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How many square roots of the matrix:

$$A = \left[\begin{array}{c c c} -1 & 0 & 0 & 0\\ 0 & -1 & 0 & 0 \\ 0 & 0 & -1 & 0 \\ 0 & 0 & 0 & -1 \end{array}\right]$$

where the square root $Q$ satisfies:

$$ QQ = A $$

exist?

Edit: I'm only considering the case of real numbers.

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    It partly depends on the context - are the entries for the matrices supposed to be real numbers or complex numbers or integers or one of the many other possibilities? – 2017-01-17
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    As a start you could analyse the $2\times 2$ situation - you can get square roots for the $4\times 4$ case from that. – 2017-01-17

2 Answers 2

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Hint: if $Q$ is one square root, then $R Q R^{-1}$ is another, where $R$ is any invertible $4 \times 4$ matrix. And (if $Q$ is not a multiple of the identity) there are lots of $R$ that don't commute with $Q$.

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All eigenvalues of $Q$ must be either $i$ or $-i$. Furthermore, $Q$ must be diagonalizable (because the square of non-diagonalizable invertible matrices can be shown to be non-diagonalizable, but $A$ is diagonalizable). If we allow complex numbers, all matrices which are similar to a diagonal matrix $D$ with $i$ and $-i$ as diagonal elements (i.e. all matrices $RDR^{-1}$ with an invertible $R$) can be chosen as $Q$.

If we allow only real numbers in $Q$, then the eigenvalues of $Q$ must show up as conjugate pairs, i.e. $Q$ has the eigenvalue $i$ with multiplicity $2$ and the eigenvalue $-i$ with multiplicity $2$. Each real matrix, which is diagonalizable in $\mathbb{C}$ and has this configuration of eigenvalues, is similar to $$B = \left(\begin{array}{cccc} 0 & -1 & 0 & 0 \\ 1 & 0 & 0 & 0 \\ 0 & 0 & 0 & -1 \\ 0 & 0 & 1 & 0 \end{array}\right)$$ which means that each real $Q$ can be written as $RBR^{-1}$ with an invertible matrix $R$.