proof of Jordan canonical form theorem

This theorem can be proved combining the cyclic decomposition theorem and the primary decomposition theorem. By hypothesis, the characteristic polynomial of $T$ factorizes completely over $F$ , and then so does the minimal polynomial of $T$ (or its annihilator polynomial). This is because the minimal polynomial of $T$ has exactly the same factors on $F[X]$ as the characteristic polynomial of $T$ . Let’s suppose then that the minimal polynomial of $T$ factorizes as $m_{T}=(X-\lambda_{1})^{\alpha_{1}}\ldots(X-\lambda_{r})^{\alpha_{r}}$ . We know, by the primary decomposition theorem, that

V=\bigoplus_{i=1}^{r}\ker((T-\lambda_{i}I)^{\alpha_{i}}).

Let $T_{i}$ be the restriction of $T$ to $\ker((T-\lambda_{i}I)^{\alpha_{i}})$ . We apply now the cyclic decomposition theorem to every linear operator

(T_{i}-\lambda_{i}I)\colon\ker(T-\lambda_{i}I)^{\alpha_{i}}\to\ker(T-\lambda_{% i}I)^{\alpha_{i}}.

We know then that $\ker(T-\lambda_{i}I)^{\alpha_{i}}$ has a basis $B_{i}$ of the form $B_{i}=B_{1,i}\bigcup B_{2,i}\bigcup\ldots\bigcup B_{d_{i},i}$ such that each $B_{s,i}$ is of the form

B_{s,i}=\{v_{s,i},(T-\lambda_{i})v_{s,i},(T-\lambda_{i})^{2}v_{s,i},\ldots,(T-% \lambda_{i})^{k_{s,i}}v_{s,i}\}.

Let’s see that $T$ in each of this “cyclic sub-basis” $B_{s,i}$ is a Jordan block: Simply notice the following fact about this polynomials:

$\displaystyle X(X-\lambda_{i})^{j}$	$\displaystyle=$	$\displaystyle(X-\lambda_{i})^{j+1}+X(X-\lambda_{i})^{j}-(X-\lambda_{i})^{j+1}$
	$\displaystyle=$	$\displaystyle(X-\lambda_{i})^{j+1}+(X-X+\lambda_{i})(X-\lambda_{i})^{j}$
	$\displaystyle=$	$\displaystyle(X-\lambda_{i})^{j+1}+\lambda_{i}(X-\lambda_{i})^{j}$

and then

T(T-\lambda_{i}I)^{j}(v_{s,i})=(T-\lambda_{i})^{j+1}(v_{s,i})+\lambda_{i}(T-% \lambda_{i}I)^{j}(v_{s,i}).

So, if we also notice that $(T-\lambda_{i}I)^{k_{s,i}+1}(v_{s,i})=0$ , we have that $T$ in this sub-basis is the Jordan block

\begin{pmatrix}\lambda_{i}&0&0&\cdots&0&0\\ 1&\lambda_{i}&0&\cdots&0&0\\ 0&1&\lambda_{i}&\cdots&0&0\\ \vdots&\vdots&\vdots&\ddots&\vdots&\vdots\\ 0&0&0&\cdots&\lambda_{i}&0\\ 0&0&0&\cdots&1&\lambda_{i}\end{pmatrix}

So, taking the basis $B=B_{1}\bigcup B_{2}\bigcup\ldots\bigcup B_{r}$ , we have that $T$ in this basis has a Jordan form.

This form is unique (except for the order of the blocks) due to the uniqueness of the cyclic decomposition.

Title	proof of Jordan canonical form theorem
Canonical name	ProofOfJordanCanonicalFormTheorem
Date of creation	2013-03-22 14:15:36
Last modified on	2013-03-22 14:15:36
Owner	CWoo (3771)
Last modified by	CWoo (3771)
Numerical id	11
Author	CWoo (3771)
Entry type	Proof
Classification	msc 15A18