Find invariant factors of a power of a matrix, given a matrix in Jordan canonical form

Question

Suppose \begin{align} A&=(I_4+(N_1 \oplus N_3))\oplus(N_2\oplus N_4 \oplus N_5)\oplus(-I_3+N_3) \\ &= \begin{pmatrix} 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 1 & 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 1 & 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 1 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 1 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 1 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 1 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & -1 & 1 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & -1 & 1 \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & -1 \\ \end{pmatrix}, \end{align} which is in Jordan canonical form. How can I find the invariant factors and hence the Jordan canonical form of $A^2$?

I know that $$A^2 = (I_4+2(N_1 \oplus N_3)+(0_2 \oplus N_3^2)) \oplus (N_2^2 \oplus N_4^2 \oplus N_5^2) \oplus (I_3-2N_3+N_3^2).$$ The characteristic polynomial of $A^2$ is $\chi_{A^2}(x)(x-1)^7 x^{11}$ and the minimal polynomial of $A^2$ is $\mu_{A^2}(x)=(x-1)^3x^3$. I know that I have to find the invariant factors $d_1,d_2,\ldots,d_{18}$ of $A^2$. And $d_1 \mid d_2 \mid \cdots \mid d_{18}$, $d_{18}=\mu_{A^2}(x)$, and $d_1 d_2 \cdots d_{18} = \chi_{A^2}(x)$.

So obviously I know what $d_{18}$ is, but what is $d_{17}$? There are many factors that divide $d_{18}(x)$, so it seems I have a "choice" in "choosing" $d_{17}$, but I don't. I do know thought I must consider the ranks of powers of the matrix $A^2$.

Anyway, as soon as I find all the invariant factors, then I have everything I need to find the Jordan canonical form of $A^2$. So my question is really just finding the invariant factors of $A^2$.

Answer 1

You can easily calculate $A^2$ by blocks instead, and use this result for each block:

If $\lambda$ is an eigenvalue of $A$, consider the sequence of subspaces: $$0\varsubsetneq \ker(A-\lambda I)\varsubsetneq \ker(A-\lambda I)^2\varsubsetneq\dotsm\varsubsetneq \ker(A-\lambda I)^r=\ker(A-\lambda I)^{r+1}=\dotsm$$ and set $d_k=\dim\ker(A-\lambda I)^k$. Then

$d_k-d_{k-1}={}$ number of Jordan blocks for the eigenvalue $\lambda$ of size $\ge k$.