Let A be an m × n matrix. Show that (a) R = {y ∈ Rm | Ax = y for some x ∈ Rn } is a subspace of Rm (b) K = {x ∈ Rn | Ax = 0} is a subspace of Rn . 42 Solution: (a) Note that 0 ∈ R since A0 = 0. If y1 , y2 ∈ R then ∃ x1 , x2 such that Ax1 = y1 and Ax2 = y2 . Then A(x1 + x2 ) = Ax1 + Ax2 = y1 + y2 so y1 + y2 ∈ R. Also, A(cx) = cAx = cy so cy ∈ R. Therefore R is a subspace of Rm . (b) Obviously 0 ∈ K. Note that if x1 , x2 ∈ K then A(x1 + x2 ) = Ax1 + Ax2 = 0 + 0 = 0 so x1 + x2 ∈ K. Also, if Ax = 0 then A(cx) = c(Ax) = c0 = 0 so cx ∈ K. Therefore K is a subspace of Rn . 72. Suppose that U, V ⊂ Rn are subspaces of Rn . Prove that the intersection U ∩ V is a subspace of Rn . Solution: Since U and V are subspaces of Rn we have 0 ∈ U and 0 ∈ V . Therefore 0 ∈ U ∩ V . Suppose that x ∈ U ∩ V and y ∈ U ∩ V . Then x, y ∈ U and x, y ∈ V . Therefore x + y ∈ U and x + y ∈ V since U and V are subspaces. Hence x + y ∈ U ∩ V . Finally, suppose that x ∈ U ∩ V and c ∈ R. Then cx ∈ U and cx ∈ V since U and V are subspaces. Therefore cx ∈ U ∩ V . Hence U ∩ V is a subspace. 73. Consider the following two subsets of R4 : U = {(x, x + 1, y + z, 2x)| x, y, z ∈ R} V = {(x, x + y + 1, z, 2x)| x, y, z ∈ R} Although these sets look similar, one is a subspace and one is not. (a) Explain why U is not a subspace of R4 . (b) Show that V = span{v1 , v2 , v3 }, where v1 = (1, 1, 0, 2), v2 = (0, 1, 0, 0) and v3 = (0, 0, 1, 0). Deduce that V is a subspace of R4 . Solution: (a) U is not a subspace because it does not contain the zero vector 0: the system of equations x = 0, x + 1 = 0, y + z = 0, 2x = 0 is not consistent, therefore 0 ∈ / U. (b) We have (x, x + y + 1, z, 2x) = x(1, 1, 0, 2) + (y + 1)(0, 1, 0, 0) + z(0, 0, 1, 0). Therefore any vector in V is a linear combination of v1 , v2 and v3 . On the other hand, as x, y and z range over all real numbers, and so (x, x + y + 1, z, 2x) ranges over all the vectors in V , we get every such linear combination (in more detail, r(1, 1, 0, 2) + s(0, 1, 0, 0) + 43 t(0, 0, 1, 0) = (r, r + (s − 1) + 1, t, 2r) ∈ V ). Therefore V is the set of all linear combinations of v1 , v2 and v3 , i.e. V = span{v1 , v2 , v3 }. Hence V is a subspace since the span of a set of vectors is always a subspace. Basis 74. What constraints must be imposed on a, b ∈ R, so that a 0 , is 0 b a basis for R2 . Solution: We require that the two vectors are linearly independent. Clearly this x will be the case if a, b 6= 0. Any vector ∈ R2 can now be formed y from the linear combination y 0 x a x , + = y a 0 b b which is well defined if a, b 6= 0 — the same restriction. x y 75. Find a basis for W = x − 2y + w = 0, 3x − z + 3w = 0 . z w Solution: W is the solution space of the two linear homogenous equations x − 2y + w = 0 and 3x − z + 3w = 0. (Boxes in the following designate where we will pivot.) 0 1 −2 3 0 −1 1 −2 0 0 6 −1 1 0 −1/3 0 1 −1/6 1 3 1 0 1 0 0 0 0 0 0 0 The solution is of the form s/3 − t 1/3 −1 x y s/6 1/6 0 = z s = s 1 + t 0 w t 0 1 44 The expression above shows that the two vectors span W , and the two vectors are linearly independent they are not multiples of each because 1 −1 / 3 1 0 / 6 , . other. So a basis for W is 1 0 1 0 76. Find a basis for the subspace U = {x ∈ R4 | Ax = 0} where 1 1 1 1 A = 2 0 2 0 . 3 1 3 1 Solution: 1 0 1 0 Row reducing A gives the matrix 0 1 0 1. Let x3 = s and x4 = t 0 0 0 0 be the free variables. Then x1 = −s and x2 = −t so the solutions are of the form x1 −s −1 0 x2 −t 0 −1 x= x3 = s = s 1 + t 0 x4 t 0 1 for s, t ∈ R. Thus we see that U is spanned by −1 0 0 −1 , . 1 0 0 1 Since these two vectors are not multiples of each other they are linearly independent, hence they are a basis for U . 