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Math 308 C Final Exam Your Name Spring 2015, 6-8-2015 Your Signature Student ID # 1. 2. 3. 4. 5. 6. 7. 8. Form Bonus P 35 25 17 17 20 7 12 5 3 (10) 141 Points of • No books are allowed. But you are allowed one sheet (10 x 8) of handwritten notes (back and front). You may use a calculator. • Place a box around your final answer to each question. • If you need more room, use the back of each page and indicate to the grader how to find the logic order of your answer. • Raise your hand if you have questions or need more paper. • For TRUE/FALSE problems, you just need to cross the right box. For each correct answer, you will get 1 point, for each incorrect answer, -1 point is added. For no answer you will get zero points. In each subsection of the TRUE/FALSE part, you can never get less than zero points. • In order to get points for formal correctness, underline vectors, use {}-brackets for sets, declare parameters, mark equivalent matrices properly and keep a reasonable order and neatness. Do not open the test until everyone has a copy and the start of the test is announced. GOOD LUCK! 1.(35 points) For each correct answer in the TRUE/FALSE part, you will get 1 point, for each incorrect answer, there will be one point subtracted, i.e. you get -1 point. For no answer, you get 0 points. You can not get less than 0 points out of one subproblem (which are the problems, (a)-(g)) (a) A homogeneous system can either have a unique solution or infinitely many solutions. The Gauss-Jordan algorithm leads to equivalent matrices. A homogeneous system with at least one free variable has infinitely many solutions. Let A be a (4, 6)-matrix. Then any solution to Ax = 0 is a vector in R6 . TRUE FALSE TRUE TRUE FALSE FALSE TRUE FALSE Let 0 6= u. Then the span of {u} has infinitely many elements. Let S1 $ S2 ⊆ Rn . Then span(S1 ) $ span(S2 ). If Ax = b has a solution, then b ∈ row(A). The zero vector is always an element of the span of vectors. Let m > n and let S = {u1 , u2 , . . . , un } ⊆ Rm . Then S is linearly dependent. For any vector u and c ∈ R, the set {u, cu} is linearly dependent. If [A|0] has exactly one solution, the columns of A are linearly independent. TRUE FALSE (b) TRUE TRUE TRUE TRUE FALSE FALSE FALSE FALSE TRUE FALSE TRUE FALSE TRUE FALSE TRUE FALSE TRUE TRUEE FALSE FALSE TRUE FALSE TRUE FALSE (c) Let S = span{u1 , u2 } and let u1 , u2 be linearly independent. Then dim(S) = 2. Let A be an (n, n)-matrix with rank(A) = n. Then nullity(A) = 0. {u1 , u2 , u3 , u4 } ⊆ R3 is a linearly dependent set. Let S ⊆ Rn be a subspace of dimension m. Then any basis of S contains exactly m elements. Let S1 , S2 ⊆ R3 be subspaces with dim(S1 ) = 1, dim(S2 ) = 2. Then S1 ⊆ S2 . If A is a matrix in echelon form with no zero rows, then the rows form a basis of row(A). Let A be an (n, m)-matrix. Then nullity(A) ≤ m. (d) For any homomorphism T there is a corresponding matrix Let T : Rm → Rn be a homomorphism and let A be its corresponding matrix. Then A is an (n, m)-matrix. Let m > n and let T : Rm → Rn be a homomorphism. Then T can be surjective, but not injective. Let T be a homomorphism with corresponding (m, n)matrix A. Then {Au | u ∈ Rn } = range(T ). Let T