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 $m_T={(X-{\lambda }_1)}^{{\alpha }_1}\mathrm{\dots }{(X-{\lambda }_r)}^{{\alpha }_r}$.We know, by the primary decomposition theorem, that

$$V=\underset{i=1}{\overset{r}{\oplus }}\ker ({(T-{\lambda }_{i}I)}^{{\alpha }_i}).$$

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

$$(T_i-{\lambda }_{i}I):\ker {(T-{\lambda }_{i}I)}^{{\alpha }_i}\to \ker {(T-{\lambda }_{i}I)}^{{\alpha }_i}.$$

We know then that $\ker {(T-{\lambda }_{i}I)}^{{\alpha }_i}$ has a basis $B_i$ of the form$B_i=B_{1,i}\bigcup B_{2,i}\bigcup \mathrm{\dots }\bigcup B_{d_i,i}$such that each $B_{s,i}$ is of the form

$$B_{s,i}=\{v_{s,i},(T-{\lambda }_i)v_{s,i},{(T-{\lambda }_i)}^2{v}_{s,i},\mathrm{\dots },{(T-{\lambda }_i)}^{k_{s,i}}{v}_{s,i}\}.$$

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:

$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$

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

So, taking the basis $B=B_1\bigcup B_2\bigcup \mathrm{\dots }\bigcup B_r$, 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.