Raising a square matrix to a negative half power

If you look at Powers of diagonal matrices, it would be the reciprocal of the square root of each term in the diagonal.

Lets do an example for a diagonal matrix:

$$A=\begin{bmatrix}2 & 0 \\ 0 & 3\end{bmatrix}$$

$$A^{-1/2} = \begin{bmatrix} \frac{1}{\sqrt{2}} & 0 \\ 0 & \frac{1}{\sqrt{3}}\end{bmatrix} = \frac{1}{6}\begin{bmatrix}3 \sqrt{2} & 0 \\ 0 & 2 \sqrt{3}\end{bmatrix}$$

Clear?

The general idea (I think so) is Matrix function and Taylor expansion. Let $R$ is square matrix, we may write it in the form $R = I+X$ ($I$ - identity matrix), then using Taylor expansion: $$ R^{-1/2} = (I+X)^{-1/2} = I -\frac{1}{2}X + \frac{3}{8}X^2 + \cdots $$ I would be nice if $X$ is small.

So this goes along with Gr3gT's answer, but I'll chime in anyway.

Any $m \times n$ matrix $A$ can be written as:

$A = U \Sigma V^{H}$

Where $U$ is an $m\times m$ matrix whose columns are the left eigenvectors, $V$ is an $n\times n$ matrix whose columns are the right eigenvectors, and $\Sigma$ is a diagonal matrix of singular values.

Since $U$ and $V$ are unitary, we have:

$A^{\frac{1}{2}} = U \Sigma^{\frac{1}{2}} V^{H}$

So then:

$A^{\frac{-1}{2}} = (A^{\frac{1}{2}})^{-1} = (U \Sigma^{\frac{1}{2}} V^{H})^{-1} = V \Sigma^{\frac{-1}{2}} U^{H}$

Now when $A$ is not square, we can simply compute:

$A^{\frac{-1}{2}} = (A^{+})^{\frac{1}{2}} = U_{+} \Sigma^{\frac{1}{2}}_{+} V^{H}_{+}$

By taking the SVD of $A^{+}$, the pseudo-inverse of $A$.