Download Math 115AH: Homework 9 Solutions - D-invariant Subspaces and Characteristic Polynomials and more Assignments Linear Algebra in PDF only on Docsity! Math 115AH: Homework 9 Solutions (Artin, Sections 4.3, 4.4, 4.6) Problem 4.3.9: (a) I’ll leave it to you to determine the matrix. Nilpotence is immediate; we have dn+1 dxn+1 p(x) = 0 for all p ∈ P . (b) Define the subspaces Pk for 0 ≤ k ≤ n, where Pk = {p ∈ P | deg(p) ≤ k}. (deg(p) is the degree of p, i.e. the highest power of x that appears in p(x).) Each Pn is clearly D- invariant; I claim these are the only D-invariant subspaces. To prove this, take W to be a k-dimensional D-invariant subspace of P . Let m = max{deg(f) | f ∈ W}, and take a monic polynomial f ∈ W , deg(f) = m. (“Monic” means the coefficient of xm is 1). Then the set {Dmf = 1, Dm−1f, . . . , Df, f} is linearly independent (why?) and contains m + 1 elements, so dim(W ) = k ≥ m + 1. But by definition of m, we have W ⊂ Pm =⇒ dim(W ) = k ≤ m + 1 = dim(Pm). So k = m + 1, and hence W = Pm (it is an m + 1-dimensional subspace of the m + 1-dimensional vector space Pm). Problem 4.4.3: Let A ∈ M3(R); then its characteristic polynomial pA is a cubic with real coefficients. The idea here is that any cubic with real coefficients has at least one real root. You can approach this either by: • noticing that, for any such cubic f , one has lim x→−∞ f(x) = − lim x→∞ f(x) and applying the intermediate value theorem, or • using the fact that the roots of polynomials come in complex conjugate pairs with multiplicity. Problem 4.4.7: (a) Choose two linearly independent eigenvectors v1 and v2 for λ, and extend them to a basis B = {v1, . . . , vn} for V . Then the matrix of T with respect to B will be of the form [T ] = λ 0 ∗ . . . ∗ 0 λ ∗ . . . ∗ 0 0 ∗ . . . ∗ ... ... ... . . . ... 0 0 ∗ . . . ∗ 1 Using expansion by minors, it is clear that the characteristic polynomial has the form pT (t) = det(tI − [T ]) = (t− λ)2f(t) for some polynomial f , which shows that λ is a multiple root of pT . (b) No. Consider the linear transformation LA ∈ End(R2), where A = [ 1 1 0 1 ] . Problem 4.4.9: A and At have the same eigenvalues, because pA(x) = det(xI −A) = det(xI −A)t = det(xIt −At) = det(xI −At) = pAt(x). They do not necessarily have the same eigenvectors; once again, consider A = [ 1 1 0 1 ] . Problem 4.4.10: (a) Take A ∈ M2(R), A = [ a b c d ] , a, b, c, d > 0. Then we have pA(t) = det(tI −A) = (t− a)(t− d)− bc = t2 − (a + d)t− bc. The discriminant of this quadratic is D = (a + d)2 + 4bc > 0, so the quadratic has two distinct real roots, λ1 = (a + d) + √ D 2 λ2 = (a + d)− √ D 2 , and these are the eigenvalues. (b) (Due credit goes to Lincoln Atkinson for this solution.) From the above formula, we see that |λ1| > |λ2|, so that λ1 is the larger eigenvalue. Suppose [ x y ] is a λ1-eigenvector. It is easy to see that x 6= 0 and y 6= 0. Then we obtain the system of equations ax + by = λ1x cx + dy = λ1y. 2
