Lecture 21: The decomposition theorem into generalized eigenspaces; multiplicity of eigenvalues and upper-triangular matrices (1) Travis Schedler Tue, Nov 29, 2011 (version: Tue, Nov 29, 1:00 PM) Goals (2) The decomposition theorem into generalized eigenspaces Multiplicity of eigenvalues and the characteristic polynomial for uppertriangular matrices Read Chapter 8, do the midterm review problems and take last year s Midterm 2 as practice; prepare for Thursday s midterm! Warm-up exercise (3) Let A = λ 1 λ 2... be upper-triangular with entries λ 1,..., λ n 0 λ n on the diagonal. Recall: the λ i are the eigenvalues of A. For each eigenvalue λ, let d λ denote the number of times it occurs on the diagonal of A. Let the characteristic polynomial of A be χ A (x) = (x λ 1 ) (x λ n ) = λ (x λ) d λ = det(xi A). Note that, if you multiply two upper-triangular matrices, the diagonal entries multiply. Hence, A m has diagonal entries λ m 1,..., λ m n for all m 1. Warm-up exercise continued (4) (a) Show that rk(a) is at least the number of entries λ i that are nonzero. (Hint: Show that the columns with nonzero λ i are linearly independent.) (b) Show that, for all m 1, dim null((a λi) m ) d λ ; hence dim V (λ) d λ. (Hint: A λi is also upper-triangular! Apply the formula for the diagonal of (A λi) m, and also (a).) 1
(c) Show that χ A (A) := (A λ 1 I) (A λ n I) = 0. [Cayley-Hamilton theorem!] (Hint: Induct on n. The inductive hypothesis implies (A λ 1 I) (A λ n 1 I) is all zero except for the rightmost column. Now, since the bottom row of (A λ n I) is zero, deduce that this product is zero.) Today we will prove that the inequality in (b) is an equality, and moreover, V = i V (λ i). Solution to warm-up exercise (5) (a) The columns with nonzero λ i form a matrix which, after performing row operations, is in row-echelon form with number of nonzero rows equal to the number of columns (exercise!). So these columns are linearly independent. (b) For all m 1, (A λi) m is upper-triangular with diagonal entries (λ i λ) m. So the number of zero entries of (A λi) m is d λ. By (a), rk((a λi) m ) n d λ, and the rank-nullity theorem implies dim null((a λi) m ) d λ. For the hence statement, take m large enough that (A λi) m V (λ) = 0 (e.g., take a basis (v 1,..., v k ) of V (λ) and m large enough that (A λi) m v i = 0 for all i). (c) We follow the hint. Perform induction on n, with base case n = 0 or n = 1. By hypothesis, the upper-left n 1 n 1 part of (A λ 1 I) (A λ n 1 I) is zero, i.e., all but the rightmost column, since this is upper-triangular. Now, the bottom row of A λ n I is zero. The product of any matrix with only the rightmost column nonzero by a matrix with zero bottom row is zero. Multiplicity of eigenvalues (6) Theorem 1 (Theorem A). If A and A are two upper-triangular matrices for T, then the number of times each eigenvalue appears on the diagonal is equal in each. One way to prove this, which we do next week, uses determinant. Today we prove Theorem A using generalized eigenspaces: Theorem 2 (Theorem B). Let A be upper-triangular and let λ appear on the diagonal d λ times. Then dim V (λ) = d λ. In the warm-up, we proved dim V (λ) d λ. Now, given T, dim V (λ) for T is independent of basis and equals dim V (λ) for any matrix M(T ), independent of basis. So d λ is independent of the basis. 2
