(VI.D) Generalized Eigenspaces

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1 (VI.D) Generalized Eigenspaces Let T : C n C n be a f ixed linear transformation. For this section and the next, all vector spaces are assumed to be over C ; in particular, we will often write V for C n. In what follows, I will write S for an arbitrary transformation, which could be T, or σi T, or its restriction to a subspace, etc. We are looking for forms A = [T]ê can be put into (via S 1 B AS B) even if it is not diagonalizable. The structure underlying the rational canonical form was a decomposition of V = C n into T-cyclic subspaces in 1-1 correspondence with the nontrivial invariant factors r (λ),..., n (λ) of A. In the present section we describe the structure beneath the Jordan canonical form which, unlike the rational form, actually reduces to D when A is diagonalizable (= S B DS 1 B ). We can forget about most of the F[λ] stuff here; the theory is fortunately easier than that in the last two sections. Recall that if A is diagonalizable with eigenvalues σ 1,..., σ m, 1 then V is the sum of the corresponding eigenspaces and in fact the geometric multplicities add to n : In the language of direct sums, dim E σi (A) = n. V = E σ1 (A) E σm (A). What we claim is that there are generalized eigenspaces E s σ i such that V = E s σ 1 (A) E s σ m (A) 1 Here we mean the list of distinct eigenvalues, i.e. not repeated according to multiplicity. 1

2 2 (VI.D) GENERALIZED EIGENSPACES even if A is not diagonalizable. The proof will require a few facts about stable image/kernel, and nilpotent transformations (S : U U such that S k is the zero transformation for some k ). Throughout it is important to remember that if W V is closed under the action of T then the restriction of T to W makes sense as a linear transformation and is written T W (and read T on W ). Stable Image and Kernel. Given a transformation S : V V, the series of subspaces of V and 0 = keri kers kers 2... V = imi ims ims 2... both level off at some point (since V is finite dimensional). Let K be sufficiently large that ims K = ims K+1 =... kers K = kers K+1 =... ; these are called the stable image and stable kernel of S. An equivalent definition of these objects (subspaces of V ) is: ker s S = w V S k w = 0 for some k im s S = w V for every k, v V s.t. w = S k v. REMARK 1. The v such that S k v = w in the second definition are in general different for each k (even for k K ). We claim that (i) im s S ker s S, (ii) im s S + ker s S = V. To see (i), let w im s S ker s S ; that is, w = S K v and S K w = 0, so tht 0 = S K (S K v) = S 2K v. But then v kers 2K = ker s S = kers K, so that ( w =) S k v = 0.

3 (1) (VI.D) GENERALIZED EIGENSPACES 3 To see (ii), apply rank-nullity to S K to get dim V = dim(ims K ) + dim(kers K ) = dim(im s S) + dim(ker s S), and the modular law dim(w 1 + W 2 ) + dim(w 1 W 2 ) = dim W 1 + dim W 2 (cf. Exercise II.C.2) for subspaces W 1, W 2 V to get (im s S) + dim(ker s S) = dim(im s S ker s S) + dim(im s S + ker s S) (i) = dim(im s S + ker s S). Combining this with (1), dim(im s S + ker s S) = dim V and (ii) follows. We rewrite (i) and (ii) as V = im s S ker s S. This is always true, for any S : V V. Moreover, since S respects this decomposition (as you can check), one may speak of the restrictions S ker s S and S im s S. By definition some power k of S annihilates ker s S, and so S ker s S is nilpotent. On the other hand, ker (S im s S ) = kers im s S ker s S im s S = 0 by (i) above, and thus S im s S is invertible. We have proved PROPOSITION 2. Given any S : V V, there is a direct-sum decomposition V = U 0 W 0 respected by S, such that S W0 is nilpotent and S U0 is invertible. Now let s look more generally at the situation where S respects a (possibly different) direct sum decomposition V = U W. We claim that (a) kers = (U kers) + (W kers), and (b) (U kers) (W kers) = 0. Now (b) is immediate since U W = 0. To see (a): take any v kers and write it v = u + w (possible because V = U W ); clearly 0 = S v = S u + S w. Since S respects U W, S u U and S w W, but then S u = S w is a problem since U W = 0. So we must

4 4 (VI.D) GENERALIZED EIGENSPACES have S u = S w = 0! That means u U kers, w W kers, and since v is their sum we have proved (a). Of course (a)+(b) = kers = (U kers) (W kers), so applying this to S K we get PROPOSITION 3. Given S : V V respecting some direct-sum decomposition V = U W, one has ker s S = (U ker s S) (W ker s S). Nilpotent Transformations. Every S : V V has an eigenvalue (unless V = 0 ), since the characteristic polynomial f S (λ) has a root in C. (This is where we really need V = C n.) This eigenvalue has at least one nonzero eigenvector. What if zero is the only one? PROPOSITION 4. S is nilpotent 0 is its only eigenvalue. PROOF. ( ) Suppose 0 = only eigenvalue of S = only root of f S (λ). That is, f S (λ) = λ n. By Cayley-Hamilton, S satisfies its own characteristic polynomial, so S n = 0. ( ) Suppose S k = 0, and also suppose λ is an eigenvalue of S. There is a nonzero v such that S v = λ v, and thus 0 = S k v = λ k v = λ k = 0 = λ = 0. Stable Eigenspace. Given λ and eigenvalue of S : V V ( λ any root C of f S (λ) ), recall the definition E λ (S) := ker(λi S) = v V (λi S) v = 0 of the eigenspace of λ. Define the generalized or stable eigenspace E s λ (S) := kers (λi S) = Clearly E s λ (S) E λ(s). v V (λi S) k v = 0 for some k.

