HOMEWORK 7 solutions

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1 Math 4377/6308 Advanced Linear Algebra I Dr. Vaughn Climenhaga, PGH 651A Fall 2013 HOMEWORK 7 solutions Due 4pm Wednesday, October 16. You will be graded not only on the correctness of your answers but also on the clarity and completeness of your communication. Write in complete sentences. 1. Let β = {f 1, f 2, f 3 } and γ = {g 1, g 2, g 3 } be the bases for P 2 given by f 1 (x) = 2 + 2x x 2 g 1 (x) = 2 + x f 2 (x) = 1 + x g 2 (x) = 1 + x + 2x 2 f 3 (x) = 1 + x 2 g 3 (x) = 1 + x + x 2 Let α = {e 1, e 2, e 3 } be the standard basis for P 2, with e 1 (x) = 1, e 2 (x) = x, and e 3 (x) = x 2. (a) Compute the change-of-coordinates matrix I γ β that turns β-coordinates into γ-coordinates, either directly or using the following steps. (i) Find Iβ α and Iα γ. (This requires almost no computation.) (ii) Draw a commutative diagram showing that I γ β is the matrix satisfying Iγ α I γ β = Iα β. (iii) Recall from your introductory linear algebra course that the matrix equation AB = C can be solved for B by row reducing the augmented matrix [A C] to the form [I B]. (iv) Keeping this in mind, row reduce [Iγ α Iβ α] to obtain [I Iγ β ]. Once you have computed I γ β, verify directly that Iα γ I γ β = Iα β. Solution. [15 points] To do this directly, recall that J = I γ β is the matrix whose coefficients are determined by These equations become f 1 = J 11 g 1 + J 21 g 2 + J 31 g 3, f 2 = J 12 g 1 + J 22 g 2 + J 32 g 3, f 3 = J 13 g 1 + J 23 g 2 + J 33 g x x 2 = J 11 (2 + x) + J 21 ( 1 + x + 2x 2 ) + J 31 (1 + x + x 2 ), 1 + x = J 12 (2 + x) + J 22 ( 1 + x + 2x 2 ) + J 32 (1 + x + x 2 ), 1 + x 2 = J 13 (2 + x) + J 23 ( 1 + x + 2x 2 ) + J 33 (1 + x + x 2 ).

2 Math 4377/6308 Homework 7 solutions Page 2 of 6 Comparing coefficients in the first of these gives 2 = 2J 11 J 21 + J 31 (constant term) 2 = J 11 + J 21 + J 31 (linear term) 1 = 2J 21 + J 31 (quadratic term) which can be solved. One can find the remaining entries of J similarly, by comparing coefficients in the second and third equations above. Instead of giving details we describe the alternate solution sketched in the problem, which amounts to carrying out the same computations but in a more organised way. First, note that the columns of Iβ α are the coordinate representations of f 1, f 2, f 3 relative to the standard basis α, which are easy to determine because they are given by the coefficients of the polynomials. We have [f 1 ] α = 2 2, [f 2 ] α = 1 1, [f 3 ] α = 1 0, Iβ α = Similarly, we get Iγ α = One possible commutative diagram relating the various change-ofcoordinates matrices is K n (α) I α β I γ β K n (β) I α γ K n (γ) I α I β I γ P 2 where we have written K n (α) to denote the vector space K n interpreted as coordinate representations relative to α, and similarly for K n (β) and K n (γ). This diagram illustrates that I α β = Iα γ I γ β. Thus finding J = I γ β amounts to solving the matrix equation AJ = B, where A = Iγ α and B = Iβ α. Writing J in column form as

3 Math 4377/6308 Homework 7 solutions Page 3 of 6 J = [x 1 x 2 x n ], where x j K n, and writing B similarly as B = [y 1 y n ], we see that this amounts to solving the equations Ax j = y j for 1 j n. Each of these can be solved by row reducing the augmented matrix [A y j ] to get [I x j ]. (Note that we end up with the identity on the left because A = Iγ α is invertible.) We use the same sequence of row reductions for each j (because the left part of the matrix does not depend on j), and so we can write all n of these processes at once as [A B] = [A y 1 y n ] [I x 1 x n ] = [I J], where the arrow signifies row reduction. Thus in the present problem, to find J it suffices to row reduce [Iγ α Iβ α ] = where we have omitted the intermediate steps in the row reduction. We conclude that J = I γ β = , and verify that Iγ α I γ β = = = Iβ α (b) Let p(x) = x 2 +x and find [p] α. Compute I β α = (I α β ) 1, and use this together with I γ β to find [p] β and [p] γ. Verify that I α γ [p] γ = [p] α. Solution. [10 points] We can invert Iβ α by carrying out the row reduction [Iβ α I] [I (Iα β ) 1 ], which yields