77. Show that if {u, v} is a basis for a subspace W , then so is {u + v, v}. Solution: Let x ∈ W , then x = c1 u + c2 v for some c1 , c2 ∈ R, since {u, v} spans W . We have x = c1 u + c2 v = c1 (u + v) + (c2 − c1 )v so x ∈ span{u + v, v}. That is, still spans W . To see that {u+v, v} is linearly independent, consider c1 (u+v)+c2 v = 0, so c1 u + (c1 + c2 )v = 0, but since {u, v} is linearly independent this implies that c1 = 0 and c1 + c2 = 0, so c1 = c2 = 0 and hence {u + v, v} is linearly independent. 45 78.* Let v1 , . . . , vr be a sequence of vectors in a subspace W of Rn with the property that every vector in W can be written uniquely as a linear combination of the vectors v1 , . . . , vr . Prove that the vectors v1 , . . . , vr form a basis for W . Solution: To show that the sequence of vectors v1 , . . . , vr is a basis for W we need to show that (i) it spans W and (ii) it is linearly independent. We are told that every vector in W is a linear combination of v1 , . . . , vr , therefore the sequence of vectors spans W . We need to show that it is linearly independent. Suppose that c1 v1 + · · · + cr vr = 0. We need to show that c1 = · · · = cr = 0. We’re told that every vector in W can be written uniquely as a linear combination of the vectors v1 , . . . , vr . Well, 0 is a vector in W (W is a subspace) and certainly 0 = 0v1 + · · · + 0vr . But this must be the only way in which we can write 0 as a linear combination of v1 , . . . , vr . Therefore we must have c1 = 0, . . . , cr = 0. Hence the vectors v1 , . . . , vr are linearly independent and form a basis for W . 79. Explain why the following are not bases for the indicated subspaces. x −6 6 x + 6y = 0 ; for V = , (a) y 1 −1 x 4 (b) 1 for W = y x − 2y − 2z = 0 ; z 1 2 4 (c) 1 , 0 for W as in part (b). 1 −1 5 4 1 (d) 4 , 2 , 6 for R3 . 1 3 2 Solution: (a) (b) (c) (d) 6 −6 0 The vectors are linearly dependent: + = . −1 1 0 The equation is clearly that of a plane, and so the space W is 2 dimensional, and hence should have 2 basis vectors. 2 The vector 0 doesn’t satisfy the equation, so isn’t even in −1 W . 5 1 4 6 = 4 + 2 so these vectors are linearly dependent. 3 2 1 46 5. Eigenvalues and Eigenvetors Eigenvalue Problem 80. Let A be an n × n matrix. (a) Define what it means for x ∈ Rn to be an eigenvector of A with eigenvalue λ. (b) Define the characteristic polynomial of A. Solution: (a) An eigenvector of A with eigenvalue λ is a vector x 6= 0 satisfying Ax = λx. (b) The characteristic polynomial of A is |λI − A|. 81. Find all eigenvalues and eigenvectors for the following matrices. 0 3 (a) A1 = 6 −3 2 1 0 (b) A2 = 1 3 1 0 1 2 1 5 5 (c) A3 = 0 3 2 0 2 3 Solution: (a) First find the characteristic polynomial of A1 : |λI − A1 | = λ −3 −6 λ + 3 =λ(λ + 3) − 18 =λ2 + 3λ − 18 =(λ + 6)(λ − 3) So the two eigenvalues of A2 are λ = −6, 3. To compute the eigenspaces, solve (λI − A1 )x = 0. λ = −6 -6 -6 1 0 -3 -3 1/2 0 47 λ=3 0 0 0 0 3 -6 1 0 -3 6 -1 0 0 0 0 0 −1/2 = s s ∈ R and E3 = 1 So the eigenspaces are: E−6 1 s s∈R , 1 (b) The characteristic polynomial is: |λI − A2 | = λ−2 −1 0 −1 λ − 3 −1 0 −1 λ − 2 −1 −1 λ − 3 −1 +1 0 λ−2 −1 λ − 2 = (λ − 2) [(λ − 3)(λ − 2) − 1] − (λ − 2) = (λ − 2) = (λ − 2) [(λ − 3)(λ − 2) − 2] = (λ − 2) λ2 − 5λ + 4 = (λ − 2)(λ − 1)(λ − 4) So the three eigenvalues of A2 • λ = 2: 0 −1 0 -1 0 0 1 0 0 1 0 0 are λ = 2, 1, 4. −1 0 −1 −1 −1 0 −1 −1 −1 0 −1 0 1 1 0 -1 −1 0 0 1 1 0 0 0 −1 So the eigenspace is E2 = s 0 1 • λ = 1: −1 −1 0 −1 −2 −1 0 −1 −1 1 0 −1 0 1 1 0 0 0 1 So the eigenspace is E1 = s −1 1 • λ = 4: 2 −1 0 −1 1 −1 0 −1 2 1 0 −1 0 1 −2 0 0 0 48 0 0 0 0 0 0 0 0 0 0 0 0 s∈R . 