The shear has too few eigenvectors to be diagonalizable. To decompose such operators we enlarge the eigenspaces, admitting vectors that are not eigenvectors but are killed by some power of . It is simplest to start with the eigenvalue , where the relevant operator is itself.
7.6.1 Generalized Kernels
Definition 7.6.1 (Generalized kernel). The generalized kernel of is These are all the vectors that are eventually sent to by repeatedly applying .
Note. The notation is Auroux’s; it is not standard.
The difference between the kernel and the generalized kernel can be summed up as follows: a vector is in the kernel if one application of sends it to , and in the generalized kernel if some number of applications does. If , then also , so the kernels of the powers of form an increasing chain and is their union. The union of an increasing chain of subspaces is a subspace, so is a subspace of .
Proposition 7.6.2 (Stabilization of kernels). Let .
- If for some , then for all , and .
- If is finite-dimensional with , then , and hence .
Proof. (a) We show ; repeating the argument gives the rest. The inclusion always holds. If , then , so . Hence . So the chain is constant from on, and its union is .
(b) Let . If for some , then (a) gives . Otherwise each inclusion for is strict, so , whence and ; then also . ∎
Example 7.6.3. Let be given by , , with matrix . Then , but . Note also that , so here.
Definition 7.6.4 (Nilpotent operator). An operator is nilpotent if for some positive integer . Then . When is finite-dimensional the converse holds, because by Proposition 7.6.2(b). If , the smallest such is called the index of nilpotency of .
As the example shows, in general . Replacing by a high enough power repairs this:
Lemma 7.6.5 (Kernel–image decomposition for a power). Let be finite-dimensional, , and let satisfy (for example ). Then both summands are -invariant, and restricts to an isomorphism of onto itself. In particular, .
Proof. Both summands are -invariant by Proposition 7.3.3.
Intersection is trivial: Suppose . Write for some . Then , so by Proposition 7.6.2(a). Thus .
Sum is all of : By Rank–Nullity applied to , . Since the intersection is trivial, the sum is direct and has dimension . Hence it equals .
Restriction to the image: The kernel of is (here ). An injective operator on a finite-dimensional space is an isomorphism (Corollary 6.1.16).
The last sentence follows from Proposition 7.6.2(b). ∎
Note. This decomposition is the key structural result for understanding non-diagonalizable operators. The generalized kernel captures the “nilpotent part” of : the restriction of to it is nilpotent. On the complementary invariant subspace , the restriction of is an isomorphism.
7.6.2 Generalized Eigenspaces
Now we apply the same idea to for an arbitrary scalar . From now on in this section, is finite-dimensional with .
Definition 7.6.6 (Generalized eigenspace). Let and . Set the last equality by Proposition 7.6.2(b). A nonzero vector of is a generalized eigenvector of with eigenvalue . When is an eigenvalue of , is called the generalized eigenspace of for .
Note. Every eigenvector with eigenvalue is a generalized eigenvector, since ; but may be strictly larger. Its elements are the vectors that are “eventually killed” by .
Example 7.6.7 (Generalized eigenspace of the shear). Recall the shear with matrix . The only eigenvalue is , and the eigenspace is , which is one-dimensional.
However, .
So the generalized eigenspace is . The vector is a generalized eigenvector that is not an eigenvector: , but .
Proposition 7.6.8 (Basic properties of generalized eigenspaces). Let and . Then:
- , and if and only if is an eigenvalue of .
- is -invariant.
- The restriction of to is nilpotent.
- If , the restriction of to is invertible. Consequently , and if , the only eigenvalue of is .
Proof. (a) The inclusion is clear. If is nonzero, let be the largest integer with ; then , so is an eigenvector with eigenvalue . Conversely, an eigenvector lies in .
(b) and is a polynomial in , so this is Proposition 7.3.3.
(c) By definition is zero on , and is invariant under by (b).
(d) Let with . Then , so . As , we get . Thus is injective on the invariant finite-dimensional subspace , hence invertible there (Corollary 6.1.16). Then is injective on too; since it kills , we get . Finally, an eigenvalue of is a scalar for which is not injective on , so . ∎
Proposition 7.6.9 (Invariant splitting). Let and , and put . Then both summands are -invariant, and restricts to an isomorphism of onto itself. In particular, is not an eigenvalue of .