be a surjective homomorphism. Then the range of T is equal to the domain of T . Let T : Rn → Rn be a homomorphism with corresponding matrix A. If [A|0] has infinitely many solutions then T is not surjective. TRUE FALSE A with T (u) = Au. TRUE FALSE TRUE FALSE TRUE FALSE TRUE FALSE TRUE FALSE (e) (f) Let A be an (n, m)-matrix and B be an (m, k)-matrix. Then AB is an (n, k)-matrix. It is possible for a non-zero (n, n)-matrix that Ak = 0n,n for some k ∈ N. Let A be an (n, n)-matrix. Then det(A2 ) = (det(A))2 . A square matrix A is singular, if and only det(A) = 0. Let A be an (m, n)-matrix. Then rank(A) ≤ m. Matrices in this part have dimension (n, n). An eigenvector cannot be the zero vector, but an eigenvalue may be 0. Let A be a matrix, u ∈ Rn , λ ∈ R such that Au = λu. Then u is an eigenvector of A with eigenvalue λ. Let A be a matrix with eigenvalue λ. Then the matrix A − λIn is not invertible. Let Eλ be an eigenspace for a matrix A. Then 1 ≤ dim(Eλ ). Let A be a matrix with characteristic polynomial (λ − 1)2 (λ + 1).Then A is invertible. TRUE FALSE TRUE FALSE TRUE TRUE TRUE FALSE FALSE FALSE TRUE FALSE TRUE FALSE TRUE FALSE TRUE FALSE TRUE FALSE The zero vector is orthogonal to every other vector of the same Euclidean space. Two non-zero vectors with all components being negative cannot be orthogonal. TRUE FALSE TRUE FALSE (g) 2.(1+4+1+6+2+(1+1)+(2+1+1+1+4)=25 points) Let h be a real number. Consider the following linear homomorphism: u1 hu1 + 4u2 + 2hu3 . 2u2 + hu3 Th : u2 → u3 (4 − h)u1 + 5u2 + 3u3 (a) Find the corresponding matrix Ah , such that Th (u) = Au for all u ∈ R3 . h 4 2h 0 2 h Ah = 4−h 5 3 (b) Determine h such that Th /Ah is invertible. h 4 2h 0 2 h ) = 6h − 5h2 det( 4−h 5 3 and det(A) = 0 ⇐⇒ h(6 − 5h) = 0 ⇐⇒ h = 0 or h = 6/5. So Th /Ah is invertible if and only if h 6= 0 or h 6= 6/5. For the rest of this problem, i.e. for (c)-(g), we set h = 1, so that Th becomes T1 . (c) Write down the matrix A1 that corresponds to T1 . 1 4 2 0 2 1 3 5 3 (d) Find the inverse of T1 . 1 4 2 1 0 0 0 2 1 0 1 0 3 5 3 0 0 1 1 4 2 1 0 0 2 1 0 1 0 → 0 0 −7 −3 −3 0 1 1 4 2 1 0 0 0 2 1 0 1 0 0 0 1 −6 7 2 1 4 0 13 −14 −4 6 −6 −2 → 0 2 0 0 0 1 −6 7 2 1 0 0 1 −2 0 0 1 0 3 −3 −1 0 0 1 −6 7 2 So T −1 1 −2 0 : u 7→ 3 −3 −1 u −6 7 2 (e) Find u such that T1 (u) = [1, 0, −2]t . 1 −2 0 1 1 3 −3 −1 0 = 5 −6 7 2 −2 −10 (f) Let R : R3 → R4 be a homomorphism that has 2 0 2 0 −1 2 B= 1 −1 −1 2 2 1 as matrix. (i) What is the domain of R? R3 (ii) What is the codomain of R? R4 (g) Consider the concatenation R ◦ T1 , which maps u ∈ R3 to R(T1 (u)) (first T1 , then R). (i) What is the domain of R ◦ T1 . Because we first apply T1 , the domain of R must be the same as the domain of T1 which is R3 . (ii) What is the codomain of R ◦ T1 ? As we lastly apply R, the codomain of R ◦ T1 is the same as the codomain of R which is R4 . (iii) Can R ◦ T1 be surjective? Justify briefly. As the domain is R3 and the codomain is R4 , i.e. 3 < 4, the homomorphism cannot be surjective. (iv) Find the associated matrix C such that R ◦ T1 (u) = Cu for every u ∈ R3 . We have T1 (u) = Au and R(v) = Bv, we get the following: R ◦ T1 (u) = R ◦ (Au) = BA(u). So we need to multiply the two matrices: 2 0 2 8 18 10 1 4 2 0 −1 8 5 2 6 BA = 1 −1 −1 · 0 2 1 = −2 −3 −2 3 5 3 2 2 1 5 17 9. 