4 Math 4377/6308 Homework 7 solutions Page 4 of 6 ( ) 01 Now we see that [p] α =, and so [p] β = Iα[p] β α = = This allows us to compute [p] γ by [p] γ = I γ β [p] β = = Finally, we check that Iγ α [p] γ = = 0 1 = [p] α (c) Let T L(P 2 ) be the differentiation operator, and find [T ] α. Use I β α and I α β (which you computed in the previous parts) to find [T ] β. Solution. [10 points] We have T e 1 = 0, T e 2 = e 1, and T e 3 = 2e 2, so [T ] α = Thus [T ] β = Iα[T β ] α Iβ α = = =

5 Math 4377/6308 Homework 7 solutions Page 5 of 6 2. Let A, B, C be invertible n n matrices. (a) Show that AB and BA are always conjugate. Solution. [5 points] It suffices to observe that A(BA)A 1 = AB, so that AB can be obtained from BA via a conjugation by A. (b) Use this to show that ABC and CAB are always conjugate. Solution. [5 points] Similarly, C(ABC)C 1 = CAB. (c) Give an example of invertible matrices A, B, C such that ABC and BAC are not conjugate. Hint: Choose A, B such that AB BA, and then let C = (AB) 1. Solution. [5 points] Let A = ( ) and B = ( ). Then AB = ( ) and BA = ( ). Let C = (AB) 1 = ( ), then ABC = ( ) and BAC = ( ) ( ) = ( ), which is not conjugate to ABC = I because for any invertible J we have JIJ 1 = JJ 1 = I, so the only matrix conjugate to the identity is the identity itself. 3. Recall that a matrix A M n n is strictly upper triangular if A ij = 0 whenever i j. Show that every strictly upper triangular matrix is nilpotent that is, there exists k such that A k = 0. Hint: use induction to show that for every k = 1, 2, 3,..., we have (A k ) ij = 0 whenever j i + k 1. Solution. [10 points] As suggested, we use induction. For k = 1 the induction claim is exactly the hypothesis that A is strictly upper triangular. Now suppose A k has the property that (A k ) ij = 0 whenever j i + k 1. We prove the analogous property for A k+1 : given any i, j with j i + (k + 1) 1 = i + k, we have (A k+1 ) ij = n (A k ) il A lj. l=1

6 Math 4377/6308 Homework 7 solutions Page 6 of 6 If l j, then the fact that A is strictly upper triangular implies that A lj = 0, and so the terms with l j in the above sum vanish, and we get j 1 (A k+1 ) ij = (A k ) il A lj. l=1 But for l j 1 we use j i + k to deduce that l i + k 1, which by the inductive hypothesis implies that (A k ) il = 0, and hence these terms all vanish as well. Thus (A k+1 ) ij = 0 whenever j i + k, and by induction we have the result for all k = 1, 2, 3,.... In particular, with k = n we see that j i+n 1 for every 1 j n and 1 i n, and thus (A n ) ij = 0 for all such i, j, meaning that A n is the zero matrix. 4. Suppose T L(V ) satisfies the equation T 2 = T. Prove that V = R T N T. Hint: this requires showing that V = R T + N T, and that R T N T = {0}. To show the first, it may help to prove that v T (v) N T for all v V. Solution. [10 points] First we show that V = R T + N T. That is, for every v V we produce w R T and x N T such that v = w + x. Let w = T (v), then w R T. Let x = v w, then v = w + x, and it remains to show that x N T. To this end, observe that T x = T (v w) = T v T w = T v T (T v) = T v T 2 v = 0 where the last equality uses the fact that T = T 2. Thus x N T and we have shown that V = R T + N T. It remains to show that R T N T = {0}. To this end, suppose w R T N T. Because w R T, we have w = T v for some v V. Then T w = T (T v) = T v = w, where the second equality uses the fact that T 2 = T. But w N T implies that T w = 0, and hence w = 0.