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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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1 Answer 1

up vote 2 down vote accepted

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

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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
    
Sorry, I wrote that matrix wrong. At any rate, I've edited the end of my answer; take a look. –  Omnomnomnom Nov 16 '13 at 1:14

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