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 $$V=\bigoplus_{i=1}^{r}\ker((T-\lambda_iI)^{\alpha_i}).$$ Let $T_{i}$ be the restriction of $T$ to $\ker((T-\lambda_iI)^{\alpha_{i}})$ . We apply now the cyclic decomposition theorem to every linear operator $$(T_{i}-\lambda_iI) \colon \ker(T-\lambda_iI)^{\alpha_{i}}\to \ker(T-\lambda_iI)^{\alpha_{i}}.$$ We know then that $\ker(T-\lambda_iI)^{\alpha_i}$ 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: \begin{eqnarray*} X(X-\lambda_i)^{j} &=& (X-\lambda_i)^{j+1}+X(X-\lambda_i)^{j}-(X-\lambda_i)^{j+1} \\ &=& (X-\lambda_i)^{j+1}+(X-X+\lambda_i)(X-\lambda_i)^{j} \\ &=& (X-\lambda_i)^{j+1}+\lambda_i(X-\lambda_i)^{j} \end{eqnarray*}and then $$T(T-\lambda_iI)^{j}(v_{s,i})=(T-\lambda_i)^{j+1}(v_{s,i})+\lambda_i(T-\lambda_iI)^{j}(v_{s,i}).$$ So, if we also notice that $(T-\lambda_iI)^{k_{s,i}+1}(v_{s,i})=0$ , 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.