The decomposition theorem (7) Theorem B (hence also Theorem A) will follow from: Theorem 3 (Theorem C: strengthened decomposition theorem). Let V be f.d. and λ 1,..., λ k be some eigenvalues of T L(V ). Let m i be such that (T λ i I) mi V (λi) = 0. Then ( k V = V (λ 1 ) V (λ k ) range (T λ i I) ). mi If λ 1,..., λ k are all the eigenvalues of T, then the last factor above has no eigenvectors. Hence: Corollary 4 (Theorem 8.23.(a): dec. thm. for F = C). If F = C and V is f.d., then V = λ V (λ). Note: The reason the m i exist in the theorem is because, if (v 1,..., v k ) is a basis of V (λ), then we can take m i to be such that (T λ i I) mi v j = 0 for all j (cf. PS 10, #8d). Decomposition theorem for upper-triangular matrices (8) Recall ( k V = V (λ 1 ) V (λ k ) range (T λ i I) ). mi (0.1) Corollary 5. Let V be f.d. T admits an upper-triangular matrix (for any F) if and only if V = λ V (λ). Proof: : Take λ 1,..., λ k to be the diagonal entries of M(T ) and m i = d λi to be the number of times λ i occurs. Then the last factor of (0.1) is zero, by (c) of the warm-up. : It suffices to find a basis of each V (λ) in which T is upper triangular. Putting them together gives a basis of V for which T is upper-triangular (and in fact block-diagonal with these upper-tri matrices as blocks). For each V (λ), the proof is just like the case of F = C; see PS 11. (Note that V (λ) always has an eigenvector.) Proof of Theorem B (9) Theorem 6 (Theorem B). Let A be upper-triangular (F arbitrary) and let λ appear on the diagonal d λ times. Then dim V (λ) = d λ. Proof. By the last theorem, V = i V (λ i). Hence dim V = i dim V (λ i). But dim V = d λi, and part (b) of the warm-up showed that dim V (λ i ) d λi for all i. Hence dim V (λ i ) = d λi for all i. Corollary (Cayley-Hamilton theorem: 8.20): If T admits an upper-tri matrix (equiv., V = λ V (λ)), then λ (T λi)dim V (λ) = 0. This is always true for F = C and V f.d. (where dim V (λ) = d λ ). 3
Base case of dec. thm: k = 1 (10) Proposition 0.2. Let V be f.d. Suppose (T λi) m is zero on V (λ). Then V = V (λ) range((t λi) m ). Proof. Note that V (λ) = null(t λi) m. Hence, the rank-nullity theorem implies dim V = dim V (λ) + dim range((t λi) m ). It suffices to show that V (λ) range((t λi) m ) = {0}. Let v be in this intersection. Let v = (T λi) m u. Then (T λi) m v = 0 = (T λi) 2m u. Thus u V (λ), so (T λi) m u = 0. Thus v = 0. Inductive step of dec. thm (on board)(11) By inductive hypothesis, V = V (λ 1 ) V (λ k 1 ) range k 1 (T λ i I) mi. Let U be the last factor. The base case applied to U yields U = U(λ k ) range((t λ k I) m k U ). Hence: V = V (λ 1 ) V (λ k 1 ) U(λ k ) range k (T λ i I) mi. It remains to show U(λ k ) = V (λ k ). This is equivalent to V (λ k ) U = range k 1 (T λ ii). It suffices to show that (T λ i I) restricts to an isomorphism on V (λ k ) for all i k. Then all of V (λ k ) is in the range. If not, there exists a nonzero v null(t λ i I) V (λ k ), i.e., an eigenvector of e.v. λ i λ k in V (λ k ). But then (T λ k ) m k v = 0 = (λ i λ k ) m k v 0, a contradiction. Preview of next time: Jordan form (12) We see that, when T admits an upper-triangular matrix, it actually is block diagonal with blocks of the form λ λ... = M(T V (λ)), 0 λ 4
i.e., upper triangular with a single diagonal entry. These blocks are not unique: they depend on the bases of V (λ). Next time: we will prove Jordan form: that in fact we can make M(T ) canonically block-diagonal with (generally smaller) blocks λ 1 0 0 0 λ 1 0 0........... λ 1 0 λ 1 0 λ 5