Let $J_{m\times 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?
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 $i+k\le n$ and $Ae_i=0$ otherwise. By a similar argument as above, we get $$\ker A^k={\rm span}(e_{n-k+1},\dots,e_n)\,.$$