3.(5+2+5+2+2+1=17 points) Consider the following matrix: −1 1 −2 0 2 1 2 −4 −2 0 . A = 0 −1 2 −1 2 4 −2 0 (a) Find the null space of A. −1 1 −2 0 2 1 −1 1 −2 0 2 1 0 0 −1 2 −4 −2 0 0 → 0 −1 2 −4 −2 0 [A|0] = 2 −1 2 4 −2 0 0 0 0 0 0 0 1 The backward substitution gives then: x6 = 0, x5 = t1 , x4 = t2 , x3 = t3 , x2 = 2t3 − 4t2 − 2t1 and x1 = −4t2 . Hence the of A is 0 −4 0 −2 −4 2 0 0 1 null(A) = span{ , 1 , 0 }. 0 1 0 0 0 0 0 0 0 . 0 null space (b) Verify that the vector v = [−4, −8, 0, 1, 2, 0]t is an element of null(A).Any element u of null(A) satisfies Au. So −4 −8 −1 1 −2 0 2 1 0 = [0, 0, 0]t . 0 −1 2 −4 −2 0 1 2 −1 2 4 −2 0 2 0 (c) Find a basis of null(A), that contains the vector v = [−4, −8, 0, 1, 2, 0]t . We can use Algorithm II for finding a basis. −4 0 −4 0 1 0 1 0 −8 −2 −4 2 −4 −1 −2 0 0 0 0 0 1 0 0 1 1 → 0 0 1 0 0 0 0 2 2 1 0 0 1 0 0 0 0 0 0 0 0 0 0 1 0 1 0 0 −1 2 0 0 0 0 1 0 0 0 0 0 1 −2 0 0 0 0 0 → 1 0 1 0 0 −1 2 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 We see that the first, second and fourth columns are pivot columns, hence we choose accordingly the basis vectors from the original vectors. Thus, a basis is −4 0 0 −8 −2 2 0 0 1 { , 0 , 0 } 1 2 1 0 0 0 0 (d) What is the nullity of A? What is the rank of A? As the basis we computed has three vectors, the nullity of A is 3. By the rank-nullity theorem, the rank is 6 − 3 = 3. (e) Consider the homomorphism T : R6 → R3 , u 7→ Au. (i) Interpret null(A) in terms of T . The nullity of A is just the kernel of T . (ii) Is T surjective? Justify your answer by using (d).Yes, T is surjective. The reason is that the rank of A is 3, which means that the dimension of the column space is 3. But the column space of A is equal to the range of T . The codomain of T is R3 , so that the range which has dimension 3, must be equal to R3 , hence surjective. 4.(6+3+5+2+1=17 points) Consider the following matrix: 1 0 0 0 0 −2 −1 0 0 0 1 1 0 −2 −2 −1 (a) Determine the characteristic polynomial χA of A. Show all your work! (Key, so that you can continue: The characteristic polynomial is χA = λ4 + λ3 − λ2 − λ.) We use cofactor expansion, which gives: χA = = = = = = (λ − 1)[(−2 − λ)(1λ)(−1λ) + 2 + 2(−2λ)] (λ − 1)[(−2 − λ)(1λ)(−1λ) − 2 − 2λ] (λ − 1)[(−2 − λ)(1λ)(−1λ) + 2(−1 − λ)] (λ − 1)(−1 − λ)[(−2 + 2λ − λ + λ2 + λ)] (λ − 1)(−1 − λ)(1 + λ) −(λ − 1)(1 + λ)2 λ (b) Determine the eigenvalues for A. The eigenvalues of A are the roots of χA , which are 0,-1,1. (c) Compute the eigenspace for the eigenvalue λ = −1. What is the dimension of this eigenspace? We need to find the nullspace of [A − (−1)I|0]. 2 0 0 0 0 2 0 0 0 0 0 −1 −1 0 0 → 0 −1 −1 0 0 0 0 0 2 1 0 0 2 1 0 0 −2 −2 0 0 0 −2 −2 0 0 Backward substitution gives then x4 = t, x3 = −1/2t, x2 = 1/2t, x1 = 0 (x1 is not a free variable but simply zero!). Hence 0 0 1/2 1 E−1 = span{ −1/2 } = span{ −1 }. 1 2 Because this span is linearly independent, it is a basis at the same time. From that we see that dim(E−1 ) = 1. (d) What is the only possible dimension of the eigenspace with eigenvalue λ = 0? Answer this question with the help of χA and justify your answer. The multiplicity of λ = 0 as a root of χA is 1. Hence dim(E0 ) ≤ 1. But any eigenspace must have at least dimension one (because the zero vector never is an eigenvector!) , the dimension of E0 must be equal to 1. (e) Based on the knowledge about the eigenvalues of this matrix, what can be said about the determinant of A? Justify your answer.As A has eigenvalue 0, the negation of the big theorem implies, that det = 0. 