# Square root of Diagonal matrix

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?

• It is sufficient to observe that $\begin{pmatrix}\sqrt{a_1} & 0 \\ 0 & \sqrt{a_2}\end{pmatrix}\begin{pmatrix}\sqrt{a_1} & 0 \\ 0 & \sqrt{a_2}\end{pmatrix}= \begin{pmatrix}a_1 & 0 \\ 0 & a_2\end{pmatrix}$ – user247327 Oct 21 '18 at 19:50

## 1 Answer

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 $$B = \begin{pmatrix} 0 & 1 \\ -1 & 0 \\ \end{pmatrix}.$$

• Thanks for your reply! Does field $\mathbb{F}$ imply any special properties, or is it an arbitrary field? – litmus Oct 22 '18 at 7:00
• 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. – Paul Frost Oct 22 '18 at 12:21