Proof. Apply Lemma 7.6.5 to the operator with . This gives and the statement about on . Both summands are kernels or images of a polynomial in , so they are -invariant (Proposition 7.3.3). ∎
Proposition 7.6.10 (Independence of generalized eigenspaces). Let be distinct. If with , then for all . That is, the sum is direct.
Proof. We show ; the same argument applies to every . Suppose , and let be the largest integer with . Then , i.e., . Consider the operator Its factors commute (Proposition 7.2.2). For , , because we may apply first and . On the other hand, , so Applying to gives , a contradiction. Hence . ∎
7.6.3 The Generalized Eigenspace Decomposition
The following theorem is the main structural result for operators on finite-dimensional vector spaces over algebraically closed fields. Its proof is an induction on the dimension built on the invariant splitting, and it does not use the Cayley–Hamilton Theorem.
Theorem 7.6.11 (Generalized eigenspace decomposition). Let be a finite-dimensional vector space over an algebraically closed field , and let . Let be the distinct eigenvalues of . Then In particular, .
Proof. By Proposition 7.6.10, the sum of the is direct, so it remains to show that they span . We use induction on . If , there is nothing to prove.
Let and assume the result for all spaces of smaller dimension. By Theorem 7.5.1, has at least one eigenvalue, say . By Proposition 7.6.9, with invariant under . Since , we have . By the induction hypothesis applied to , where are the eigenvalues of and are its generalized eigenspaces. An eigenvector of is an eigenvector of , so each is one of ; and . Therefore so the generalized eigenspaces span . The dimension formula follows because the sum is direct. ∎
Corollary 7.6.12 (The image of ). In the situation of Theorem 7.6.11, with ,
Proof. For , the operator maps isomorphically onto itself (Proposition 7.6.8(d)), so . Hence . By Proposition 7.6.9 and Theorem 7.6.11, both sides have dimension , so they are equal. ∎
Combining the decomposition with triangular form gives a matrix in which each eigenvalue occupies its own block.
Corollary 7.6.13 (Block-triangular form). Let be algebraically closed, , and let be the distinct eigenvalues of , with . There is a basis of in which the matrix of is block diagonal, where each is upper triangular with all diagonal entries equal to .
Proof. Each is a nonzero invariant subspace. By Theorem 7.5.2, it has a basis in which has an upper-triangular matrix . The diagonal entries of are eigenvalues of (Corollary 7.5.4), and the only such eigenvalue is (Proposition 7.6.8(d)). Concatenating these bases gives a basis of , in which the matrix is block diagonal by Theorem 7.3.4. ∎
Theorem 7.6.14 (Algebraic multiplicity equals ). Let be algebraically closed, , and let be the distinct eigenvalues of . Then Hence the algebraic multiplicity of each eigenvalue is , and the algebraic multiplicities add up to .
Proof. The matrix of Corollary 7.6.13 is upper triangular, and appears on its diagonal exactly times. Apply Corollary 7.5.4(b). ∎
Corollary 7.6.15 (Geometric and algebraic multiplicity). Let be algebraically closed and an eigenvalue of . Then The operator is diagonalizable if and only if the geometric multiplicity equals the algebraic multiplicity for every eigenvalue, i.e., if and only if for every eigenvalue .
Proof. The inequalities follow from and Theorem 7.6.14. By Theorem 7.4.13, is diagonalizable if and only if , and by Theorem 7.6.11. Since for each , the two sums agree if and only if every term agrees. ∎
Example 7.6.16 (Decomposition for a matrix). Consider with matrix The matrix is upper triangular, so : the eigenvalue has algebraic multiplicity 2 and the eigenvalue has algebraic multiplicity 1.
For : and . So , which has dimension 2, while the eigenspace has dimension 1.
For : the powers of all have kernel , so , of dimension 1.
Indeed, , the dimensions agree with the algebraic multiplicities, and is not diagonalizable.