Linear Algebra Exercises-n-Answers.pdf

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202 Linear Algebra, by Hefferon (c) The minimal polynomial m(t) sends every vector in the domain to zero and so it is not one-to-one (except in a trivial space, which we ignore). By the first item of this question, since the composition m(t) is not one-to-one, at least one of the components t − λ i is not one-to-one. By the second item, t − λ i has a nontrivial nullspace. Because (t − λ i )(⃗v) = ⃗0 holds if and only if t(⃗v) = λ i · ⃗v, the prior sentence gives that λ i is an eigenvalue (recall that the definition of eigenvalue requires that the relationship hold for at least one nonzero ⃗v). Five.IV.1.30 This is false. The natural example of a non-diagonalizable transformation works here. Consider the transformation of C 2 represented with respect to the standard basis by this matrix. ( ) 0 1 N = 0 0 The characteristic polynomial is c(x) = x 2 . Thus the minimal polynomial is either m 1 (x) = x or m 2 (x) = x 2 . The first is not right since N − 0 · I is not the zero matrix, thus in this example the minimal polynomial has degree equal to the dimension of the underlying space, and, as mentioned, we know this matrix is not diagonalizable because it is nilpotent. Five.IV.1.31 Let A and B be similar A = P BP −1 . From the facts that A n = (P BP −1 ) n = (P BP −1 )(P BP −1 ) · · · (P BP −1 ) = P B(P −1 P )B(P −1 P ) · · · (P −1 P )BP −1 = P B n P −1 and c · A = c · (P BP −1 ) = P (c · B)P −1 follows the required fact that for any polynomial function f we have f(A) = P f(B) P −1 . For instance, if f(x) = x 2 + 2x + 3 then A 2 + 2A + 3I = (P BP −1 ) 2 + 2 · P BP −1 + 3 · I = (P BP −1 )(P BP −1 ) + P (2B)P −1 + 3 · P P −1 = P (B 2 + 2B + 3I)P −1 shows that f(A) is similar to f(B). (a) Taking f to be a linear polynomial we have that A − xI is similar to B − xI. Similar matrices have equal determinants (since |A| = |P BP −1 | = |P | · |B| · |P −1 | = 1 · |B| · 1 = |B|). Thus the characteristic polynomials are equal. (b) As P and P −1 are invertible, f(A) is the zero matrix when, and only when, f(B) is the zero matrix. (c) They cannot be similar since they don’t have the same characteristic polynomial. The characteristic polynomial of the first one is x 2 − 4x − 3 while the characteristic polynomial of the second is x 2 − 5x + 5. Five.IV.1.32 Suppose that m(x) = x n + m n−1 x n−1 + · · · + m 1 x + m 0 is minimal for T . (a) For the ‘if’ argument, because T n + · · · + m 1 T + m 0 I is the zero matrix we have that I = (T n +· · ·+m 1 T )/(−m 0 ) = T ·(T n−1 +· · ·+m 1 I)/(−m 0 ) and so the matrix (−1/m 0 )·(T n−1 +· · ·+m 1 I) is the inverse of T . For ‘only if’, suppose that m 0 = 0 (we put the n = 1 case aside but it is easy) so that T n + · · · + m 1 T = (T n−1 + · · · + m 1 I)T is the zero matrix. Note that T n−1 + · · · + m 1 I is not the zero matrix because the degree of the minimal polynomial is n. If T −1 exists then multiplying both (T n−1 + · · · + m 1 I)T and the zero matrix from the right by T −1 gives a contradiction. (b) If T is not invertible then the constant term in its minimal polynomial is zero. Thus, T n + · · · + m 1 T = (T n−1 + · · · + m 1 I)T = T (T n−1 + · · · + m 1 I) is the zero matrix. Five.IV.1.33 (a) For the inductive step, assume that Lemma 1.7 is true for polynomials of degree i, . . . , k − 1 and consider a polynomial f(x) of degree k. Factor f(x) = k(x − λ 1 ) q1 · · · (x − λ l ) q l and let k(x − λ 1 ) q 1−1 · · · (x − λ l ) q l be c n−1 x n−1 + · · · + c 1 x + c 0 . Substitute: k(t − λ 1 ) q1 ◦ · · · ◦ (t − λ l ) q l (⃗v) = (t − λ 1 ) ◦ (t − λ 1 ) q1 ◦ · · · ◦ (t − λ l ) q l (⃗v) = (t − λ 1 ) (c n−1 t n−1 (⃗v) + · · · + c 0 ⃗v) = f(t)(⃗v) (the second equality follows from the inductive hypothesis and the third from the linearity of t). (b) One example is to consider the squaring map s: R → R given by s(x) = x 2 . It is nonlinear. The action defined by the polynomial f(t) = t 2 − 1 changes s to f(s) = s 2 − 1, which is this map. x s2 −1 ↦−→ s ◦ s(x) − 1 = x 4 − 1 Observe that this map differs from the map (s − 1) ◦ (s + 1); for instance, the first map takes x = 5 to 624 while the second one takes x = 5 to 675.