0 0 0 0 0 0 s∈R . 0 0 0 0 0 0 1 So the eigenspace is E4 = s 2 s ∈ R . 1 (c) Using the first column to compute the determinant: λ−1 −5 −5 0 λ−3 −2 0 −2 λ − 3 |λI − A3 | = λ−3 −2 −2 λ − 3 = (λ − 1) [(λ − 3)(λ − 3) − 4] = (λ − 1) λ2 − 6λ + 5 = (λ − 1) = (λ − 1)(λ − 1)(λ − 5) So the three eigenvalues of A3 are λ = 1, 1, 5. • λ = 1: 0 -5 −5 0 0 −2 −2 0 0 −2 −2 0 0 1 1 0 0 0 0 0 0 0 0 0 0 1 Eigenspace is E1 = s 0 + t −1 s, t ∈ R . 0 1 • λ = 5: −5 −5 0 4 0 2 −2 0 0 −2 2 0 1 −5/4 −5/4 0 0 2 −2 0 0 −2 2 0 1 0 −10/4 0 0 1 −1 0 0 0 0 0 10/4 So the eigenspace is E5 = s 1 s ∈ R . 1 82. Show that if A has eigenvalues λ1 , λ2 , . . . , λn then (a) cA has eigenvalues cλ1 , cλ2 , . . . , cλn for any constant c ∈ R. (b) If A−1 exists, then A−1 has eigenvalues 1/λ1 , 1/λ2 , . . . , 1/λn m m (c) Am has eigenvalues λm 1 , λ2 , . . . , λn for m = 1, 2, 3, . . .. Solution: 49 (a) If 0 = |λi I − A| then 0 = cn |λi I − A| = |c(λi I − A)| = |(cλi )I − (cA)|. Hence cA has eigenvalues cλi , i = 1, 2, . . . , n. (b) If 0 = |λi I − A| then λi 6= 0 (otherwise det A = 0 and so A−1 doesn’t exist). Hence 0 = |λi I − A| = |A|.|λi .A−1 − I| = (λi )n |A|.|A−1 − since λi 6= 0. Hence 0 = | λ1i I − A−1 | so for all i = 1, 2, . . . , n. 1 λi 1 I| λi is an eigenvalue of A−1 (c) Ax = λi x so A2 x = A(Ax) = A.λi x = λi (Ax) = λ2i x. Then, if Ak x = λki x, we get Ak+1 x = A.Ak x = A.λki x = λki .(Ax) = λki .(λi x) = λk+1 x. i So by mathematical induction, if λi is an eigenvalue for A, then m m Am x = λ m i x i.e. A has eigenvalues λi for all i. Note they have the same eigenvectors. 83. Show that if 0 < θ < π then the matrix cos θ − sin θ A= sin θ cos θ has no real eigenvalues. Give a geometric interpretation of this fact, given that the vector Av is obtained from the vector v by a counterclockwise rotation about 0 through an angle θ. Solution: The characteristic polynomial of A is given by the determinant |λI − A| and hence is equal to λ − cos θ sin θ − sin θ λ − cos θ = (λ − cos θ)2 + sin2 θ = λ2 − 2λ cos θ + cos2 θ + sin2 θ = λ2 − 2λ cos θ + 1 The discriminant of the quadratic λ2 − 2λ cos θ + 1 is 4 cos2 θ − 4 = −4 sin2 θ < 0, since 0 < θ < π. Therefore the quadratic has no real roots and hence A has no real eigenvalues. Given that Av is obtained from v by a rotation about the origin, it is clear that the only way that Av can be a multiple of v is if Av = v (a rotation through 0 radians) or if Av = −v (a rotation through π radians). Both of these possibilities are excluded by the constraint on θ. 50 Properties of eigenvalues 84. For each of the following matrices: 5 0 0 A = 0 4 1 , 0 1 4 3 1 1 B = 1 3 1 ; 1 1 3 (a) determine all eigenspaces, Eλ , and state whether the multiplicity of λ equals the dimension of Eλ ; (b) check that the determinant equals the product of the eigenvalues, and the trace is their sum. Solution: (a) Characteristic polynomial: |λI − A| = λ−5 0 0 0 λ − 4 −1 0 −1 λ − 4 λ − 4 −1 −1 λ − 4 =(λ − 5) (λ − 4)2 − 1 =(λ − 5) λ2 − 8λ + 15 =(λ − 5) =(λ − 5)(λ − 3)(λ − 5) So the roots, and hence the eigenvalues are λ = 3, 5, and the multiplicities are 1 and 2 respectively. λ=3 λ=5 −2 0 0 0 0 −1 −1 0 0 −1 −1 0 1 0 0 0 0 1 1 0 0 0 0 0 0 E3 = s−1 s ∈ R 1 0 0 0 0 0 1 −1 0 0 −1 1 0 0 0 0 0 0 1 −1 0 0 0 0 0 0 1 E5 = s 0 + t 1 s, t ∈ R 0 1 The dimensions of E3 , and E5 are 1 and 2 respectively, so they are equal to the multiplicity of the corresponding eigenvalues. det(A) = 75 = 3 × 5 × 5, 51 tr(A) = 13 = 3 + 5 + 5 (b) Characteristic polynomial: |λI − B| = λ − 3 −1 −1 −1 λ − 3 −1 −1 −1 λ − 3 λ − 3 −1 −1 −1 −1 λ − 3 +1 −1 −1 λ − 3 −1 λ − 3 −1 −1 =(λ − 3) (λ − 3)2 − 1 − 2(λ − 2) =(λ − 3) =(λ − 3)(λ2 − 6λ + 8) − 2(λ − 2) =(λ − 3)(λ − 2)(λ − 4) − 2(λ − 2) =(λ − 2) [(λ − 3)(λ − 4) − 2] =(λ − 2)(λ2 − 7λ + 10) =(λ − 2)2 (λ − 5) So the roots, and hence the eigenvalues are λ = 2, 5 with multiplicity 2 and 1 respectively. λ=5 λ=2 −1 −1 −1 0 −1 −1 −1 0 −1 −1 −1 0 1 1 1 0 0 0 0 0 0 0 0 0 −1 −1 E2 = s 1 + t 0 0 1 2 −1 −1 0 −1 2 −1 0 −1 −1 2 0 1 −1/2 −1/2 0 0 3/2 −3/2 0 0 −3/2 3/2 0 1 0 −1 0 0 1 −1 0 0 0 0 0 1 E5 = s 1 1 The dimensions of E2 , and E5 are 2 and 1 respectively, so they are equal to the multiplicity of the corresponding eigenvalues. det(B) = 20 = 2 × 2 × 5, tr(B) = 9 = 2 + 2 + 5 85. Answer the following true or false and give reasons. (a) λ4 + λ2 − 1 cannot be the characteristic polynomial of a 3 × 3 matrix. (b) An eigenvalue of multiplicity 7 can have an eigenspace of dimension 3. (c) A real n × n matrix always has a real eigenvalue when n is odd. (d) If tr(A) = 0 then A cannot be invertible. (e) A polynomial with real coefficients must have real roots. 52 (f) If λ is an eigenvalue of an n × n matrix A with eigenvector x, then for any scalar µ, λ − µ is an eigenvalue of A − µIn with x as a corresponding eigenvector. Solution: (a) True: a 3 × 3 matrix can have at most 3 eigenvalues, whereas a degree 4 polynomial can have 4 roots. Alternatively, if A is an n × n matrix then its characteristic polynomial is of degree n, hence the degree of the polynomial λ4 + λ2 − 1 is too large. (b) True: the dimension of an eigenspace is equal to or less than its multiplicity. (c) True: The characteristic polynomial of an n × n matrix is of degree n, and a polynomial of odd degree always has a real root (complex roots of polynomials come in complex conjugate pairs). 0 1 0 1 1 0 (d) False: e.g. = . 1 0 1 0 0 1 (e) False: consider λ2 + 1 = 0, which has roots λ = ±i. (f) True: if Ax = λx, then (A − µIn )x = λx − µx = (λ − µ)x; therefore x is an eigenvector of A − µIn with eigenvalue λ − µ. 86. (a) Show that λ = 0 is an eigenvalue of A if and only if A is not invertible. (b) Without calculation find one eigenvalue and two linearly independent eigenvectors of 1 2 3 A = 1 2 3 . 1 2 3 Solution: (a) The determinant of A is equal to the product of eigenvalues, which is equal to zero if and only if one of the eigenvalues equals zero. The matrix is invertible if and only if the determinant is not equal to zero. Hence the result. (b) The determinant of A is zero, so λ = 0 must be an eigenvalue. The vectors 3 2 0 and −1 −1 0 when pre-multiplied by A, take combinations of columns which result in zero, and therefore give the zero vector, thus they are eigenvectors corresponding to the eigenvalue λ = 0. 53 87. (a) Let A be an n × n matrix. Show that A and its transpose At have the same characteristic polynomial and hence the same eigenvalues. (b) Beware! Even though A and At have the same eigenvalues, the eigenspace for A at an eigenvalue λ need not be equal to the eigenspace for At at λ. Investigate this phenomenon with the matrix 1 1 A= 0 1 Solution: (a) The characteristic polynomial of A is det(λI − A). The characteristic polynomial of At is det(λI − At ). The claim is that these two polynomials are equal. Why? First we need to recall the following property of the determinant: the determinant of a square matrix is equal to the determinant of its transpose, i.e. det(B) = det(B t ) for an n × n matrix B. Since the matrix I is equal to its own transpose we have λI − At = (λI − A)t . Therefore det(λI − At ) = det((λI −A)t ) = det(λI −A). Therefore the characteristic polynomials of A and At are equal (and so the matrices have exactly the same eigenvalues). (b) The given matrix A has characteristic polynomial equal to (λ−1)2 . Therefore A has one eigenvalue λ = 1 of multiplicity 2. Solving the homogenous system (λI − A)x = 0 when λ = 1 we get 0 0 0 0 -1 0 1 0 0 0 0 0 1 and so we find that the λ = 1 eigenspace of A is span . On the 0 other hand At has exactly the same characteristic polynomial, but if we solve the homogenous system (λI − At )x = 0 when λ = 1 we get 0 -1 1 0 0 0 0 0 0 0 0 0 which shows that the λ = 1 eigenspace of At is span 88. Suppose that for some 3 × 3 matrix A, we have 1 2 1 2 A 1 = 2 and A 0 = 0 1 2 1 2 54 0 . 