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Math 2280 Spring 2008 Test 2 You will have 50 minutes to complete this test. You may not use a calculator or any other electronic device on this test. You will be graded on your work and not necessarily on your answer, so be sure to show as much of your reasoning as possible in your responses. Each question is worth 10 points. If there are two parts, each part will be graded out of five points. Good luck! 1. Find bases for the image and kernel of the matrix 2 4 8 A = 4 4 1 . 7 9 3 We have that 1 0 0 2 4 8 1 2 4 1 0 − 27 ∼ 0 1 0 = rref(A). A = 4 4 1 ∼ 0 −4 −15 ∼ 0 1 15 4 25 0 0 4 0 0 1 7 9 3 0 −5 −25 This means that A is invertible and therefore of A is 2 4 , 7 im(A) = R3 . We have that a basis for the image 4 8 4 , 1 . 9 3 We also know that ker(A) = {~0}, so technically there is no basis for ker(A), but if you put that ~0 is a basis for ker(A), that will be okay. 2. Find the coordinate vector [~x]B of ~x with respect to the basis B = {~v1 , ~v2 }, where 3 1 0 ~x = 1 , ~v1 = −1 , ~v2 = 1 . −4 0 −1 We have that [~x]B = where c1 c2 , c1~v1 + c2~v2 = ~x, which is equivalent 1 0 −1 1 0 −1 to the augmented system . .. 1 0 .. 3 . 3 .. .. . 1 ∼ 0 1 . 4 . .. 0 −1 .. −4 . −4 .. 1 0 . 3 ∼ 0 1 ... 4 ⇒ c1 = 3, c2 = 4. .. 0 0 . 0 1 Therefore, [~x]B = 3 4 . 3. Find a basis for all polynomials f (t) in P3 such that f (1) = 0 and Z 1 f (t) dt = 0. −1 If f (t) ∈ P3 , then we can write f (t) = a + bt + ct2 + dt3 for some choice of constants a, b, c, and d. If f (1) = 0, then a + b + c + d = 0. If Z 1 f (t) dt = 0 −1 then 1 1 2 1 3 1 4 0= (a + bt + ct + dt ) dt = at + bt − ct + dt 2 3 4 −1 −1 1 1 1 1 1 1 2 = a + b + c + d − −a + b − c + d = 2a + c. 2 3 4 2 3 4 3 Z 1 2 3 If we put these two equations into matrix form, then the coefficient matrix A is 1 1 1 1 1 0 53 0 1 1 1 1 ∼ ∼ . A= 0 −2 − 34 −2 0 1 − 23 1 2 0 23 0 The solution vector to the corresponding homogeneous system is 5 0 − 35 c a −3 −1 b 2c − d 2 = 3 = c 3 +d 0 1 c c 1 d 0 d , where c and d can be chosen arbitrarily. Therefore, Z 1 5 2 2 3 f (t) ∈ P3 : f (1) = f (t) dt = 0 = − c + c − d t + ct + dt : c, d ∈ R , 3 3 −1 i.e. we can write any polynomial in our set as 2 5 2 5 2 3 2 c − d t + ct + dt = c − + t + t + d −t + t3 . − c+ 3 3 3 3 Notice that the coefficients of the polynomials here are exactly the entries of the vectors above. The vectors above are a basis for ker(A) and they are also the coordinate vectors of these polynomials with respect to the basis {1, t, t2 , t3 }. This means that these two polynomials, 2 − 35 + 23 t + t2 and −t + t3 are linearly independent (because their coordinate vectors are) and they obviously span the given set, so they are a basis and the dimension of our set is 2. 4. Determine if the linear transformation T : P2 → P2 defined by T (a + bt + ct2 ) = a − bt + ct2 is an isomorphism. One way to do this problem is to think in terms of coordinates. If f (t) = a + bt + ct2 is a polynomial in P2 , then its coordinate vector with respect to the standard basis B = {1, t, t2 } for P2 is a [f ]B = b . c The mapping in question takes a b c a −b , c to i.e. it changes the sign of the second component of the coordinate vector. It is easy to see that the matrix for this mapping is 1 0 0 0 −1 0 . 0 0 1 Clearly this matrix is invertible (it is diagonal with nonzero diagonal entries), so T is also invertible and therefore and isomorphism. 5. In the plane V defined by the equation 2x1 + x2 − 2x3 = 0, consider the bases 2 1 U = {~u1 , ~u2 } = 2 , −2 2 1 and 3 1 2 , 0 . B = {~v1 , ~v2 } = 2 3 Find the change of basis matrix S from B to U. From the lecture, we have that S= Since ~v1 = ~u1, it is clear that [~v1 ]U [~v2 ]U [~v1 ]U = 1 0 To find the U coordinates of the vector ~v2 , we solve .. . ~u ~u ... ~v = 2 1 2 .. . 3 . . the augmented matrix equation .. 1 2 . 3 . 2 −2 .. 0 .. 2 1 . 3 .. . . 3 1 0 .. 1 .. ∼ 0 1 ... 1 ∼ 0 −6 . −6 .. . 0 −3 . −3 0 0 .. 0 Therefore, 1 2 S= 1 1 0 1 ⇒ [~v2 ]U = 1 1 . . 6. Find the QR factorization of the matrix 5 3 4 6 A= 2 7 . 2 −2 Let us denote the columns of A by ~v1 and ~v2 . Recall that k~v1 k ~u1 · ~v2 , A = QR = ~u1 ~u2 0 k~v2 k where ~u1 and ~u2 are obtained form ~v1 and ~v2 through the Gramm-Schmidt process. First we do the Gramm-Schmidt portion of the work: 5 5 1 ~v1 4 = 1 4 =√ ~u1 = k~v1 k 25 + 16 + 4 + 4 2 7 2 2 2 3 −2 5 3 5 6 1 1 4 6 4 2 ~v2⊥ = ~v2 − (~u1 · ~v2 )~u1 = 7 − 7 (15 + 24 + 14 − 4) 7 2 = 7 − 2 = 5 −2 −4 2 −2 2 −2 −2 2 1 2 ~v2⊥ 1 = . ~u2 = ⊥ = √ k~v2 k 4 + 4 + 25 + 16 5 7 5 −4 −4 Therefore, 5 −2 1 4 2 , Q= 7 2 5 2 −4 R= 7 7 0 7 7. (a) If A and B are orthogonal, is B −1 AB orthogonal as well? (b) If A and B are symmetric, is AB 2 A symmetric as well? 4 . Justify your answer on both parts. (a) We have (B −1 AB)(B −1 AB)T = B −1 ABB T AT (B −1 )T = B −1 (B −1 )T = B T (B T )T = B T B = I. This shows that B −1 AB is orthogonal. (b) We have (AB 2 A)T = AT (B 2 )T AT = AT (BB)T AT = AT B T B T AT = ABBA = AB 2 A. This shows that AB 2 A is symmetric. 8. Find the least squares solution of the 1 A= 1 0 system A~x = ~b, where 1 3 0 and ~b = 3 . 1 3 To find the least squares solution ~x∗ , we solve the that 1 1 1 0 1 AT A = 1 0 1 0 and AT~b = 1 1 0 1 0 1 normal equations AT A~x∗ = AT~b. We have 1 2 1 0 = 1 2 1 3 3 = 6 . 6 3 The augmented matrix for the normal equations is # " " # " . . 1 0 2 1 .. 6 1 12 .. 3 ∼ ∼ . .. 0 1 0 32 .. 3 1 2 . 6 # .. . 2 2 ∗ ⇒ ~x = . .. 2 . 2 9. Find the determinant of the matrix 0 0 A= 1 0 2 0 2 0 3 0 3 3 4 4 . 4 4 To compute the determinant, we will expand along the third row, since it is almost completely zeros. 0 2 3 0 2 det(A) = 4 · det 1 2 3 = 4 · 3 · det = 4 · 3 · (0 − 1) = −24. 1 2 0 0 3 5 10. Find the area of the parallelogram defined by 8 3 . and 2 7 We form a matrix A with these two vectors as columns, 3 8 A= . 7 2 We proved in class that the volume V of the parallelogram is given by V = |det(A)| = |6 − 56| = | − 50| = 50. 6