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Let $A$ be an $n\times n$ matrix with real entries such that $A^{2}=-I$ where $I$ is the identity matrix.

Then $n$ is even, and if $n=2k$, then $A$ is similar over the field of real numbers to a matrix of the block form $$ \begin{bmatrix}‎ ‎0 &‎ -l_{k} \\‎ ‎l_{k} & 0 ‎\end{bmatrix} $$ Where $l_{k}$ is the $k\times k$ identity matrix.

Why? How?

el_tenedor
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444
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    $\text{det}(A^2)=\text{det}(-I)=(-1)^n\text{det}(I)=(-1)^n$. Since $\text{det}(A^2) \geq 0$, $n$ must be even in order to get $(-1)^n=1$ – Itay4 Apr 09 '17 at 16:07
  • That's right. Thanks. – 444 Apr 09 '17 at 16:13
  • Duplicate: https://math.stackexchange.com/questions/949189/a-is-n-times-n-real-matrix-with-a2-i-to-prove-that-n-is-even, https://math.stackexchange.com/questions/2007643/let-a-be-an-n-times-n-matrix-with-real-entries-such-that-a2-i-0-then, https://math.stackexchange.com/questions/471445/prove-properties-of-a2-i – mdcq Apr 10 '18 at 16:24

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