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A given $n\times n$ matrix $B$ of real entries obeys $B^{2}=-I$.

I know that $B$ has only 2 eigenvalues $i,-i$ and that $n$ has to be even. Therefore each eigenvalue must have $n/2$ multiplicity. How do I prove that there are $n$ linearly independent eigenvectors?

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    $\begingroup$ Observe that $x^2+1$ is the characteristics polynomial as well as minimal polynomial of $B$ which has distinct linear factors and hence $B$ is diagonalizable and therefore it has $n$ linearly independent eigen vectors. $\endgroup$ – little o Feb 28 '19 at 15:49
  • $\begingroup$ Just to clarify: $x^2 +1$ is not the characteristic polynomial of $B$ unless $n=2$. But it is the minimal polynomial. $\endgroup$ – cs47511 Feb 28 '19 at 16:05
  • $\begingroup$ @Dbchatto67 Any way of showing that there are $n$ linearly independent eigenvectors without invoking that $B$ is diagonalizable? $\endgroup$ – user2175783 Feb 28 '19 at 16:06
  • $\begingroup$ @csprun you are correct. $\endgroup$ – little o Feb 28 '19 at 16:07
  • $\begingroup$ Those two conditions are equivalent, and @Dbchatto67 hasn't invoked it, he has proved it (albeit concisely). One can prove that if a matrix $B$ satisfies a polynomial with no repeated roots, e.g. $x^2+1$ over $\mathbb{C}$, then it's diagnalizable (in an algebraic closure of the ground field). Again, you can prove this easily with JNF. $\endgroup$ – cs47511 Feb 28 '19 at 16:07
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This is essentially the same as what @Dbchatto67 says in the comments, but if you know about Jordan normal form (very much worth learning if not), then you can solve this easily by thinking in JNF. A $k\times k$ Jordan block (over $\mathbb{C}$) $$J = \begin{pmatrix} \lambda & 1 & 0 & \dots & 0 & 0 \\ 0 & \lambda & 1 & 0 & \dots & 0 \\ \vdots & \vdots & \ddots & \ddots & \ddots & \vdots \\ 0 & 0 & 0 & 0 & \dots & \lambda \end{pmatrix}$$ does not satisfy $J^2 = -I$ unless $\lambda = \pm i$ (as you already saw) and $k=1$. So the JNF of $B$ over $\mathbb{C}$ is diagonal and there are $n$ linearly independent eigenvectors over $\mathbb{C}$.

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    $\begingroup$ That is precisely what I was trying to say. Very nice explanation @csprun. $\endgroup$ – little o Feb 28 '19 at 16:10

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