I am unable to find a real matrix $A$ that satisfies $A^2=B$ for the matrix given below by any methods I know, since it is already diagonalized. Does there exist such a real $A$ at all, and more importantly, is there any way to prove whether it does? $$B=\begin{bmatrix} 1 & 0 & 0 \\ 0 & -1 & 0 \\ 0 & 0 & -1 \end{bmatrix}$$
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There does not exist a real matrix but if you you use the field of complex numbers then it is simply the square root of those eigenvalues – Ziad Fakhoury May 28 '17 at 13:19
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1@Zaid there does exist a real matrix. – Ben Grossmann May 28 '17 at 13:26
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Yes there is. The eigenvalues of $A$ must be square roots of $1$, $-1$ and $-1$. They could be $1$, $i$ and $-i$, or $-1$, $i$ and $-i$. (They can't be $1$, $i$ and $i$ since that set is not closed under conjugation.)
Angina Seng
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Yes, there indeed exists such a real matrix. In general, a real square root will exist so long as all negative eigenvalues appear with even multiplicity. As for finding a square root, it's simple in this case
Hint: The transformation induced by $B$ is a $180^\circ$ rotation about the $x$-axis. We can take $A$ to be another rotation.
Ben Grossmann
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