Let me sketch a proof of existence of the Jordan canonical form which, I believe, makes it somewhat natural.

Let us say that a linear endomorphism $f:V\to V$ of a nonzero finite dimensional vector space is **decomposable** if there exist *proper* subspaces $U_1$, $U_2$ of $V$ such that $V=U_1\oplus U_2$, $f(U_1)\subseteq U_1$ and $f(U_2)\subseteq U_2$, and let us say that $f$ is **indecomposable** if it is not decomposable. In terms of bases and matrices, it is easy to see that the map $f$ is decomposable iff there exists a basis of $V$ such that the matrix of $f$ with respect to which has a non-trivial diagonal block decomposition (that it, it is block diagonal two blocks)

Now it is not hard to prove the following:

**Lemma 1.** *If $f:V\to V$ is an endomorphism of a nonzero finite dimensional vector space, then there exist $n\geq1$ and nonzero subspaces $U_1$, $\dots$, $U_n$ of $V$ such that $V=\bigoplus_{i=1}^nU_i$, $f(U_i)\subseteq U_i$ for all $i\in\{1,\dots,n\}$ and for each such $i$ the restriction $f|_{U_i}:U_i\to U_i$ is indecomposable.*

Indeed, you can more or less imitate the usual argument that shows that every natural number larger than one is a product of prime numbers.

This lemma allows us to reduce the study of linear maps to the study of *indecomposable* linear maps. So we should start by trying to see how an indecomposable endomorphism looks like.

There is a general fact that comes useful at times:

**Lemma.** *If $h:V\to V$ is an endomorphism of a finite dimensional vector space, then there exists an $m\geq1$ such that $V=\ker h^m\oplus\def\im{\operatorname{im}}\im h^m$.*

I'll leave its proof as a pleasant exercise.

So let us fix an indecomposable endomorphism $f:V\to V$ of a nonzero finite dimensional vector space. As $k$ is algebraically closed, there is a nonzero $v\in V$ and a scalar $\lambda\in k$ such that $f(v)=\lambda v$. Consider the map $h=f-\lambda\mathrm{Id}:V\to V$: we can apply the lemma to $h$, and we conclude that $V=\ker h^m\oplus\def\im{\operatorname{im}}\im h^m$ for some $m\geq1$. moreover, it is very easy to check that $f(\ker h^m)\subseteq\ker h^m$ and that $f(\im h^m)\subseteq\im h^m$. Since we are supposing that $f$ is indecomposable, one of $\ker h^m$ or $\im h^m$ must be the whole of $V$. As $v$ is in the kernel of $h$, so it is also in the kernel of $h^m$, so it is not in $\im h^m$, and we see that $\ker h^m=V$.

This means, precisely, that $h^m:V\to V$ is the zero map, and we see that $h$ is *nilpotent*. Suppose its nilpotency index is $k\geq1$, and let $w\in V$ be a vector such that $h^{k-1}(w)\neq0=h^k(w)$.

**Lemma.** The set $\mathcal B=\{w,h(w),h^2(w),\dots,h^{k-1}(w)\}$ is a basis of $V$.

This is again a nice exercise.

Now you should be able to check easily that the matrix of $f$ with respect to the basis $\mathcal B$ of $V$ is a Jordan block.

In this way we conclude that every indecomposable endomorphism of a nonzero finite dimensional vector space has, in an appropriate basis, a Jordan block as a matrix.
According to Lemma 1, then, every endomorphism of a nonzero finite dimensional vector space has, in an appropriate basis, a block diagonal matrix with Jordan blocks.