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I am trying to solve the following problem:

Let $A$ be an $2n$ by $2n$ matrix such that $A^2 = -I_{2n}$. Prove that $A$ is similar to the matrix \begin{equation} \left( \array{0 && -I_n \\ I_n && 0} \right). \end{equation}

I have no idea how to start this problem.

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    $\begingroup$ Are you familiar with eigenvalues? $\endgroup$
    – Alex Meiburg
    Jul 25, 2017 at 23:36
  • $\begingroup$ @Alex Yes, I am. $\endgroup$
    – user109871
    Jul 25, 2017 at 23:39
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    $\begingroup$ @user109871 to what extent are you familiar with them? $\endgroup$
    – mdave16
    Jul 26, 2017 at 0:29
  • $\begingroup$ Can you ay anything useful about the eigenvalues of $A$? $\endgroup$
    – John Hughes
    Jul 26, 2017 at 0:31
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    $\begingroup$ Shouldn't we be referring to eigenvalues and eigenvectors? Seems like eigenvalues alone may not cut it. $\endgroup$
    – gary
    Jul 26, 2017 at 0:40

1 Answer 1

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This is only true over fields such as $\Bbb R$ and $\Bbb Q$ in which the equation $x^2=-1$ is not soluble. Over the complex numbers, $A=iI$ satisfies $A^2=-I$ but is not similar to the given matrix.

Over $\Bbb R$ or $\Bbb Q$ though, the condition $A^2=-I$ means that the rational canonical form of $A$ is uniquely defined, so all matrices $A$ with $A^2=-I$ are similar.

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