linear algebra - Block matrix characteristic polynomial

Let $n \in \Bbb N^*$ and $A \in \cal M_n(\Bbb R)$ a square matrix. Let the block matrix $B=\begin{pmatrix}A&A\\A&A \end{pmatrix} \in \cal M_{2n}(\Bbb R)$

1) Calculate the characteristic polynomial $\chi_B$ unsing $\chi_A$.

2) Proof that $A$ is diangonalizable $\iff$ $B$ is diagonalisable.

Answer

For the first part: we have
$$
tI_{2n} - B =
\pmatrix{tI_n - A & -A\\ -A & tI_n - A}
$$
Noting that these matrices commute (or in particular, that the lower two matrices commute), we have
$$

\begin{align}
\det(tI_{2n} - B) &= \det((tI_n - A)^2-A^2)\\
&= \det[((tI_n - A)-A)((tI_n - A)+A)]\\
&= \det((tI_n - A)-A) \det((tI_n - A)+A)\\
&= \det((tI_n - 2A) \det(tI_n)\\
&= \det(2((t/2) - A)) \det(tI_n)\\
&= 2^n \det((t/2) - A) t^n\\
&= 2^n t^n \chi_A\left(\frac t2\right)
\end{align}
$$

Second part: I haven't put it all together nicely, but here are some potentially helpful thoughts

Thought 1: If $B$ is diagonalizable, then $B$ has $2n$ linearly independent eigenvectors. Now, let
$$
v = \pmatrix{v_1\\v_2}
$$
be an eigenvector of $B$, with $v_1,v_2 \in \mathbb{C}^n$. What can we say about the vectors $v_1$ and $v_2$? Remember that $B$ has the eigenvector $0$ of multiplicity $n + \operatorname{null}(A)$.

Thought 2:

$$
\pmatrix{I & I\\ I& -I} \pmatrix{A & A\\ A & A}\pmatrix{I& I\\I & -I} =
\pmatrix{4A & 0\\0 & 0}
$$
Or, in particular,
$$
\frac{1}{2}\pmatrix{I & I\\ I& -I} \pmatrix{A & A\\ A & A}
\pmatrix{I& I\\I & -I} = \\
\left(\frac{1}{\sqrt{2}}\pmatrix{I & I\\ I& -I}\right)^{-1} \pmatrix{A & A\\ A & A} \frac{1}{\sqrt{2}}\pmatrix{I& I\\I & -I} =
\pmatrix{2A & 0\\0 & 0}

$$