Let
$
M=\begin{bmatrix}
A & A \\
0 & A
\end{bmatrix}.
$
We want to find when $M$ is diagonalizable.
Since $M$ is a block upper triangular matrix, its eigenvalues are exactly the eigenvalues of $A$, each counted twice.
Now let $\lambda$ be an eigenvalue of $A$, and let $v \ne 0$ be an eigenvector of $A$ such that
$Av=\lambda v.$
Then
$M\begin{bmatrix} v \\ 0 \end{bmatrix}=\begin{bmatrix} Av \\ 0 \end{bmatrix}=\begin{bmatrix} \lambda v \\ 0 \end{bmatrix}=\lambda \begin{bmatrix} v \\ 0 \end{bmatrix}.$
So $\begin{bmatrix} v \\ 0 \end{bmatrix}$ is an eigenvector of $M$ corresponding to $\lambda$.
Now consider the vector $\begin{bmatrix} 0 \\ v \end{bmatrix}$. We have
$
M\begin{bmatrix} 0 \\ v \end{bmatrix}
=
\begin{bmatrix} Av \\ Av \end{bmatrix}
=
\begin{bmatrix} \lambda v \\ \lambda v \end{bmatrix}.
$
Hence
$
(M-\lambda I)\begin{bmatrix} 0 \\ v \end{bmatrix}
=
\begin{bmatrix} \lambda v \\ 0 \end{bmatrix}
=
\lambda \begin{bmatrix} v \\ 0 \end{bmatrix}.
$
If $\lambda \ne 0$, then
$
(M-\lambda I)\begin{bmatrix} 0 \\ v \end{bmatrix} \ne 0,
$
but
$
(M-\lambda I)^2\begin{bmatrix} 0 \\ v \end{bmatrix}=0.
$
So for every nonzero eigenvalue $\lambda$ of $A$, the matrix $M$ has a nontrivial Jordan block corresponding to $\lambda$. Therefore $M$ cannot be diagonalizable.
So, if $M$ is diagonalizable, then $A$ cannot have any nonzero eigenvalue. Hence the only eigenvalue of $A$ is $0$. Therefore $A$ is nilpotent.
But then $M$ is also nilpotent, because all its eigenvalues are $0$. A diagonalizable nilpotent matrix must be the zero matrix. So we must have $M=0.$
From
$
M=\begin{pmatrix}
A & A \\
0 & A
\end{pmatrix},
$
$\Rightarrow A=0.$
Conversely, if $A=0$, then
$
M=\begin{pmatrix}
0 & 0 \\
0 & 0
\end{pmatrix},
$
which is diagonalizable.
Therefore,
$
\begin{pmatrix}
A & A \\
0 & A
\end{pmatrix}
\text{ is diagonalizable if and only if } A=0.
$
So the correct answer is $\boxed{\text{A. } A=0}.$