1 (a) Give one eigenvalue of A. 0 (b) Explain why 1 is an eigenvector for this eigenvalue and hence 0 0 find A 1. 0 (c) If det(A)=12, then what is the multiplicity of the eigenvalue from (a)? Solution: (a) 2 is an eigenvalue of A. (b) Since [1 1 1]t , [1 0 1]t ∈ E2 and E2 is a subspace then [0 1 0]t = [1 1 1]t − [1 0 1]t ∈ E2 . Hence A[0 1 0]t = [0 2 0]t . (c) The multiplicity m of an eigenvalue is greater than or equal to the dimension of the corresponding eigenspace. As the two eigenvectors are linearly independent, the dimension of E2 is at least 2, and so the multiplicity of the eigenvalue 2 is at least 2. As A is a 3 × 3 matrix, the multiplicity of the eigenvalue 2 can be at most 3. The product of the three eigenvalues is equal to the determinant, so 2×2×λ3 = 12, and so λ3 = 3, and the eigenvalue 2 has multiplicity 2. Diagonalisation 89. Consider the matrices −12 7 A1 = −7 2 4 2 2 A2 = 2 4 2 2 2 4 −1 3 9 A3 = 0 −7 −18 . 0 2 5 For each matrix Ai above (a) Determine the eigenvalues and eigenvectors. (b) For each eigenvalue, state its multiplicity and give the dimension of its associated eigenspace. (c) Hence determine whether the matrix Ai is diagonalisable, stating the reason for your answer. 55 (d) For each diagonalisable matrix Ai , determine a matrix P such that P −1 Ai P = D, where D is a diagonal matrix. What is D? (For the purposes of this exercise, order your eigenvalues from smallest to largest i.e. λ1 ≤ λ2 . . . ≤ λn .) (e) Determine if each matrix Ai is invertible and, where possible, use your results from (d) to write the inverse matrix A−1 i in the form P ∆P −1 , where ∆ is a diagonal matrix. (You do not need to find A−1 if not possible by this method.) Solution: A1 : (a) A1 has eigenvalue −5 with multiplicity 2. Eigenvectors are x = t(1, 1), t 6= 0. (b) E1 = span{(1, 1)}, dim(E1 ) = 1. (c) A1 is not diagonalisable as it does not have two linearly independent eigenvectors. (d) No answer required as A1 is not diagonalisable. (e) det(A) = 25 6= 0 so A is invertible though not diagonalisable. A2 : (a) A2 has eigenvalues 2 and 8 with multiplicities 2 and 1 respectively. For λ = 2 the eigenvectors are x = s(−1, 1, 0) + t(−1, 0, 1), s and t not both zero. For λ = 8 the eigenvectors are x = t(1, 1, 1), t 6= 0. (b) E2 = span{(−1, 1, 0), (−1, 0, 1)}, dim(E2 ) = 2. E8 = span{(1, 1, 1)}, dim(E8 ) = 1. (c) A2 is diagonalisable as it has 3 linearly independent eigenvectors. 2 0 0 −1 −1 1 0 1 and D = 0 2 0. (d) P = 1 0 0 8 0 1 1 1/2 0 0 (e) det(A2 ) = 32 6= 0 so A2 is invertible. ∆ = 0 1/2 0 0 0 1/8 with P as above. A3 : (a) A3 has eigenvalue -1 with multiplicity 3. The eigenvectors are x = t(1, 0, 0) + s(0, −3, 1), s and t not both zero. (b) E3 = span{(1, 0, 0), (0, −3, 1)}, dim(E3 ) = 2. (c) A3 is not diagonalisable as it does not have three linearly independent eigenvectors. (d) No answer required as A3 is not diagonalisable. (e) det(A3 ) = −1 6= 0 so A3 is invertible though not diagonalisable. 2 3 90. Let A = . 