Show that a triangular block matrix is non diagonalizable Let $A$ be a $n$-square matrix. Consider the upper triangular matrix :
\begin{equation} M = \begin{pmatrix} A & A \\ 0 & A \end{pmatrix} \end{equation}
I need to show that $M$ is diagonalizable if and only if $A$ is the zero matrix.
I have thought of considering the characteristic polynomial of $M$ that equals to $\chi_A^2$.
I'm thinking of using the result that a matrix is diagonalizable if and only if its minimal polynomial is of the form $\prod_i (X-\lambda_i)$ where the $\lambda_i$ are distinct, and using the fact that :
\begin{equation} P(M)= \begin{pmatrix} P(A) & (XP')(A) \\ 0 & P(A) \end{pmatrix} \end{equation}
but I don't really see how.
 A: First, we observe that if $M$ is diagonalizable, then $A$ cannot have any non-zero eiegnvalues.
Indeed, suppose for contradiction that $x \in \Bbb F^n$ is non-zero and satisfies $Ax = \lambda x$, with $\lambda \neq 0$. It follows that
$$
(M-\lambda I) \pmatrix{0\\ x} = \pmatrix{\lambda x\\ 0}, \quad (M-\lambda I)^2 \pmatrix{0\\x} = (M-\lambda I) \pmatrix{\lambda x\\ 0} = 0.
$$
Thus, $(0,x)$ is an element of $\ker(M-\lambda I)^2$ but not an element of $\ker(M-\lambda I)$, but the existence of such an element implies that $M$ is non-diagonalizable.
With that, we deduce that $A$ has zero as its only eigenvalue, which implies that $A$ is nilpotent. Verify that this implies that $M$ is also nilpotent. However, if $M$ is nilpotent and diagonalizable, then it must be the zero matrix, which means that $A$ must be the zero matrix.
A: The first direction, that $A=\mathbf 0\implies M$ is diagonalizable holds in a trivial way.
For the second direction: suppose $M$ diagonalizable
$\implies$ it is annihilated by a polynomial $s$ that splits linearly with no repeated roots
$\implies A$ is annihilated by $s$
$\implies A$ is diagonalizable.
$\implies M= Z^{-1}\left[\begin{matrix}A & \mathbf 0\\\mathbf{0} & A\end{matrix}\right]Z$
(because both are similar to the same diagonal matrix $D$)
Let $q$ be the minimal polynomial of $A$.  Via diagonalizability this means $q$ splits linearly and has no repeated roots.  It also means $q$ is the minimal polynomial of $\displaystyle \left[\begin{matrix}A & \mathbf 0\\\mathbf{0} & A\end{matrix}\right]$.
$\implies q$ is minimal polynomial of $M$ (via similarity to A).  Applying $q$ to $M$ gives
$\left[\begin{matrix}q\big(A\big)& *\\\mathbf{0} & q\big(A\big)\end{matrix}\right] = \mathbf 0$
suppose $q$ has degree $d\geq 2$ and examine the $*$:
$\mathbf 0 = d\cdot A^d+c_{d-1}\cdot (d-1)\cdot A^{d-1}+\cdots + c_1 \cdot A =g\big(A\big)$
If $\text{char }\mathbb K = p \gt 0$ and $d$ is a multiple of $p$ i.e. if $p\%d =0$ then the result is immediate because $\mathbf 0 = c_{d-1}\cdot (d-1)\cdot A^{d-1}+\cdots + c_1 \cdot A $
i.e. $A$ is annihilated by a polynomial of degree $\lt d$ which is a contradiction. (Note: technically the case of $g = 0$ is dealt with separately, essentially via the argument in the below paragraph: $g=0\implies$ $q$ and its derivative $q'$ have a common root $\mu$, hence $\mu$ is a repeated root of $q$, which would be a contradiction.)
The remaining case is when $d\neq 0$ (i.e. it is a unit in $\mathbb K$)
$t(x) := g(x) -d\cdot q(x)$. This has degree $\lt d$ and must be the zero polynomial (degree $=-\infty$) -- otherwise, again, $A$ is annihilated by a polynomial of degree lower than $d$, a contradiction.
Finally utilize (i) that $g(x) = x\cdot q'(x)$ and (ii) since we have assumed $d\geq 2$ there is some $\mu\in \mathbb K-\{0\}$  such that $q(\mu)=0$. But this means $g(\mu) = 0= \mu\cdot q'(\mu)$.  Thus $q$ and $q'$ have a common root, $\mu$, which is thus a repeated root of $q$, a contradiction.
We conclude that the degree of $q$ must be $d=1 \implies M\propto I_{2n}\implies A = \mathbf 0$.
A: Your idea works. Suppose $M$ is diagonalisable over some field. Then the minimal polynomial $m(x)$ of $M$ is a product of distinct linear factors. Since
$$
0=m(M)=\pmatrix{m(A)&Am'(A)\\ 0&m(A)},
$$
we have $m(A)=Am'(A)=0$. In particular, $m(A)=0$ and $m(x)$ is divisible by the minimal polynomial $a(x)$ of $A$. As $m$ is a product of distinct linear factors, it has not any repeated zeroes. Therefore $m$ and $m'$ do not share any zero. In turn, $a$ and $m'$ do not share any zero. Thus $m'(A)$ is invertible. So, from $Am'(A)=0$ we infer that $A=0$.
