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

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?

• Do you have the minimal and characteristic polynomials for $A$? Commented Oct 17, 2021 at 19:27
• Dear @AHusain, I can confidently say that the characteristic polynomial is trivial here as $A$ has a single unique eigenvalue equal to $\lambda$. Commented Oct 17, 2021 at 21:18

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.