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Given the following matrix, I was asked to tell whether the matrix is the square of some matrix or not.

$$ \begin{pmatrix} 1 & 0 & 0\\ 0 & -1 & 0\\ 0 & 0 & -1 \end{pmatrix} $$

Since this matrix is diagonal, a square root is

$$ \begin{pmatrix} 1 & 0 & 0\\ 0 & i & 0\\ 0 & 0 & i \end{pmatrix} $$ Am I right?

José Carlos Santos
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Sachchidanand Prasad
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    Right since $i^2=-1$. – Wuestenfux May 21 '17 at 08:10
  • The context can influence the correctness of your answer, but you're right. – N74 May 21 '17 at 08:11
  • One can interpret this geometrically: The original matrix is the transformation matrix of $\Bbb R^3$ defined by reflection through the $x$-axis, or just as well a rotation of $\pi$ about the $x$-axis. Thus, a real square root of the original matrix is the matrix encoding a rotation of $\pi / 2$ about the $x$-axis---this is just the rotation in Jose Carlos Santos' answer (for a certain choice of rotation direction). – Travis Willse Feb 18 '19 at 23:34

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You are right if the question was about matrices with complex coefficients, but wrong if it was about matrices with real coefficients. In the latter case, an answer would be$$\begin{pmatrix}1&0&0\\0&0&-1\\0&1&0\end{pmatrix}.$$

José Carlos Santos
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  • How you have calculated? – Sachchidanand Prasad May 21 '17 at 08:21
  • Rotation by 90° is the square root of rotation by 180°. -- Also, this is not just about complex vs. real coefficients: José's solution works for arbitrary fields (or even rings) of coefficients – Hagen von Eitzen May 21 '17 at 08:22
  • Just a correction: rotation by 90° is a square root of rotation by 180°, not the square root. – José Carlos Santos May 21 '17 at 08:25
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    Sorry, @HagenvonEitzen, but I do not get it. Can I prove this that the square root is the matrix provided in the answer? – Sachchidanand Prasad May 21 '17 at 08:25
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    Your matrix is $\left(\begin{smallmatrix}1&0&0\0&\cos(\pi)&-\sin(\pi)\0&\sin(\pi)&\cos(\pi)\end{smallmatrix}\right)$ and you want a square root of it. All you need is $2\times2$ matrix whose square is $\left(\begin{smallmatrix}\cos(\pi)&-\sin(\pi)\\sin(\pi)&\cos(\pi)\end{smallmatrix}\right)$. But the later matrix is the matrix of a rotation of $\pi$ radians. So, take a rotation of $\pi/2$ radians or a rotation of $-\pi/2$ radians. – José Carlos Santos May 21 '17 at 08:30