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I cannot find an answer to if it is generally possible to take the square root of a diagonal matrix $A$ by taking the square root of each individual component along the main diagonal, e.g. for a 2-by-2 matrix $$ \sqrt{A} = \begin{pmatrix} \sqrt{a_1} & 0 \\ 0 & \sqrt{a_2} \\ \end{pmatrix}. $$ Is this OK to do provided that it is a (square) diagonal matrix?

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    $\begingroup$ It is sufficient to observe that $\begin{pmatrix}\sqrt{a_1} & 0 \\ 0 & \sqrt{a_2}\end{pmatrix}\begin{pmatrix}\sqrt{a_1} & 0 \\ 0 & \sqrt{a_2}\end{pmatrix}= \begin{pmatrix}a_1 & 0 \\ 0 & a_2\end{pmatrix}$ $\endgroup$ – user247327 Oct 21 '18 at 19:50
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I assume that you consider matrices with entries in a field $\mathbb{F}$. If square roots $\sqrt{a_i}$ exist in $\mathbb{F}$, then it is ok. However, a diagonal matrix $A$ may have a square root even if the $a_i$ do not square roots in $\mathbb{F}$. An example for $\mathbb{F} = \mathbb{R}$ is $$ A = \begin{pmatrix} -1 & 0 \\ 0 & -1 \\ \end{pmatrix}. $$ In fact, a square root of $A$ is given by $$ B = \begin{pmatrix} 0 & 1 \\ -1 & 0 \\ \end{pmatrix}. $$

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  • $\begingroup$ Thanks for your reply! Does field $\mathbb{F}$ imply any special properties, or is it an arbitrary field? $\endgroup$ – litmus Oct 22 '18 at 7:00
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    $\begingroup$ The field is arbitrary. But the properties of $\mathbb{F}$ determine whether every (diagonal) matrix $A$ has a square root, and whether there exist square roots in diagonal form. For example, the matrix $A$ in my answer has a square root, but no diagonal square root. For $\mathbb{F} = \mathbb{C}$ it has a diagonal square root. $\endgroup$ – Paul Frost Oct 22 '18 at 12:21

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