4 1 56 1 (a) Verify that v1 = is an eigenvector of A. What is the corre1 sponding eigenvalue λ1 ? (b) Use the trace of A to find a second eigenvalue λ2 . (c) Find the eigenspace for λ2 . (d) Write down an invertible matrix P such that P −1 AP = D, where D = diag(λ1 , λ2 ). Solution: 2 3 1 5 1 (a) = = 5 . Thus (1, 1)t is an eigenvector of A 4 1 1 5 1 corresponding to the eigenvalue 5. (b) A is a 2 × 2 matrix so it has two eigenvalues. We know λ1 = 5. But tr(A) = 2 + 1 = 3 = λ1 + λ2 so λ2 = −2. −4 −3 (c) −2I − A = . Solving (−2I − A)x = 0 gives x = −4 −3 s[−3/4 1]t . Thus E−2 = span{[−3/4 1]t }. 5 0 1 −3/4 (d) D = and P = . 0 −2 1 1 3 0 0 91. Show that the matrix A = 0 2 0 is not diagonalizable. 0 1 2 Solution: The characteristic polynomial of A is λ−3 0 0 0 λ−2 0 0 −1 λ − 2 = (λ − 3)(λ − 2)2 and so A has eigenvalues λ = 3 (multiplicity 1) and λ = 2 (multiplicity 2). By Theorem 5.6, A is diagonalizable if and only if it has 3 linearly independent eigenvectors v1 , v2 and v3 . Since λ = 3 has multiplicity 1 the λ = 3 eigenspace has dimension 1 (since the dimension of an eigenspace is always ≤ to the multiplicity of the corresponding eiegenvalue). So if the linearly independent vectors v1 , v2 and v3 exist then only one of them could be an eigenvector for λ = 3. So the only way A can be diagonalizable is if the λ = 2 eigenspace has 2 linearly independent eigenvectors and hence is of dimension 2. To find the λ = 2 eigenspace E2 we solve (λI − A)x = 0 when λ = 2: -1 0 0 1 0 0 0 0 -1 0 1 0 57 0 0 0 0 0 0 0 0 0 0 0 0 0 0 has dimension 1. Therefore A is not diagHence E2 = span 1 onalizable. 92.* (a) A square matrix A is called nilpotent if Ak = 0 for some integer k > 1. Show that 0 is the only eigenvalue of a nilpotent matrix. (b) A square matrix A is called idempotent if A2 = A. Show that the only possible eigenvalues of A are 0 or 1. (c) Suppose that A is a diagonalizable matrix such that every eigenvalue of A is 0 or 1. Prove that A is idempotent. Solution: (a) Let Ax = λx. Note that, since x is an eigenvector, x 6= 0. Observe that A2 x = Aλx = λ2 x. More generally (you can prove this by induction), Ak x = λk x. Since Ak = 0, this implies λk x = 0, and hence (given x 6= 0) we can see that λk = 0 and hence λ = 0. (b) Let Ax = λx. As before A2 x = λ2 x, but as A is idempotent A2 = A, and so A2 x = λx. Thus λ2 x = λx, so (λ2 − λ)x = 0, and as before x 6= 0 so (λ2 − λ) = 0, which can be rewritten λ(λ − 1) = 0. Thus the possible values of λ are 0 or 1. (c) Since A is diagonalizable there is an invertible matrix P such that P −1 AP = D, where D is a diagonal matrix whose diagonal entries are all either 0 or 1 (since the eigenvalues of A are all either equal to 0 or 1). Observe that D2 = D (since 02 = 0 and 12 = 1). We have (P −1 AP )2 = (P −1 AP )(P −1 AP ) = P −1 A2 P and so P −1 A2 P = P −1 AP . Therefore A2 = A, i.e. A is idempotent. 93. Let s be a real number, and let s −1 s−1 0 1 1 A= 0 1 1 2 2 0 s −s s −s 0 0 . 0 2 (a) For which values of s is A diagonalisable? (b) For s = 0, find an invertible matrix P such that P −1 AP is a diagonal matrix. 58 Solution: (a) First, find the eigenvalues of A. The characteristic polynomial is λ−s 1 1−s 0 0 λ−1 −1 0 det(λI − A) = det 0 −1 λ−1 0 0 s − s2 s − s2 λ − 2 λ−1 −1 0 λ−1 0 = (λ − s) det −1 2 2 s−s s−s λ−2 λ − 1 −1 = (λ − s)(λ − 2) det −1 λ − 1 = (λ − s)(λ − 2) (λ − 1)2 − 1 = (λ − s)(λ − 2)2 λ. So the eigenvalues of A are 2, 0 and s. For A to be diagonalisable, there must exist four linear independent eigenvectors of A (Theorem 5.6 in the lecture). Since the eigenvalue 2 has multiplicity > 1 (2 if s 6= 2 or 3 if s = 2), it is sensible to first find out which dimension the eigenspace E2 has. Remark: If dim E2 is less than the multplicity of the eigenvalue 2, then we know we