The matrix $B = \left(\begin{smallmatrix}A& 0\\A & A\end{smallmatrix}\right)$ is diagonalizable, if only if $A$ is diagonalizable. Let $A\in\mathcal {M}_{n}(\mathbb {C}), $ I am trying to prove that:
The matrix $B = \left(\begin{smallmatrix}A& 0\\A & A\end{smallmatrix}\right)$ is diagonalizable, if only if $A$ is diagonalizable.
My effrot :
$(\Rightarrow) $ we suppose that $B $ is diagonalizable then $m_B$ is a product of distinct linear polynomials.  Let $m_A(x)$, $m_B(x)$,  be the minimal polynomials of $A$, and $B$, respectively.
We have
$$
0 = m_B(B) = \pmatrix{m_B(A) & 0\\(*)&m_B(A)}
$$
so, $m_B(A) = 0$.  It follows that $m_A$ divides $m_B$, thus $m_A $  is a product of distinct linear polynomials. Then $A $ is diagonalisable.
For $(\Leftarrow )$ I have no idea
An idea please.
 A: This is not true: $\pmatrix{ 1 & 0 \\ 1& 1}$.
If $A$ is diagonalizable, then $B$ is similar to a matrix of the form $\pmatrix{D & 0 \\ D& D}$ where $D$ is diagonal.
By permuting, rows and columns, we get a block-diagonal matrix  with diagonal blocks of type $\pmatrix{d_i& 0 \\ d_i & d_i}$. These blocks are diagonalizable if and only if $d_i=0$. Then $B$ is diagonalizable if and only if $D=0$ which is equivalent to $A=0$.
A: You can prove inductively that $$\forall m\in\Bbb{N},\quad B^m=\begin{pmatrix}A^m&0\\mA^m&A^m\end{pmatrix}$$ so that for every polynomial $P$ it holds that $$P(B)=\begin{pmatrix}P(A)&0\\AP'(A)&P(A)\end{pmatrix}.$$ Therefore, letting $\mu_A$ be $A$'s minimal polynomial and $P\in\Bbb{C}[X]$ arbitrary we get that $P$ annihiliates $B$ if and only if $\mu_A|P$ and $\mu_A|XP'$. This implies that if we let $$\mu_A=X^\nu\prod_{i=1}^r(X-\lambda_i)^{m_i}$$ with the product running over the nonzero eigenvalues of $A$ (and $\nu\in\Bbb{N}$ possibly $=0$) we get that $\mu_B=X^\nu\prod_{i=1}^r(X-\lambda_i)^{m_i+1}$.
The only way $\mu_B$ has simple roots is to have $\mu_A=X$, i.e. $A=0$. In other words, if $B$ is diagonalizable then $A$ is the zero matrix.

EDIT. Let $P=X^\nu\prod_{i=1}^r(X-\lambda_i)^{m_i+1}$. We prove that $P=\mu_B$ in two steps:

*

*$P$ divides $\mu_B$: this we prove by showing that $0$ has multiplicity $\geq\nu$ as a root in $\mu_B$ and if $\lambda$ has multiplicity $m\geq 1$ as a root in $\mu_A$ then it has multiplicity $\geq m+1$ in $\mu_B$

*$P(B)=0$ and so $\mu_B | P$
This implies $P=\mu_B$.

Suppose $0$ is an eigenvalue of $A$ i.e. $\nu\geq 1$. Then $\mu_A|\mu_B$ implies that $0$ has multiplicity $\geq \nu$ in $\mu_B$ and the extra $X$ in $X\mu_B'$ implies that $X^\nu$ automatically satisfies $X^\nu | X\mu_B'$ so $0$ will have the multiplicity $\geq \nu$ as a root of $\mu_B$.

Suppose $\lambda$ is a nonzero eigenvalue of $A$ (if there is one) with multiplicity $m\geq 1$ as a root of $\mu_A$. Then $(X-\lambda)^m|\mu_B$ and $(X-\lambda)^m|\mu_B'$ since $(X-\lambda)^m$ is coprime with $X$. Thus $\lambda$ is a root of multiplicity $\geq m+1$ of $\mu_B$.

Combining the two results above yields $P|\mu_B$. Conversely, $P(A)=0$ and $AP'(A)=0$ since $\mu_A$ divides $P$ by construction and $\mu_A$ divides $XP'$ because of multiplicities. Thus $\mu_B|P$ and so $\mu_B=P$.
