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Find all real-valued $n\times n$ matrices $X$ such that $$ X^2 + E = 0, $$

here $E$ --- identity matrix, $n$ --- odd.

Nikita Evseev
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    This question bears remarkable similarity to this one: http://math.stackexchange.com/questions/791333/find-all-3-times-3-matrices-x-such-that-x2e-0 – Robert Lewis May 16 '14 at 03:54
  • $\det (-E) = 1$ of $n$ is even. So his proof only applies to the case $n$ odd, as it should. – Robert Lewis May 16 '14 at 06:16

2 Answers2

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Let $X^2=-E$ and let $D$ be the determinant of $X$. Now, justify the following: $D^2=-1$. What do you conclude?

Ittay Weiss
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Since $X^2+E=0$ you have that the if $f(x)=x^2+1$ then $f(X)=0$. Hence the minimal polynomial of $X$ divides $f(x)$ and so $m_X(x)=x^2+1$.

This means that the eigenvalues of $X$ are complex. But the characteristic polynomial of $X$ is of degree $n=odd$ so it has a real root. Contradiction. So there are no real matrices satisfying the given equation.

Kal S.
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