cannot find four linearly independent eigenvectors, since the combined dimension of the eigenspaces is at most 3. So compute E2 by solving (2I − A)x = zero. Apply row reductions: 2−s 0 2−s 2−s 1 1−s 0 0 0 1 −1 1 −1 0 (2I − A) = 0 0 0 2(s − s2 ) −1 1 0 2 2 0 0 0 0 s−s s−s 0 If s 6= 0, 1, 2, then this linear system has only one free parameter, hence dim E2 = 1. So A is clearly not diagonalisable if s 6= 0, 1, 2. Now check if A is diagonalisable if s either 0, 1 or 2: s = 2: Continue the row reduction above for s = 2: 0 0 0 0 0 1 0 0 0 1 −1 0 0 0 1 0 0 0 −4 0 0 0 0 0 . 0 0 0 0 0 0 0 0 So the eigenspace has dimension dim E2 = 2. However, for s = 2, the multiplicity of the eigenvalue 2 is 3, so dim E2 is less than the multiplicity, and thus A is not diagonalisable for s = 2. s = 1: For s = 1, the row reduction is already complete: 1 0 1 0 0 1 −1 0 0 0 0 0 . 0 0 0 0 59 0 0 . 0 0 The solution is −1 0 n 1 0o E2 = span 1 , 0 . 0 1 So the eigenspace E2 has dimension dim E2 = 2, which equals the multiplicity of the eigenvalue 2. The other eigenvalues are 0 and s = 1, and both eigenspaces E0 and E1 are at least one-dimensional, so in total we find four linearly independent eigenvectors by picking an eigenvector from E0 and E1 each and combining them with the two basis vectors of E2 . Hence A is diagonalisable for s = 1. s = 0: Continue the row reduction above for s = 0: 1 0 1 0 2 0 2 0 0 1 −1 0 0 1 −1 0 0 0 0 0 . 0 0 0 0 0 0 0 0 0 0 0 0 The solution is, as in the previous case, 0 −1 n 1 0o E2 = span 1 , 0 . 1 0 So the eigenspace E2 has dimension dim E2 = 2, which equals the multiplicity of the eigenvalue 2. But now the eigenvalue 0 = s also has multiplicity 2, so we need to determine the dimension of the eigenspace E0 in order to find out if we can obtain four linearly indepdenent eigenvectors. Solve (0I − A)x = zero: 0 1 1 0 0 1 1 0 0 0 0 1 0 −1 −1 0 0 0 0 0 . 0 −1 −1 0 0 0 0 0 0 0 0 −2 The solution is, 1 0 n0 −1o E0 = span 0 , 1 . 0 0 So E0 is indeed of dimension dim E0 = 2, and we can obtain four linearly independent eigenvectors for A by choosing two basis vectors from E0 and combining them with two basis vectors from E2 . Hence A is diagonalisable for s = 0. 60 (b) By part (a) we know that A is diagonalisable for s = 0, and we can find a matrix P by choosing its columns to be four linearly independent eigenvectors of A (Theorem 5.7 in the lecture). Use the basis vectors for E0 and E2 from part (a): 1 0 −1 0 0 −1 1 0 . P = v 1 v2 v3 v4 = 0 1 1 0 0 0 0 1 Then we know that 0 0 P −1 AP = 0 0 0 0 0 0 0 0 2 0 0 0 = diag(0, 0, 2, 2). 0 2 Cayley-Hamilton Theorem 94. Use the Cayley-Hamilton theorem to compute the inverse of the matrix 1 2 5 A = 0 2 −1 . 0 0 1 Solution: Expanding around the bottom row, the characteristic polynomial for A is |λI − A| = (λ − 1) λ − 1 −2 = (λ − 1)2 (λ − 2) = λ3 − 4λ2 + 5λ − 2. 0 λ−2 By the Cayley-Hamilton Theorem A3 − 4A2 + 5A − 2I = O and therefore 2A−1 = A2 − 4A + 5I 1 6 8 4 8 20 5 0 0 = 0 4 −3 − 0 8 −4 + 0 5 0 0 0 1 0 0 4 0 0 5 2 −2 −12 1 . = 0 1 0 0 2 Finally A−1 1 −1 −6 = 0 1/2 1/2 . 0 0 1 61 1 1 0 95. Let A = 1 0 1. Use the Cayley-Hamilton theorem to compute 0 1 1 3 4 A and A . Solution: |λI − A| = First we find the characteristic equation. λ − 1 −1 0 −1 λ −1 0 −1 λ − 1 = (λ − 1) λ −1 −1 −1 + −1 λ − 1 0 λ−1 = (λ − 1)[λ(λ − 1) − 1] − (λ − 1) = (λ − 1)[λ(λ − 1) − 2] = λ3 − 2λ2 − λ + 2 Therefore A3 − 2A2 − A + 2I = calculating 4 2 2 1 2A2 + A − 2I = 2 4 2 + 1 2 2 4 0 0 and hence A3 = 2A2 + A − 2I, 1 0 2 0 0 3 3 2 0 1 − 0 2 0 = 3 2 3 1 1 0 0 2 2 3 3 and then A4 = A(A3 ) = A(2A2 + A − 2I) = 2A3 + A2 − 2A, 6 6 4 2 1 1 2 2 0 6 5 A4 = 6 4 6 + 1 2 1 − 2 0 2 = 5 6 4 6 