Let $n \in \Bbb N^*$ and $A \in \cal M_n(\Bbb R)$ a square matrix. Let the block matrix
$$B = \begin{pmatrix}A&A\\A&A \end{pmatrix} \in \cal M_{2n}(\Bbb R)$$
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Calculate the characteristic polynomial $\chi_B$ using $\chi_A$.
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Proof that $A$ is diagonalisable $\iff$ $B$ is diagonalisable.
Best 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, $$ \frac{1}{2}\pmatrix{I & I\\ I& -I} \pmatrix{A & A\\ A & A} \pmatrix{I& I\\I & -I} = \\ \left(\frac{1}{\sqrt{2}}\pmatrix{I & I\\ I& -I}\right)^{-1} \pmatrix{A & A\\ A & A} \frac{1}{\sqrt{2}}\pmatrix{I& I\\I & -I} = \pmatrix{2A & 0\\0 & 0} $$