[Math] Jordan canonical form of a squared matrix

jordan-normal-formlinear algebra

I was wondering: if I were given a matrix $A$, I could calculate its Jordan canonical form. If I considered then $A^2$, I could say that if $\lambda$ is an eigenvalue of $A$, then $\lambda^2$ is an eigenvalue of $A^2$. However I couldn't say anything about the Jordan canonical form of $A^2$. Are there any connections between the two canonical forms? Given the one of $A$, could I guess the one of $A^2$, or at least have some hints except the eigenvalues? Thanks for the help.

Best Answer

There is indeed a formula.

To begin with, if the JCF of $A$ is $$\begin{pmatrix} J(\lambda_1,m_1)&&0\\&\ddots&\\0&&J(\lambda_t,m_t)\end{pmatrix}$$ then $A^2$ is obviously similar to $$\begin{pmatrix} (J(\lambda_1,m_1))^2&&0\\&\ddots&\\0&&(J(\lambda_t,m_t))^2\end{pmatrix}$$

The power of a Jordan block is rarely a Jordan block. But, if the eigenvalue is $\ne 0$, it is always similar to a Jordan block. Specifically, if $J(\lambda, m)$ is a Jordan block of size $m$ and eigenvalue $\lambda$, then $$(J(\lambda,m))^s\sim\begin{cases} J(\lambda^n,m)&\text{if }\lambda\ne0\\ \begin{pmatrix}J(0,m-s+1) &0\\ 0 &\bf{0}_{(s-1)\times (s-1)}\end{pmatrix}&\text{if }\lambda=0\text{ and } m\ge s-1\\ \bf0_{m\times m}&\text{if }\lambda=0\text{ and } m< s-1\end{cases}$$

So, the visual procedure for $s=2$ is:

  • the non-nilpotent blocks remain unchanged, but the eigenvalues are squared.

  • for each nilpotent block of size $m\ge3$, you put there a nilpotent block of size $m-1$

  • the rest goes into a $\ker A$.

Added reference: A full proof of the lemma about the JCF of $(J(\lambda,m))^s$ can be found in Oscar Cunningham's answer to this older question.

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