Raising a square matrix to a negative half power I want to implement the following formula (taken from Kaiser, 1970) in R where $R$ is square matrix of correlations:
$$S = (\textrm{diag } R^{-1})^{-1/2}$$
I understand the diagonal and inverse operations, but I am unclear on the meaning of raising a square matrix to a negative half power. Thus, my questions


*

*What does it mean to raise a square matrix to a negative half power?

*What general ideas of linear algebra does this assume?

*(if this is in scope) How would this be implemented in R?


References


*

*Kaiser, H. F. (1970). A second generation little jiffy. Psychometrika, 35(4), 401-415.

 A: 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.
A: 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 $A$.
A: 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?
A: At a book of pr. Ng i saw that $A^{\frac{1}{2}}=U\Lambda^{\frac{1}{2}}U^{*}$, where A is a hermitian matrix, $U$ is a unitary matrix with columns the eigenvectors of matrix $A$, $U^{*}$ is the conjugate transpose of matrix $U$ and $\Lambda$ is a diagonal matrix with entries the eigenvalues of matrix $A$. So, $A^{-\frac{1}{2}}=(A^{\frac{1}{2}})^{-1}$. Of course here your matrix is diagonal, so i agree with Amzoti.