2 1 1 0 5.(5+3+5+4+3=20 points) Let S = span{s1 = 2 , s2 = −1 }. (a) Find a 1 −1 ⊥ basis B for S . We must find the null space of t s1 0 . st2 0 So let’s solve: 2 1 2 1 0 2 1 2 1 0 → 1 0 −1 −1 0 0 1 4 3 0 Backward substitution gives x3 = t1 , x4 = t2 , x2 = −4t1 − 3t2 , x1 = t1 + t2 . Hence 1 1 −4 −3 B = { 1 , 0 }. 0 1 (b) Show that the vector v = [−1, −1, 4, −5]t is orthogonal to S. We calculate the dot product of v with every vector of the defining span: 2 −1 1 −1 2 4 = 0 1 5 and 1 −1 0 −1 −1 4 = 0, −1 5 hence orthogonal. (c) Write v as a linear combination of the vectors in B from (a). We need to find the coefficients of the linear system given by 1 1 −1 −4 + c2 −3 = −1 , c1 1 0 4 0 1 5 in other words, we need to solve 1 1 −1 −4 −3 −1 1 0 4 0 1 5 This leads to the solution c1 = 4, c2 1 −4 4 1 0 = −5. In summary, 1 −1 −3 −1 − 5 0 = 4 1 5 (d) Show that for any vector u and any real number c ∈ R the equality ||cu|| = |c| · ||u|| √ √ √ holds.We have ||cu|| = cu · cu = c2 u · u = |c| u · u = |c| · ||u||. (e) Find the condition on two vectors w1 , w2 ∈ Rn such that their distance is equal to 0. The only vector that has length 0 is the zero vector. Hence: 0 = ||w1 − w2 || if and only if w1 − w2 = 0, which is when w1 = w2 . 6. (3+4=7 points) Let 2 −2 A= . −1 1 (a) Verify that the characteristic polynomial χA of A is χA = λ2 − 3λ. det(A) = (2 − λ)(1 − λ) − 2 = λ2 − 3λ. (b) λ by A and compute A2 − 3 · A = Compute χA (A), i.e. substitute the term. 2 −2 2 −2 2 −2 6 −6 6 −6 0 0 −3 = − = −1 1 −1 1 −1 1 −3 3 −3 3 0 0 7.(4+1+1+4+1+1=12 points) Consider the Euclidean space R3 . (a) List all possible dimensions of the subspaces of R3 . Any subspace has dimension smaller or equal to the overlaying subspace, hence the possible dimensions are 0,1,2,3 (b) How many different subspaces of R3 of dimension 3 are there?In R3 only R3 has dimension 3, hence there is 1 such space. (c) How many different subspaces of R3 of dimension 1 are there?All lines have dimension 3, hence there are infinitely many. (d) Give an example of two different one dimensional subspaces of R3 . 1 span{ 0 } 0 and 0 span{ 1 }. 0 (e) As which kind of object can we geometrically interpret one dimensional subspaces in R3 . It is the lines that have dimension 1 in R3 . (f) Which subspace(s) has/have only finitely many elements?There is only one subspace that has only finitley many elements, and this is the zero space. 8.(2+3=5 points) Let u1 , u2 , u3 , u4 , u5 be vectors in R6 and consider the set of vectors U := {u1 , u2 , 2u1 − u2 , u3 , u4 − 3u5 , u5 }. Now let A be the matrix that has the vectors of U as columns. (a) What are the dimensions of the matrix A? It is a 6x6-matrix, hence the dimensions are (6, 6). (b) What is the determinant of A. Justify your answer. As the second and the fourth columns are linear combinations of the other columns, we have linear dependent columns, which means by the Big Theorem, that det(A) = 0.