Jordan canonical form

This is a basic fact of Linear Algebra.

In general, we have that for an endomorphism $f:V\mapsto V$

and so $dim(V)=dim(Ker(f))+dim(Im(f))$.

But we don't always have that $V=Ker(f)\oplus Im(f)$. Now, if $f\circ f=f$, i.e., $f$ is idempotent, the previous formula holds: for every $v\in V$ we have that $v=v-f(v)+f(v)=(v-f(v),f(v))$ and conversely, for $(h,w)\in Ker(f)\oplus Im(f)$ we can take $h+w\in V$ such that $(h+w-f(h+w),f(h+w))=(h,w)$

This is the same as being a projection.

More generally, we have
Proposition
If $f$ is such that $Im(f)=Im(f^2)$, we have

Proof
First, let's observe that $f{|}_{Im(f)}$ is an isomorphismm. This is so because it is surjective and dimensions are equals.
We can construct an isomorphism $V\to Ker(f)\times Im(f)$ at the following way:

where $x$ is such that $f^2(x)=f(v)$ ($x$ exists because $Im(f)=Im(f^2)$).

It is easy to prove injectivity and surjectivity. Only rest to show that is well defined. Let us suppose that, for $v\in V$ we get $x\ne y$ such that $f^2(x)=f^2(y)=f(v)$. Then, since the restriction of $f$ is an isomorphism, $f(x)=f(y)$.
$◼$

At the same way, we can prove the following proposition
Propostion
For any endomorphism $f:V\to V$ there exists $m\ge 1$ such that

Proof
Observe that $Im(f)\supseteq Im(f^2)\supseteq Im(f^3)\supseteq \cdots$
And therefore, there exist $m$ such that $Im(f^m)=Im(f^{m+1})=\cdots =Im(f^{2m})=\cdots$.
At this point, we only have to apply the previous proposition to the transformation $h=f^m$.
$◼$

Now, we are ready to show the Jordan canonical form. Let $V$ be a complex vector space, and $f\in end(V)$. If $f$ is decomposable, we take a decomposition $V=\underset{i}{⨁}{U}_i$, with $f(U_i)\subseteq U_i$ (invariant subspace) and proceed with every $f_i=f{|}_{U_i}$. So assume $f$ is indecomposable.
Consider $\lambda \in \mathbb{C}$ such that $f(v)=\lambda v$. It exists, because we can choose a
basis and contruct the polynomic equation $det(A-\lambda I)=0$, and $\mathbb{C}$ is algebraically closed. So, for certain $\lambda$, $Ker(f-\lambda id)$ as, at least, dimension 1. Moreover, observe that $\lambda$ is the only (possibly multiple) solution to the previous polynomial equation

We define $h=f-\lambda id$, and apply the previous proposition, so we get

for $m\ge 1$.
Since $f$ is indecomposable and $v\in ker(h^m)$ (since $f(v)-\lambda v=0$) we have that $V=ker(h^m)$, and therefore $h^m:V\mapsto V$ is the null transformation and $h$ is \emph{nilpotent}. Let $k$ be the nilpotency index, and $w\in V$ such that $h^{k-1}(w)\ne 0$ but $h^k(w)=0$.

We can show that $\{w,h(w),h^2(w),\dots ,h^k(w)\}$ is a basis for $V$. First, observe that they are independent because if

we can apply $h$ several times to show that every $a_i=0$.

And second, we can show that they span $V$. In fact, let $E=span\{w,h(w),h^2(w),\dots ,h^k(w)\}$. Observe that:

• $f(w)\in E$, since $h(w)=f(w)-\lambda w$.

• $f^2(w)\in E$, since $h^2(w)=f^2(w)-2\lambda f(w)+{\lambda }^{2}w$. Therefore, $f(h(w))\in E$, because $f(h(w))=f^2(w)-\lambda f(w)$ and so $h^2(w)=f(h(w))-\lambda h(w)$.

• $f^3(w)\in E$ and therefore $f(h^2(w))\in E$. In fact, $f(h^2(w))=h(f(h(w)))=h(h^2(w)+\lambda h(w))=h^3(w)+\lambda h^2(w)$

• In general, $f(h^i(w))=h(f(h^{i-1}(w)))=h(h^i+\lambda h^{i-1}))=h^{i+1}(w)+\lambda h^i(w)$.

Since $f(E)\subseteq E$ and $f$ is, by hypothesis, indecomposable, we conclude that $V=E$.
It only rests to show the form of the matrix of $f$ respect to the basis $\{w,h(w),h^2(w),\dots ,h^k(w)\}$. But because of the facts of the above enumeration we conclude that

In conclusion: For any endomorphism $f$ of $V$ we can find a partition $U_i$ of $f$-stables vector subspaces of $V$.
In each of one, $f_i$ has the form of the matrix above (in a particular basis). This is so because, either $f_i$ is a escalar matrix ($f_i=\lambda Id$) or $f_i-\lambda Id$ is nilpotent, and for nilpotent operators we have the special basis $\{h^k(w),h^{k-1}(w),\dots ,w\}$.

By the way: it there exists other decomposition in $f$-stables subspaces, but not satisfying that matrix form.