# 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 ProofOfJordanCanonicalFormTheorem 2013-03-22 14:15:36 2013-03-22 14:15:36 CWoo (3771) CWoo (3771) 11 CWoo (3771) Proof msc 15A18