Partial Solution Set, Leon 4.3

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1 Partial Solution Set, Leon Let [u 1, u 2 ] and [v 1, v 2 ] be ordered bases for R 2, where u 1 (1, 1) T, u 2 ( 1, 1) T, v 1 (2, 1) T, and v 2 (1, ) T. Let L be the linear transformation defined by L(x) ( x 1, x 2 ) T, and let B be the matrix representing L with respect to [u 1, u 2 ]. (a) Find the transition matrix S corresponding to the change of basis from [u 1, u 2 ] to [v 1, v 2 ]. Solution: This part doesn t deal with L yet, rather just the change of basis matrix. The transition matrix in question is the one I ve been calling T UV, i.e., S V 1 U [ ] [ ] [ (b) Find the matrix A representing L with respect to [v 1, v 2 ] by computing A SBS 1. Solution: Note that S goes from the [u 1, u 2 ] basis to [v 1, v 2 ] basis, which is not the way the diagrams are shown in the book. ]. Now, B is the matrix ) that corresponds [ ] [ to the ][u [ 1, u 2 ] is] obtained [ ] by B U (L(e 1 1 )L(e 2 ) U U

2 Next we find S 1 1 [ ] 3 1. Then it is a simple matter to determine that [ 2 1 ] [ 1 ] [ ] [ ] A SBS V 1. Verify that L(v 1 ) a 11 v 1 + a 21 v 2 and L(v 2 ) a 12 v 1 + a 22 v 2 ) ( [ ] ) [ ] 2 2 Solution: Note that L (v 1 L and on the other hand [ ] [ ] 1 [ ] a 11 v 1 + a 21 v ( 4), so they match. 1 1 ) ( [ ] ) [ ] 1 1 Also L (v 2 L and on the other hand [ ] [ ] [ ] a 12 v 1 + a 22 v 2 + ( 1), so they match Let L be the linear transformation on R 3 given by L(x) (2x 1 x 2 x 3, 2x 2 x 1 x 3, 2x 3 x 1 x 2 ) T, and let A be the matrix representing L with respect to the standard basis for R 3. If u 1 (1, 1, ) T, u 2 (1,, 1) T, and u 3 (, 1, 1) T, then [u 1, u 2, u 3 ] is an ordered basis for R 3. (a) Find the transition matrix U corresponding to the change of basis from [u 1, u 2, u 3 ] to the standard basis. (b) Determine the matrix B representing L with respect to [u 1, u 2, u 3 ]. Solution: (a) This is simply U

3 (b) Somewhat surprisingly, B U 1 AU A. An interesting sidelight: this means that UA AU, i.e., we have an instance of a commuting pair of matrices Let L be the linear operator mapping R 3 into R 3 defined by L(x) Ax, where A Let v 1 (1, 1, 1) T, v 2 (1, 2, ) T, and v 3 (, 2, 1) T. Find the transition matrix V corresponding to a change of basis from [v 1, v 2, v 3 ] to the standard basis, and use it to determine the matrix B representing L with respect to [v 1, v 2, v 3 ]. 1 1 Solution: The transition matrix is V We want B V 1 AV Let L be the linear operator on P 3 defined by L(p(x)) xp (x) + p (x). (a) Find the matrix A representing L with respect to [1, x, x 2 ]. (b) Find the matrix B representing L with respect to basis B given by [1, x, 1 + x 2 ]. (c) Find the matrix S such that B S 1 AS. (d) Given p(x) a + a 1 x + a 2 (1 + x 2 ), find L n (p(x)). (e) Given p(x) a + a 1 x + a 2 (x 2 ), find L n (p(x)) (this is not in the book, but try it!). Solution: (a) We start by applying L to the basis vectors: L(1), L(x) x, and L(x 2 ) 2x The corresponding coordinate vectors become the columns of A 1. 2 (b) The coordinate vectors for 1 and x are unchanged, but the coordinate vector for 2x is now (,, 2) T, so B 1. 2 (c) The change of basis matrix has for its columns the vectors in R 3 corresponding to each of the vectors in [1, x, 1 + x 2 ] since we have to find the change of basis matrix from the arbitrary basis [1, x, 1 + x 2 ] to the standard basis [1, x, x 2 ]: S

4 (d) The coordinate vector of p(x) is (a, a 1, a 2 ) T B with respect to the basis B given by [1, x, 1 + x 2 ]. The nth power of L is given by the nth power of B, which is simple to compute because of the simple diagonal structure of B; B n 1. It 2 n follows that the coordinate vector for L n (p(x)) is B n (a, a 1, a 2 ) T (, a 1, 2 n a 2 ), so L n (p(x)) a 1 x + 2 n a 2 (1 + x 2 ). (e) There isn t a part (e), but I would like to emphasis how this B S 1 AS helps: The coordinate vector of p(x) is (a, a 1, a 2 ) T with respect to standard basis. The nth power of L is given by the nth power of A, which is not so simple to computer since A is not diagonal as it was in part (d). However, from B S 1 AS we can find A SBS 1, so then A 2 (SBS 1 ) 2 SBS 1 SBS 1 SB 2 S 1. Similarly, A 1 SB 1 S 1, or generally A n SB n S 1, for all natural numbers n. Thus A n 1 1 1, 1 2 n 1 which is easy to computer as a product of three matrices, instead of multiplying A by itself n times. It follows that the coordinate vector for L n (p(x)) is A n (a, a 1, a 2 ) T which is Suppose that A SΛS 1, where Λ is a diagonal matrix with main diagonal λ 1, λ 2,..., λ n. (a) Show that As i λ i s i for each 1 i n. (b) Show that if x α i s i, then A k x α i λ k i s i. (c) Suppose that λ i < 1 for each 1 i n. What happens to A k x as k? Solution: (a) For any choice of i, 1 i n, we have As i ( SΛS 1) s i SΛ ( S 1 s i ) 4 SΛe i S (Λe i ) Sλ i e i λ i Se i λ i s i.

5 The reason that S 1 s i e i is because when you multiply S 1 S I, and so when you multiply the ith column of S by S 1 you get the ith column of I, which is e i. (b) This is easily proven by induction: A x x α i s i. Assume that A k x α i λ k i s i for some k N. Then and the result follows by induction. A k+1 x AA k x A α i λ k i s i Aα i λ k i s i α i λ k i As i α i λ k i λ i s i α i λ k+1 i s i, (c) Each term in the preceding sum vanishes, since if λ < 1 then lim k λ k Suppose that A ST, where S is nonsingular. Let B T S. Show that B is similar to A. Proof: Assume that A is as described, i.e., that A ST and that S is nonsingular. Then B T S (S 1 S)T S S 1 (ST )S S 1 AS, so B is similar to A. A second way to present the proof above would be: Proof: Assume that A is as described, i.e., that A ST and since S is nonsingular T S 1 A. Then B T S (S 1 A)S S 1 AS, so B is similar to A. What s the point? Given any square S and T, with at least one of the two nonsingular, we know that it s unlikely that ST T S. But at least ST and T S are similar. And that (as we shall see) means that they have much in common (eigenvalues, for example). 5

6 Show that if A and B are similar matrices, then det(a) det(b). Solution: If A and B are similar, then B S 1 AS for some nonsingular S. Now use the fact that the determinant of a product is the product of the determinants, along with the commutativity of real multiplication: det B det(s 1 AS) det S 1 det A det S det S 1 det S det A det(s 1 S) det A det I det A det A. 6