Answers to Exercises 203 Five.IV.1.34 Yes. Expand down the last column to check that x n + m n−1 x n−1 + · · · + m 1 x + m 0 is plus or minus the determinant of this. ⎛ ⎞ −x 0 0 m 0 0 1 − x 0 m 1 0 0 1 − x m 2 ⎜ ⎝ . .. ⎟ ⎠ 1 − x m n−1 Subsection Five.IV.2: Jordan Canonical Form Five.IV.2.17 That calculation is easy. We are required to check that ( ) 3 0 = N + 3I = P T P −1 = 1 3 ( 1/2 1/2 −1/4 1/4 ) ( 2 −1 1 4 ) ( 1 −2 1 2 Five.IV.2.18 (a) The characteristic polynomial is c(x) = (x−3) 2 and the minimal polynomial is the same. (b) The characteristic polynomial is c(x) = (x + 1) 2 . The minimal polynomial is m(x) = x + 1. (c) The characteristic polynomial is c(x) = (x + (1/2))(x − 2) 2 and the minimal polynomial is the same. (d) The characteristic polynomial is c(x) = (x − 3) 3 The minimal polynomial is the same. (e) The characteristic polynomial is c(x) = (x − 3) 4 . The minimal polynomial is m(x) = (x − 3) 2 . (f) The characteristic polynomial is c(x) = (x + 4) 2 (x − 4) 2 and the minimal polynomial is the same. (g) The characteristic polynomial is c(x) = (x − 2) 2 (x − 3)(x − 5) and the minimal polynomial is m(x) = (x − 2)(x − 3)(x − 5). (h) The characteristic polynomial is c(x) = (x − 2) 2 (x − 3)(x − 5) and the minimal polynomial is the same. Five.IV.2.19 (a) The transformation t − 3 is nilpotent (that is, N ∞ (t − 3) is the entire space) and it acts on a string basis via two strings, β ⃗ 1 ↦→ β ⃗ 2 ↦→ β ⃗ 3 ↦→ β ⃗ 4 ↦→ ⃗0 and β ⃗ 5 ↦→ ⃗0. Consequently, t − 3 can be represented in this canonical form. ⎛ ⎞ 0 0 0 0 0 1 0 0 0 0 N 3 = ⎜0 1 0 0 0 ⎟ ⎝0 0 1 0 0⎠ 0 0 0 0 0 and therefore T is similar to this this canonical form matrix. ⎛ ⎞ 3 0 0 0 0 1 3 0 0 0 J 3 = N 3 + 3I = ⎜0 1 3 0 0 ⎟ ⎝0 0 1 3 0⎠ 0 0 0 0 3 (b) The restriction of the transformation s + 1 is nilpotent on the subspace N ∞ (s + 1), and the action on a string basis is given as β ⃗ 1 ↦→ ⃗0. The restriction of the transformation s − 2 is nilpotent on the subspace N ∞ (s − 2), having the action on a string basis of β ⃗ 2 ↦→ β ⃗ 3 ↦→ ⃗0 and β ⃗ 4 ↦→ β ⃗ 5 ↦→ ⃗0. Consequently the Jordan form is this ⎛ ⎞ −1 0 0 0 0 0 2 0 0 0 ⎜ 0 1 2 0 0 ⎟ ⎝ 0 0 0 2 0⎠ 0 0 0 1 2 (note that the blocks are arranged with the least eigenvalue first). )
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