# Jordan canonical form of a squared matrix

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.

• "if $\lambda$ is an eigenvalue of $A$, then $\lambda^2$ is an eigenvalue of $A^2$": yes, but what can we say about the converse? If $\mu$ is an eigenvalue of $A^2$, then either $\sqrt{\mu}$ or $-\sqrt{\mu}$ is an eigenvalue of $A$ (hint: use characteristic polynomial). Maybe only one of them is an eigenvalue (trivial example: -1 isn't an eigenvalue of $A:=I$). – paf Sep 3 '16 at 14:27

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.

• How can I prove that if the eigenvalue is $\neq 0$ then a Jordan block of $\lambda$ squared is similar to a Jordan block of $J^2$ (if I understood correctly of the same size)? – tommy1996q Sep 4 '16 at 12:28
• @tommy1996q I added a reference. If it does not suffice, feel free to ask for more details. – user228113 Sep 4 '16 at 13:38

The Jordan canonical form of a matrix would have the eigen values on its diagonal along with some additional 1's in case the some of the eigen values have algebraic multiplicity $>1$. Check this Wikipedia page for more info on this

Also, as you correctly pointed out, the eigen values of $A^{2}$ would be squares of the eigen values of $A$. So the Jordan normal form of $A^{2}$ would be an "almost diagonal" matrix (similar to what we had for the Jordan normal form of $A$), with the squares of the eigen values in place of the original ones. Note that the algebraic multiplicity of the eigen values would carry forward unchanged from $A$ to $A^{2}$.