LEC ALG4 10

More on the Jordan canonical form

When defining a Jordan block, we referred to its diagonal entry by ‘eigenvalue’. Now we see why.

Theorem 403.1(Theorem).

Let $k$ be an algebraically closed field. Let $V$ be a finite dimensional $k$-vector space. Let $\lambda \in k$. Let $T:V\to V$ be $k$-linear. Then $T$ has a JCF $J$ with a Jordan block with eigenvalue $\lambda$ $⟺$ $\lambda$ is an eigenvalue of $T$.

Proof.

$(⟹)$ Since $J$ is a lower triangular matrix, its determinant is given by the product of its diagonal entries. Thus, $(x-\lambda )$ divides the characteristic polynomial of $T$, and by Thm 84.3 $\lambda$ is an eigenvalue of $T$.

$(⟸)$ Conversely, if $\lambda$ is an eigenvalue of $T$, then $\lambda$ is a root of the characteristic polynomial of $T$, and hence must appear on the diagonal of $J$ since it is triangular.□

We now discuss the extent to which the JCF matrix of a linear operator, the existence of which is guaranteed by Thm 402.8, is unique. Recall that matrices which represent the same abstract linear operator in different bases are similar. Thus, given a matrix in JCF, we need to characterize the JCF matrices in its orbit under conjugation.

Proposition 403.2(Proposition).

Let $A$ and $B$ be matrices in JCF. Then $A$ is similar to $B$ iff $B$ can be obtained by permuting the Jordan blocks of $A$.

Proof.

This follows from the following facts:

1. All permutations of the blocks of a block diagonal matrix can be obtained by conjugating by block permutation matrices.
2. Similar matrices have the same characteristic polynomial, so they must have the same eigenvalues ${\lambda }_1,\dots ,{\lambda }_k$ ¹. Recall that the number of Jordan blocks of $A$ with eigenvalue ${\lambda }_i$ of size at least $j$ is $\dim \ker (A-{\lambda }_{i}I{)}^j-\dim \ker (A-{\lambda }_{i}I{)}^{j-1}$. Again, since $A$ and $B$ are similar, we have $\dim \ker (A-{\lambda }_{i}I{)}^j=\dim \ker (B-{\lambda }_{i}I{)}^j$ for all $i,j$.

□

We now prove the Cayley-Hamilton theorem.

[!Theorem] Cayley-Hamilton
Let $k$ be algebraically closed. Let $V$ be a finite dimensional $k$-vector space. Let $T:V\to V$ be $k$-linear and let $c_T$ be the characteristic polynomial of $T$. Then $c_T(T)=0$.

Footnotes

1. However, the characteristic polynomial does not tell us anything about their geometric multiplicities! ↩