Given a nilpotent matrix $N$ of order $n\times n$ such that $N^k=0$. Then i know that $m_{N}(x)=x^k$ and $c_{N}(x)=x^n$. This means that $N^k$ is a zero matrix. My question now is can we deduce how many zeros are there in $N^i$ for $1<i<k$. Do we need to assume any other hypothesis to tell the answer like do we need to know how many $`0'$'s were there originally in $N$?
One thing that I am sure is the after each step the number of $`0'$'s keep decreasing but the actual count is what I want to know.
Is there any bound on the number of zeros or anything like that. Can somebody help?
Think of $N$ just as a linear transformation first.
As you have noted that $m_N (x)= x^k , c_N(x)=x^n$ , then by the Theory of Jordan Canonical Forms, there exists a basis $\beta$ of $V$ (the vector space) such that w.r.t $\beta$ the matrix of $N$ , say M is a lower triangular matrix admitting a decomposition of Jordan blocks . The blocks are say, $A_l , 1 \le l \le p$ for some $p \in \Bbb N$ with $A_l =(a_{ij})_{1 \le i,j \le b_l}$ such that $a_{ij} = \delta_{i-1,j}$ with $k=b_1 \ge b_2 \ge \dots b_p \ge 1$ .
Now for the Computation part, we may start from some Rough Work : Say we consider $T_1 = \begin{pmatrix}0 & 0\\\ 1 & 0\end{pmatrix}$ then , ${T_1}^2 =0 $ so $T_1$ has $3 (= 2^2 -1)$ many zeroes and $ {T_1}^2$ has all 4 to be zero.
Now , consider $T_2 = \begin{pmatrix}0 & 0 & 0\\\ 1 & 0 & 0 \\\ 0 & 1 &0\end{pmatrix}$ then, ${T_2}^2 = \begin{pmatrix}0 & 0 & 0\\\ 0 & 0 & 0 \\\ 1 & 0 &0\end{pmatrix}$, ${T_2}^3 =0 $ , so $T_2$ has $7 (=3^2 -2 =(2+1)^2 - 2)$ zeroes, ${T_2}^2 $ has $8 = (7+1)$ many zeroes, $T_2^3$ has all $9 ( = 8+1) $ zeroes .
Since, from the first Paragraph we get that our $M$ is a direct sum of such matrices and hence comes in block forms , all other entries are zero! So if $M$ has Jordan blocks $A_l$ of size $b_l$ , then each $A_l$ consists of $b_l -1$ many 1's and $b_l^2 - b_l+1$ many zeroes, thus $M$ has precisely $\sum_{l=1}^p (b_l -1) $ many 1's and thus $n^2 -\sum_{l=1}^p (b_l -1) $ many zeroes,
$M^k$ has all it's entries to be 0 and thus has $n^2$ zeroes!
Then for each $1 < m <k$ , consider $M^m$ and the corresponding blocks $A_l ^m$. As long as ${A_l}^{m-1} \ne 0_{b_l \times b_l}$ ,each such block contains one more zero than what ${A_l}^{m-1}$ had . If ${A_l}^{m-1} \ne 0_{b_l \times b_l}$ but, ${A_l}^{m} = 0_{b_l \times b_l}$ then ${A_l}^{m+q}$ cannot give more zeroes into the mix ($\forall q \ge 1$), since ${A_l}^{m}$ is already $ 0_{b_l \times b_l}$ .
Hence, the total number of zeroes in $M^m = (n^2 -\sum_{l=1}^p (b_l -1))+ (m-1) Cardinality\{1 \le l \le p : {A_l}^{m-1} \ne 0_{b_l \times b_l}\} , \forall 1 \le m \le k $
Note that since there exists $P \in GL(V)$ such that , $N=PMP^{-1}$ we also get that $N^m= PM^m P^{-1}$ thus the argument above is said modulo similarity