5 (VI.D) GENERALIZED EIGENSPACES 5 Now we return to our original T : V V with distinct eigenvalues σ 1,..., σ m, and set W k = E s σ k (T). (These are not the W k s of VI.C!) Clearly some power of (σ k I T) annihilates W k, so that (σ k I T) Wk is nilpotent and has only eigenvalue 0. That is, if v W k satisfies (σ k I T) v = λ v, then λ = 0. Therefore, if v W k satisfies then and σ k σ must be 0, i.e. σ = σ k. T v = σ v, (σ k I T) v = (σ k σ) v Conclusion: the only eigenvalue of T Wk is σ k. Now consider for i = j the intersection of two stable eigenspaces W i W j. The only eigenvalue of T Wi is σ i, while the only eigenvalue of T Wj is σ j. Since σ i = σ j, T can have no eigenvalue. This is absurd unless W i W j = 0. Wi W j PROPOSITION 5. E s σ i (T) E s σ j (T) = 0 for all i = j. We make one further observation concerning stable eigenspaces: how to find bases for them. You know how to find bases for kernels. Working in the standard basis of C n (in terms of which [T]ê = A by definition), find bases for ker(σ i I A) ker (σ i I A) 2 ker (σ i I A) 3....

6 6 (VI.D) GENERALIZED EIGENSPACES You stop when two successive bases have the same number of elements (once kers k = kers k+1, all the remaining ones are the same as well: a nice exercise!). The Jordan Structure Theorem. Here is what holds even when T is not semisimple ( A is not diagonalizable). We emphasize that the W k have nothing to do with those in the preceding lecture. THEOREM. Let T : V V (V = C n ) be a linear transformation, with distinct eigenvalues σ 1,..., σ m and corresponding stable eigenspaces W k = E s σ k (T) = ker s (σ k I T). Then V = W 1 W m where dim W k = algebraic multiplicity of σ k. Furthermore, T respects this decomposition. PROOF. We first prove the decomposition, by induction on m. Set d k = dim W k, and A = [T]ê. Case m = 1 : σ 1 = the only eigenvalue of T on V = 0 = only eigenvalue of (σ 1 I T) on V = (σ 1 I T) nilpotent = (σ 1 I T) k = 0 = V = ker s (σ 1 I T) = W 1. Inductive Step : Assume the Theorem holds for transformations with m 1 distinct eigenvalues, and let T be as above. Apply the discussion preceding Proposition 2 to (σ m I T) to get V = ker s (σ m I T) im s (σ m I T) =: W m U m, where σ m I T respects the decomposition. Moreover, since I also respects the direct sum (or, for that matter, and direct sum!), so do T and σ k I T, k = m. So we may speak of T Um : U m U m. Since (σ m I T) is invertible on U m, σ m cannot be an eigenvalue of T there. 2 Thus T Um has eigenvalues σ 1,..., σ m 1, and by induction U m = W 1 W m 1, 2 (σm I T Um ) invertible = det (σ m I T Um ) = 0 = σ m not a root of det (λi T Um ).

7 (VI.D) GENERALIZED EIGENSPACES 7 where W k = ker s (σ k I T Um ) = ker s (σ k I T) U m = W k U k. We must show that W k = W k. Since (k = m ) σ k I T also respects the decomposition V = W m U m, we have (Prop. 3) W k = ker s (σ k I T) = W m ker s (σ k I T) U m ker s (σ k I T) = W m W k U m W k. By Prop. 4, W m W k = 0 and so as desired. W k = U m W k = W k, T respects the direct sum : We need to show T(W k ) W k. Take w ker s (σ k I T), so that for κ sufficiently large (σ k I T) κ w = 0. But then (σ k I T) κ T w = T(σ k I T) κ w = 0 = T w ker s (σ k I T). d k = algebraic multiplicity of σ k (as roots of p A (λ)) : We take B 1,..., B m to be bases for W 1,..., W m ; the collection B = B 1,..., B m yields a basis for V subordinate to the direct sum. Since T respects the direct sum, its matrix with respect to B splits into blocks down the diagonal (of dimensions d 1 d 1,..., d m d m ): [T] B =: B = SB 1 AS B = diag [T W1 ] B1,..., [T Wm ] Bm diag B 1,..., B m. Moreover, since A B, λi A λi B and f A (λ) = f B (λ). From λi B = diag λi d1 B 1,..., λi dm B m we have f B (λ) = det(λi B) = det(λi dk B k ) = f B1 (λ) f Bm (λ). k Since the only eigenvalue of T Wk is σ k (and B k = [T Wk ] Bk ) the only root of f Bk (λ) is σ k. Since B k is d k d k, it follows that deg( f Bk ) = d k

8 8 (VI.D) GENERALIZED EIGENSPACES and so f Bk (λ) = (λ σ k ) d k. But then ( f A (λ) = ) f B (λ) = (λ σ k ) d k and we are done. Exercises (1) Find the stable eigenspaces of A = (2) Suppose A is an 8 8 matrix with m A (λ) = λ(λ 1) 2 (λ 2) 3 and f A = λ 2 (λ 1) 2 (λ 2) 4. What are the dimensions of the eigenspaces and stable eigenspaces of A? (3) Show that for any transformation S : C n C n, ker(s k ) = ker(s k+1 ) implies (a) ker(s k ) = ker(s l ) for all l k, and (b) im(s k ) = im(s l ) for all l k. [Hint for (b): use Rank + Nullity and (a).]

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