Proofs Homework Set 11

Proofs Homework Set 11 M ATH 217 2011 PROBLEM 11.1. Let A be an n × n real symmetric matrix; i.e. all entries of A are real numbers and AT = A. Let v ∈ Cn be an eigenvector of A corresponding to the eigenvalue λ ∈ C, and write v̄1  ..  v̄ =  . . v̄n (Recall that the complex conjugate of a complex number z = a + bi, a, b ∈ R, is the complex number z̄ = a − bi. See Appendix B of the book for properties of the complex conjugate.) (a) Show that Av̄ = λ̄v̄ (so v̄ is an eigenvector of A with eigenvalue λ̄). Proof. Taking the complex conjugate of Av = λv, we obtain Av̄ = Av = λv = λ̄v̄. To see this, we repeatedly use the fact that w + z = w̄ + z̄ and wz = w̄z̄ holds for all w, z ∈ C. For the right hand side, we have     λ̄v̄1 λv1     λv =  ...  =  ...  = λ̄v̄. λ̄v̄n λvn For the left hand side, we have       ā11 v̄1 + · · · + ā1n v̄n a11 v̄1 + · · · + a1n v̄n a11 v1 + · · · + a1n vn       .. .. .. Av =  = =  = Av̄, . . . an1 v1 + · · · + ann vn ān1 v̄1 + · · · + ānn v̄n an1 v̄1 + · · · + ann v̄n where we used the fact that all enties of A are real (so āij = aij ). (b) Show that v̄T Av = λ̄v̄T v and that v̄T Av = λv̄T v. Proof. Taking the transpose of the equation Av̄ = λ̄v̄ from (a), we obtain v̄T A = v̄T AT = λ̄v̄T . Multiplying on the right by v gives v̄T Av = λ̄v̄T v. Multiplying the equation Av = λv on the left by v̄T gives v̄T Av = λv̄T v. (c) Show that v̄T v = v̄1 v1 + · · · + v̄n vn is a positive real number. Proof. Let us write vk = ak + ibk where ak , bk ∈ R. Then v̄1 v1 + · · · + v̄n vn = (a1 − ib1 )(a1 + ib1 ) + · · · + (an − ibn )(an + ibn ) = (a21 + b21 ) + · · · + (a2n + b2n ). Since a sum of squares is nonnegative, we know that the result is a nonnegative real number. To see that the sum is nonzero, we use the fact that v 6= 0 (since v is an eigenvector). It follows that vk = ak + ibk 6= 0 for some k, and hence the summand a2k + b2k cannot be zero either. (d) Conclude that λ = λ̄ and hence λ ∈ R. Proof. From part (b), we conclude that λ̄v̄T v = λv̄T v. Since v̄T v is a nonzero number by part (c), we can divide both sides of this equation by this number to obtain that λ̄ = λ, which is only possible when λ ∈ R. Therefore, the eigenvalues of a real symmetric matrix are always real numbers. P ROBLEM 11.2. For a polynomial p(x) and an n × n matrix A, let p(A) denote the matrix obtained by “plugging in” A for x. For example, if p(x) = x3 −2x2 +3, then p(A) = A3 −2A2 +3I. (a) Show that if λ is an eigenvalue of an n × n matrix A, prove that p(λ) is an eigenvalue of p(A). Proof. For conciseness, let us write p(x) = am xm + am−1 xm−1 + · · · + a1 x + a0 . Let v be an eigenvector of A with eigenvalue λ. Consider p(A)v. Since Ak v = λk v for every k, we see that p(A)v = am Am v + am−1 Am−1 v + · · · + a1 Av + a0 Iv = am λm v + am−1 λm−1 + · · · + a1 λv + a0 v = (am λm + am−1 λm−1 + · · · + a1 λ + a0 )v Therefore, p(A)v = p(λ)v which shows that p(λ) is an eigenvalue of the matrix p(A). (b) Show that if A is similar to B, then p(A) is simlar to p(B). Proof. Suppose that A = P −1 BP , i.e. that A is similar to B via the invertible matrix P . As above, let us write p(x) = am xm + am−1 xm−1 + · · · + a1 x + a0 . Since An = P −1 B n P , we see that p(A) = am Am + am−1 Am−1 + · · · + a1 A + a0 I = am P −1 B m P + am−1 P −1 B m−1 P + · · · + a1 P −1 AP + a0 P −1 P. Factoring P −1 on the left and factoring P on the right, we obtain that p(A) = am P −1 B m P + am−1 P −1 B m−1 P + · · · + a1 P −1 AP + a0 P −1 P = P −1 (am B m + am−1 B m−1 + · · · + a1 B + a0 I)P = P −1 p(B)P. Therefore, p(A) is similar to p(B). (c) Show that if A is diagonalizable and p(λ) is the characteristic polynomial of A, then p(A) is the zero matrix. Proof. The hypotheses say that A = P −1 DP where P is some invertible matrix and D is the diagonal matrix   λ1 0 · · · 0  0 λ2 · · · 0    D =  .. ..  . . . . . 0 0 · · · λn with λ1 , λ2 , . . . , λn being the eigenvalues of A. Note that these scalars λ1 , λ2 , . . . , λn are precisely the roots of the characteristic polynomial of A, namely p(λ) = det(A − λI) = an λn + an−1 λn−1 + · · · + a1 λ + a0 . By part (b), we see that p(A) is similar to the matrix p(D). Now we can readily compute p(D) as follows: p(D) = an Dn + · · · + a1 D + a0 I    λ1 0 · · · λn1 0 · · · 0  0 λ2 · · ·  0 λn · · · 0  2    = an  .. ..  + · · · + a1  .. .. . . . . . . . n 0 0 ··· 0 0 · · · λn   p(λ1 ) 0 ··· 0  0 p(λ ) · · · 0  2   =  .. ..  . . .  . . .  0 0 · · · p(λn )   1 0 ··· 0 0 1 · · · 0   ..  + a0  .. .. . . . λn 0 0 ···  0 0  ..  . 1 Since p(λi ) = 0 for i = 1, 2, . . . , n, we see that p(D) is simply the zero matrix. Therefore, p(A) = P −1 p(D)P is the zero matrix too.