Necessary and sufficient condition for a matrix to be diagonalizable Let $A \in M_n(\mathbb C)$ be a complex matrix. $B$ is the matrix in $M_{2n}(\mathbb C)$:
$$\begin{pmatrix}
A &  A\\
0 &  A
\end{pmatrix}$$ What is a proof that $B$ is diagonalizable if and only if $A=0$?
 A: Clearly the point is to show that $A=0$ is a necessary condition for $B$ being diagonalisable.
First off, an easy computation shows that if $C\in M_n(\Bbb C)$ commutes with $A$, then for any polynomial $P$ one has
$$
  P\left[\pmatrix{A&C\cr0&A}\right] = \pmatrix{P[A]&P'[A]C\cr0&P[A]},
$$
where $P'$ is the derivative of$~P$ (it suffices to show this by induction when $P=X^k$, and use linearity).
Now if $B$ is diagonalisable, then $A$ must be as well (as the restriction of a diagonalisable operator to an invariant subspace is diagonalisable). Now let $P\in\Bbb C[X]$ be the minimal polynomial of$~A$ (the product of a linear factors $X-\lambda$ for each distinct eigenvalue$~\lambda$ of$~A$). By above computation (for $C=A$), the kernel of $P[B]$ contains the subspace generated by the first $n$ standard basis vectors, which contains the image of $P[B]$, so $(P[B])^2=0$ and $P^2$ annihilates $B$. Since $B$ is diagonalisable, its minimal polynomial$~\mu$ is a product of distinct linear factors, and from the above $P\mid\mu\mid P^2$, whence $\mu=P$.
Now if $v$ is an eigenvector of $A$ for some $\lambda$, applying $P[B]=0$ to any vector whose "second half" is given by $v$ has as "first half", according to the above computation, $\lambda P'[\lambda]v$; this must therefore be zero. Since $P$ has simple roots $P'[\lambda]\neq0$, so $\lambda=0$. Nonzero eigenvalues being impossible, $A=0$ follows.
A: Let $C\in M_k(\mathbb{C})$ and $A\in M_n(\mathbb{C})$. Denote by $C\otimes A\in M_{kn}(\mathbb{C})$ the Kronecker product of $C$ and $A$.
Lemma: If $0\neq C\otimes A\in M_{kn}(\mathbb{C})$ is diagonalizable then $C$ is diagonalizable.
Proof: If all eigenvalues of $A$ are zero then $A$ is nilpotent and  $A^n=0$. Thus, $(C\otimes A)^n=C^{n}\otimes A^n=0$. Since $C\otimes A$ is nilpotent and diagonalizable then $C\otimes A=0$, contradiction.
So there is an eigenvalue $0\neq a$ of $A$. Let $0\neq v\in \mathbb{C}^n$ such that $Av=av$.
Notice that $C\otimes A$ leaves the subspace $\mathbb{C}^k\otimes v=\{w\otimes v, w\in \mathbb{C}^k\}$ invariant. Therefore the restriction of $C\otimes A$ to $\mathbb{C}^k\otimes v$ is diagonalizable (The restriction of a diagonalizable linear transformation to a invariant subspace is also diagonalizable). Since the matrix that represents $C\otimes A|_{\mathbb{C}^k\otimes v}$ is $aC$ (in the basis $\{e_1\otimes v,\ldots e_k\otimes v\}$ of $\mathbb{C^k}\otimes v$, where $\{e_1,\ldots e_k\}$ is the canonical basis of $\mathbb{C}^k$)  then $aC$ is diagonalizable and $C$ is too.  $\square$
Now, $\begin{pmatrix}
A &  A\\
0 &  A
\end{pmatrix}=\begin{pmatrix}
1 &  1\\
0 &  1
\end{pmatrix}\otimes A$. Since $\begin{pmatrix}
1 &  1\\
0 &  1
\end{pmatrix}$ is not diagonalizable then, by the previous lemma, $\begin{pmatrix}
1 &  1\\
0 &  1
\end{pmatrix}\otimes A$ must be zero. Thus, $A=0$.
