Finding a Jordan normal form of a matrix that is almost Jordanform-like

Question

Consider some positive integer $N$, and then a matrix $A$ of size $N^2 \times N^2 $ of the following form:

$$A = \begin{pmatrix} J & I & 0 & \ldots & 0 & 0\\ 0 & J & I & \ldots & 0 & 0\\ 0 & 0 & J & \ldots & 0 & 0\\ \ldots & \ldots & \ldots & \ldots & \ldots & \ldots \\ 0 & 0 & 0 & \ldots & J & I \\ 0 & 0 & 0 & \ldots & 0 & J \end{pmatrix}$$, where $J = \begin{pmatrix} \lambda & 1 & 0 & \ldots & 0 \\ 0 & \lambda & 1 & \ldots & 0 \\ \ldots & \ldots & \ldots & \ldots & \ldots \\ 0 & 0 & 0 & \ldots & \lambda\end{pmatrix}$ is a Jordan block of size $N\times N$, and $I$ is the unity $N\times N$ matrix. Note that all Jordan blocks are identical, and have the same eigenvalue on the diagonal.

For instance, if $N=2$, then $A = \begin{pmatrix} \lambda & 1 & 1 & 0 \\ 0 & \lambda & 0 & 1 \\ 0 & 0 & \lambda & 1 \\ 0 & 0 & 0 & \lambda \end{pmatrix}$.

From the numerical tests I know that such matrix $A$ has $N$ Jordan blocks, but how one can show this mathematically?

Answer 1

For the N=2 case, the nullity of A-$\lambda I$ is 2 hence there are two blocks. By induction you can prove the required result.