# Finding the Jordan Canonical Form for a strictly lower block triangular matrix

Let $A\in \mathbb{C}^{n \times n}$ be an arbitrary matrix, and let $M\in \mathbb{C}^{(2n) \times (2n)}$ be the following block matrix $$M=\left( \begin{matrix} 0&0 \\ A&0 \end{matrix} \right).$$

Find the Jordan canonical form $J$ of $M$ in dependence of the rank of $A$.

My attempt:

It is clear that the only possible eigenvalue of $M$ is $0$. The number of Jordan blocks of the form $\left( 0 \right)$ is precisely the dimension of the eigenspace $E_0$ of $0$. By the dimension considerations we have $$\dim E_0 = \dim \ker M = 2n - \text{rank } M.$$

Similarly, the number of blocks of the form $\left( \begin{matrix} 0&0 \\ 1&0 \end{matrix} \right)$ is $\dim \ker M^2 - \dim E_0$. (Right?)

I'm not exactly sure how to proceed, or if I'm on the right track.

This is exam preparation. Please help me. :)

## 1 Answer

Notice that $M^2=0$. So you have $rank(A)$ Jordan's blocks of order $2$.

• Thank you, but could you please elaborate on why $M^2 = 0$? – rehband Sep 6 '14 at 17:48
• @rehband $\Im(M)=\{(0_{n\times 1},v_{n\times 1}), v\in\Im(A)\}$. Notice that $\Im(M)\subset\ker(M)$, thus $M^2=0$. – Daniel Sep 6 '14 at 17:52