We now combine the generalized eigenspace decomposition with the structure theory of nilpotent operators to obtain the Jordan normal form—the “best possible” matrix representation for an operator over an algebraically closed field.
7.8.1 Jordan Blocks
Definition 7.8.1 (Jordan block). A Jordan block of size with eigenvalue is the matrix It has on the diagonal, 1’s on the superdiagonal, and 0’s elsewhere.
Example 7.8.2 (Small Jordan blocks). A Jordan block is just a scalar, acting on a line spanned by an eigenvector. Every Jordan block, of any size, contains exactly one eigenvector up to scaling, namely its first basis vector; the remaining basis vectors of a block of size are generalized eigenvectors that are not eigenvectors.
Proposition 7.8.3 (Properties of Jordan blocks). Let . Then:
- The only eigenvalue of is , with algebraic multiplicity .
- The eigenspace has dimension 1.
- The generalized eigenspace for is all of .
- is nilpotent with index .
Proof. (a) is upper triangular with all diagonal entries equal to , so (Corollary 7.5.4).
(b) sends and for , so its kernel is spanned by .
(c) Since , every vector is in .
(d) (it maps ) but . ∎
Note. A Jordan block represents the action of an operator on a single chain of generalized eigenvectors. The chain is where for and . Equivalently, for and .
Theorem 7.8.4 (Jordan normal form). Let be a finite-dimensional vector space over an algebraically closed field , and let . Then there exists a basis of with respect to which the matrix of is block diagonal: where each block is a Jordan block. This matrix is called the Jordan normal form (or Jordan canonical form) of . The scalars that occur are the eigenvalues of (possibly repeated).
Proof. By the generalized eigenspace decomposition (Theorem 7.6.11), where are the distinct eigenvalues. Each is -invariant, so by Theorem 7.3.4 we can work on each generalized eigenspace separately.
On , the operator is nilpotent (Proposition 7.6.8(c)). By the structure theorem for nilpotent operators, there is a chain basis of in which has matrix .
Since , the matrix of in this basis is
Combining the bases for all generalized eigenspaces gives a basis of in which has the desired block diagonal form. ∎
Note. The Jordan form groups blocks by eigenvalue: all blocks for a given together correspond to the decomposition of into chains. The total size of the blocks with eigenvalue equals , which is the algebraic multiplicity of (Theorem 7.6.14), and the number of these blocks is , the geometric multiplicity. In particular, is diagonalizable if and only if all its Jordan blocks have size .
Theorem 7.8.5 (Uniqueness of Jordan normal form). The Jordan normal form of is unique up to permutation of the blocks. More precisely, in any basis in which the matrix of is a direct sum of Jordan blocks, the number of blocks of size exactly with eigenvalue is where .
Proof. Fix a basis in which is a direct sum of Jordan blocks and fix . Let be the span of the basis vectors belonging to blocks with eigenvalue , and the span of the others; both are -invariant and .
We claim . On , is a direct sum of blocks , hence nilpotent, so . On , the matrix of is upper triangular with nonzero diagonal entries , so is invertible on (Lemma 7.5.3). If with , lies in , then , forcing . So .
Thus the blocks with eigenvalue form a chain basis of for the nilpotent operator , with chain lengths equal to the block sizes. For , , so ; this also holds for . By Proposition 7.7.10, the number of chains of length exactly is . These dimensions are intrinsic to , so the block structure is uniquely determined. ∎
Corollary 7.8.6 (Similarity criterion). Let be algebraically closed and . The following are equivalent:
- and are similar.
- and have the same Jordan normal form, up to the order of the blocks.
- For each and each , .
(The Jordan normal form of a matrix means that of the operator on .)
Proof. (a) (c): If , then , and maps isomorphically onto .
(c) (b): By Theorem 7.8.5, the numbers determine the Jordan form.
(b) (a): A matrix is similar to its Jordan normal form (Corollary 5.3.4), and permuting the blocks of amounts to reordering the basis, which gives a similar matrix. Similarity is an equivalence relation, so and are similar. ∎
Note. This gives a complete set of similarity invariants over an algebraically closed field. Trace and determinant alone do not distinguish all similarity classes ( and share both), but the full sequence of kernel dimensions does.
Given a matrix over an algebraically closed field, here is the procedure to find its Jordan form:
Step 1: Find eigenvalues. Compute and factor it as . The are the eigenvalues, with algebraic multiplicities ; the blocks with eigenvalue have total size .
Step 2: For each eigenvalue , determine block sizes. Compute for until (after which the sequence is constant). The number of blocks of size is . In particular:
- Number of blocks = = geometric multiplicity.
- Number of blocks of size is .
- Number of blocks of size exactly is .
Step 3: Assemble the Jordan form. List all Jordan blocks grouped by eigenvalue.
Example 7.8.7 (Computing Jordan form). Let .
Step 1: , so with multiplicity 3.
Step 2: So , , .
Number of blocks = . Size of the single block = 3.
Step 3: The Jordan form is . (The matrix is already in Jordan form.)
Example 7.8.8 (Two blocks with the same eigenvalue). Let .
Step 1: , so with multiplicity 4.
Step 2: , so . , so .
Number of blocks = 2. Blocks of size : . So both blocks have size 2.
Step 3: The Jordan form is .
Example 7.8.9 (Multiple eigenvalues). Let .
Step 1: . Eigenvalues: , , each with multiplicity 2.
Step 2 for : So and : , . One block of size 2: .
Step 2 for : So and , already the multiplicity. Two blocks of size 1: .
Step 3: The Jordan form is .
Note. In the last example, contributes two blocks because its eigenspace has dimension 2 (geometric = algebraic multiplicity). In contrast, has a block because its geometric multiplicity (1) is less than its algebraic multiplicity (2).