6 1 1 2 0 2 2 5 5 so 5 5 6 Dynamical Systems −5 9 96. Consider the dynamical system with xk+1 = Axk , where A = . −3 11 2 Show that one eigenvector of A is (2, 1)T . What is the corresponding eigenvalue? Hence determine the remaining eigenvalue and its eigenvectors. What happens to the system over a long period of time? −5 9 2 −1 2 Solution: = = (−1/2) so −3 11/2 1 −1/2 1 λ1 = −1/2 for v1 = (2, 1)T . Since λ1 + λ2 =tr(A)=1/2, we get λ2 = 1. For λ2 = 1, eigenvector s are multiples of v2 = (3, 2)T . Ak x0 gives 2 2 3 2 3 1 k k k k + c2 = c1 λ1 +c2 λ2 = c1 (− 2 ) + xk = A c1 1 2 1 2 1 62 c2 .1k . 3 2 So xk → c2 3 of 2 3 2 as k → ∞. Hence whatever x0 is, xk is a multiple 97.* (a) A square matrix B is called nilpotent if B k = 0 for some integer k > 1. Show that 0 is the only eigenvalue of a nilpotent matrix. (b) A square matrix C is called idempotent if C 2 = C. What are the possible eigenvalues of an idempotent matrix? Solution: (a) Let Bx = λx. Note that, since x is an eigenvector, x 6= 0. Observe that B 2 x = Bλx = λ2 x. More generally (you can prove this by induction), B k x = λk x. Since B k = 0, this implies λk x = 0, and hence (given x 6= 0) we can see that λk = 0 and hence λ = 0. (b) Let Cx = λx. As before C 2 x = λ2 x, but as C is idempotent C 2 = C, and so C 2 x = λx. Thus λ2 x = λx, so (λ2 − λ)x = 0, and as before x 6= 0 so (λ2 − λ) = 0, which can be rewritten λ(λ − 1) = 0. Thus the possible values of λ are 0 or 1. 98. Find the steady-state probability vectors for the Markov processes with transition matrices: 1/2 3/4 1/3 3/4 1/3 (a) (b) 1/2 1/4 1/3 1/4 2/3 0 0 1/3 1−b a (c) b 1−a (with 0 < a, b < 1) Solution: For each of the probability matrices P we must solve (I − P )x = 0. 1/4 −1/3 0 1 −4/3 0 4/3 =⇒ =⇒ x = t . (a) −1/4 1/3 0 0 0 0 1 As we need a probability vector we choose t = 1/(4/3 + 1) = 3/7 and q = (4/7, 3/7)0 . 63 1/2 −3/4 −1/3 0 (b) −1/2 3/4 −1/3 0 0 0 2/3 0 3/2 x = t 1 . 0 1 −3/2 −2/3 0 0 0 1 0 0 0 0 0 =⇒ =⇒ 3/5 Choose t = 1/(3/2 + 1) = 2/5 and q = 2/5. 0 b −a 0 1 −a/b 0 a/b =⇒ x = t . (c) =⇒ −b a 0 0 0 1 a/(a + b) Choose t = 1/(a/b + 1) = b/(a + b) and q = . b/(a + b) 99. A country is divided into four regions (A, B, C and D). It is found that each year 18% of the residents in A move to B, 12% move to C and 20% move to D. Of the residents in B, 30% move to C, 10% move to D and only 5% move to A. Of the residents in D, 50% move to A, 25% move to C and 10% move to B; while the residents of C will either stay where they are or move to A with equal probability. Write down the transition matrix for this Markov process. Solution: Using the order A, B, C, D the transition matrix is 0.50 0.05 0.50 0.50 0.18 0.55 0 0.10 P = 0.12 0.30 0.50 0.25 . 0.20 0.10 0 0.15 Note that the columns all sum to 1. 100.* Fibonacci numbers turn up in a huge variety of applications. They are determined according to the following rules: F0 = 0; F1 = 1; Fk+1 = Fk + Fk−1 ∀ k ≥ 1. 1 1 Fk+1 1 , k ≥ 0 and x0 = Let A = , xk = 1 0 Fk 0 Explain why the sequence {Fk : k = 0, 1, 2...}can be determined from xk+1 = Axk . Hence show xk+1 = Ak x0 . Show that Fk = Find F1000 . λk1 −λk2 √ 5 Show that as k → ∞, Solution: where λ1 , λ2 are eigenvalues of A, (λ1 > λ2 ). Fk+1 Fk → √ 1+ 5 2 Eigenvalues are λ1 = (Golden Ratio) . √ 1+ 5 2 , λ2 = √ 1− 5 2 . The corresponding eigenvectors are multiples of v1 = 64 λ1 , v2 = 1 λ2 . 1 Then x0 = Hence xk = 1 0 = Fk+1 Fk Therefore Fk = √1 v1 5 √1 (λk 5 1 Fk+1 = Fk = − √1 v2 5 √1 λk 5 1 λ1 1 so xk = − √1 λk 5 2 √1 Ak (v1 5 − v2 ). λ2 . 1 − λk2 ) and √1 (λk+1 5 1 √1 (λk 5 1 − λk+1 2 ) − λk2 ) as k → ∞, since |λ2 /λ1 | < 1. 65 = λ1 − λ2 (λ2 /λ1 )k → λ1 1 − (λ2 /λ1 )k