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

1. What does it mean to raise a square matrix to a negative half power?
2. What general ideas of linear algebra does this assume?
3. (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.
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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?

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Many thanks. I should have realised that. – Jeromy Anglim Mar 25 '13 at 4:45
@JeromyAnglim: no problem - I am blind to most things until I work them! Regards – Amzoti Mar 25 '13 at 4:48
Clear to me! ;-) – amWhy Apr 13 '13 at 0:27
Same here...and it can be embarrassing when I miss something in answers here on math.se! So it it does keep oneself humble. – amWhy Apr 13 '13 at 2:26

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.

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I see, that I've answered bit another question, not actually yours. The question I've answered is: what is $A^{-1/2}$ when $A$ is a square matrix. ) – Nikita Evseev Mar 25 '13 at 5:06
It is worth noting that the inverse square root of a square matrix is not unique, and that the one obtained using this particular taylor expansion is just one possible such matrix. – Glen O Mar 25 '13 at 5:31
Yes, even $\sqrt{1}$ is not unique. – Nikita Evseev Mar 25 '13 at 6:27
Thanks for the answer. The power of a diagonal diagonal matrix was my specific problem, but I was interested in reading about the more general problem involving any square matrix. – Jeromy Anglim Mar 25 '13 at 6:57

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.

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+1 I think your answer is much better than the accepted one. – Kim Jong Un May 16 '15 at 1:58