Let $A \in M_n(\mathbb{C})$, Find Canonical Jordan form of $A^2$ in correlation to Jordan form of $A$

Question

No given Information since this question relies on typical Complex matrix.

I know that $J_n(x)= xI+J_n(0)$ but how does it help me? I have no idea what to do.

Thanks in advance.

Answer 1

As you have already determined, it suffices to separately consider the Jordan blocks of $A$. For $x \neq 0$, $J_n(x)^2$ is indeed similar to $J_n(x^2)$.

The question remains: what happens when $x = 0$? We know that the only eigenvalue of $J_n(0)^2$ is zero, but it is not clear what the size of the blocks should be. I claim the following: for all integers $k \geq 1$, $J_{2k}(0)^2$ is similar to $J_{k}(0) \oplus J_k(0)$ and $J_{2k+1}(0)^2$ is similar to $J_{k}(0) \oplus J_{k+1}(0)$, where $\oplus$ denotes a direct sum.

One way to prove that this holds is by considering the Weyr characteristic (cf. also Matrix Analysis by Horn and Johnson). In other words, it suffices to note that $\operatorname{rank}((J_{n}(0)^2)^m) = n - 2m$ for all $m < n/2$ and $0$ for $m \geq n/2$. However, I find that a more satisfying approach is to construct a similarity, i.e. to find a matrix $P$ for which $P^{-1}(J_{2k+1}(0)^2)P = J_k(0) \oplus J_{k+1}(0)$.

I will only consider the case of $n = 2\cdot 2 + 1 = 5$, but the general construction is completely analogous. Note that $$ J_{5}(0)^2 = \pmatrix{0&0&1&0&0\\0&0&0&1&0\\0&0&0&0&1\\0&0&0&0&0\\0&0&0&0&0}. $$ Let $e_1,\dots,e_5$ denote the standard basis of $\Bbb R^5$. We note that the vectors $e_1,e_3,e_5$ for a Jordan chain. That is, we have $$ Je_5 = e_3, \quad Je_3 = e_1, \quad Je_1 = 0. $$ Similarly, the vectors $e_2,e_4$ form a Jordan chain since we have $$ Je_4 = e_2, \quad Je_2 = 0. $$ With that, we can conclude if we take $P$ to be the (permutation) matrix with columns $e_2,e_4,e_1,e_3,e_5$, then we have $$ P^{-1}J_5(0)^2P = \pmatrix{0&1&0&0&0\\0&0&0&0&0\\0&0&0&1&0\\0&0&0&0&1\\0&0&0&0&0} = J_2(0) \oplus J_3(0). $$