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.
I suspect you have not worked through many matrices where the Jordan form is not diagonal. As with continued fractions, the development is alphabet soup; better to do examples.
These are from a book by Evar D. Nering, second edition. I will put both examples here, check for typing errors... the eigenvalues are integers,
Made up this one,
$$ C_{jagy} =
\left(
\begin{array}{ccc}
2&7&9 \\
0&2&8 \\
0&0&2 \\
\end{array}
\right)
$$
When everything is an integer, I like a sort of backwards order: the one denominator is in finding the inverse of a square matix. Here, $(C-2I)^3 = 0.$ To get the 1's above the diagonal, find column $p_3$ with $(C-2I)^2 p_3 \neq 0.$ Then column $p_2 = (C-2I)p_3.$ Finally $p_1 = (C-2I)p_2$ is a genuine eigenvector. Make the matrix $P$ from the columns $p_1, p_2, p_3.$ The determinant of your choice of $P$ is the denominator in finding $P^{-1}.$ Finally $J=P^{-1}C P$ is in Jordan form.
$$ \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc $$
$$ A_{nering} =
\left(
\begin{array}{ccccc}
1&0&-1&1&0 \\
-4&1&-3&2&1 \\
-2&-1&0&1&1 \\
-3&-1&-3&4&1 \\
-8&-2&-7&5&4 \\
\end{array}
\right)
$$
$$ B_{nering} =
\left(
\begin{array}{ccccc}
5&-1&-3&2&-5 \\
0&2&0&0&0 \\
1&0&1&1&-2 \\
0&-1&0&3&1 \\
1&-1&-1&1&1 \\
\end{array}
\right)
$$
Best Answer
In general we have $\sigma(T|_W) \subseteq \sigma(T)$ so every eigenvalue $\lambda$ of $T|_W$ is an eigenvalue of $T$. Recall that $$\dim \ker (T-\lambda I) = \text{number of $\lambda$-blocks}$$ $$\dim \ker (T-\lambda I)^2 - \dim \ker (T-\lambda I) = \text{number of $\lambda$-blocks of size $\ge 2$}$$ $$\dim \ker (T-\lambda I)^3 - \dim \ker (T-\lambda I)^2 = \text{number of $\lambda$-blocks of size $\ge 3$}$$ $$\vdots$$
For any $w \in W$ such that $(T-\lambda I)^{j-1}|_W w \ne 0$ and $(T-\lambda I)^{j}|_W w = 0$ we also have $(T-\lambda I)^{j-1} w \ne 0$ and $(T-\lambda I)^{j} w = 0$ so clearly \begin{align} \dim \ker (T-\lambda I)|_W^{j} - \dim \ker (T-\lambda I)|_W^{j-1} &= \dim \left(\ker(T-\lambda I)|_W^{j} \dot{-}\, \ker(T-\lambda I)|_W^{j-1}\right)\\ &\ge \dim \left(\ker(T-\lambda I)^{j} \,\dot{-}\, \ker(T-\lambda I)^{j-1}\right)\\ &= \dim \ker (T-\lambda I)^j - \dim \ker (T-\lambda I)^{j-1} \end{align}
Therefore the number of $\lambda$-Jordan blocks of $T|_W$ of size $\ge j$ is less than or equal to the number of $\lambda$-Jordan blocks of $T$ of size $\ge j$.
Hence for every $\lambda$-Jordan block of $T|_W$ there is a $\lambda$-Jordan block of $T$ of greater or equal size.