Kernel of Jordan Normal Form

Question

Let $J_{mtimes m} $ be the Jordan normal form with real $c$ on the diagonal and 1 on the super diagonal. Show that
$$null (J-cI) subset null (J-cI)^2 subset ... subset null(J-cI)^{m-1} subset null (J-cI)^m $$
I was wondering if there is a shorter and alternative proofs than the one I managed to come up with.
The proof goes as follows:
Since $J$ is a Jordan normal form,
$exists x_m : (J-cI)^m x_m = 0$ and $(J-cI)^{m-1} x_m neq 0$.
By induction on m,
Base case $r=2$ then if $$k in null (J-cI) \
(J-cI)k = 0 \
(J-cI)^2= (J-cI)(J-cI) k = 0 $$
so $k in null (J-cI)^2 $
But since $exists x_m : (J-cI)^{m-1} x_m neq 0 $ then
$x_2 = (J-cI)^{m-2} x_m$ and $x_2 in null (J-cI)^2$ and $x_2 notin null (J-cI)$ thus a strict subset.
Suppose true for r=m-1 i.e
$$null (J-cI) subset null (J-cI)^2 subset ... subset null(J-cI)^{m-1}$$
Then for $k in null (J-cI)^{m-1}$, by similar procedure $k in null (J-cI)^m$ and $x_m in null (J-cI)^m, x_m notin null(J-cI)^{m-1} $ so a strict subset.
So the proposition is true by mathematical induction.
Main question is whether there's a more direct method of proving this?

Berci · Answer

Let $e_1,e_2,dots,e_n$ be the basis where the matrix of $J$ becomes a single Jordan block
$$J=pmatrix{c\ 1& c \ &ddots &ddots\ &&1&c}
text{ so }J-cI=pmatrix{0\ 1& 0 \ &ddots &ddots\ &&1&0},.$$
Call $A:=J-cI$, the values it maps to the basis elements can be read from its columns, specifically, $Ae_i=e_{i+1}$ if $i<n$ and $Ae_n=0$.
Rank of $A$ is clearly $n-1$, so explicitly $ker A={rm span}(e_n)$.
For $A^k$, observe that $Ae_i=e_{i+k}$ if $i+kle n$ and $Ae_i=0$ otherwise. By a similar argument as above, we get
$$ker A^k={rm span}(e_{n-k+1},dots,e_n),.$$

Kernel of Jordan Normal Form

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