# What is symmetric square root of matrix?

In one publication I found a following part:

$\textbf R^{-1/2}$ is a symmetric square root of matrix $\textbf R^{-1}$

I know what is a square root of matrix, but what exactly is symmetric square root of matrix?

The matrix $\textbf R$ is symmetric and invertible.

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For a symmetric real positive definite matrix $A$ there is a symmetric square root $Q$ such that $Q^2=A=Q^{1/2}Q^{1/2}$.

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So it just means that matrix Q is also symmetric? –  Gacek Jan 13 '14 at 19:38
Yes. So in your example $R^{1/2}$ would be the square root of $R$ which is symmetric (in this case) and $R^{-1/2}$ is its inverse (also symmetric). –  John U Jan 13 '14 at 19:41
Great, thank you. –  Gacek Jan 13 '14 at 19:42
Do you mean $A^{1/2}A^{1/2}$ on the right hand side or does $Q^2=Q$? –  Matta May 18 at 11:15

This is just saying that the square root is also symmetric.

For a matrix $A$, there may exist more than one matrix $Q$ such that $QQ = A$. Not all $Q$ are symmetric, but if there is a $Q$ that is symmetric, then so must be $A$, since $A^T = Q^TQ^T = QQ = A$.

One of the easy ways to find $Q$ is to use the eigendecomposition of $A$.

If $Q$ needs to be a real matrix, then $A$ has to be real and positive semidefinite.

In the case that $A$ is positive semidefinite, you can take powers of $A$ to get the inverse, or the inverse square root, etc., like for $R$ in your example.

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