# 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.

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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}
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, $$\frac{1}{2}\pmatrix{I & I\\ I& -I} \pmatrix{A & A\\ A & A} \pmatrix{I& I\\I & -I} = \\ \left(\frac{1}{\sqrt{2}}\pmatrix{I & I\\ I& -I}\right)^{-1} \pmatrix{A & A\\ A & A} \frac{1}{\sqrt{2}}\pmatrix{I& I\\I & -I} = \pmatrix{2A & 0\\0 & 0}$$
Thank's, Omnomnomnom, for this answer.It's for me of considerable interest. I have an answer in which I use matrix direct product. if $J = \begin {pmatrix} 1 & 1 \\1 & 1 \end {pmatrix}$, then $J \otimes A = B =\begin{pmatrix}A&A\\A&A\end{pmatrix}$. We have $J \sim D$ where $D = \begin{pmatrix}0 & 0 \\0 & 2 \end {pmatrix}$ and this follows that $J \otimes A \sim D \otimes A = \begin {pmatrix} 0 & 0 \\0 & 2A \end {pmatrix}$. I notice that it looks like the last part of your answer. Can we find a link between the two? – Mohamed Nov 16 '13 at 1:03
Yes, there's a link. Note that $\frac{1}{\sqrt{2}}\pmatrix{I & I\\ I&I}$ is its own inverse. Thus, taking my last answer and dividing by $2$ on both sides gives you a proof of similarity. – Omnomnomnom Nov 16 '13 at 1:07