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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?


  • Kaiser, H. F. (1970). A second generation little jiffy. Psychometrika, 35(4), 401-415.
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up vote 5 down vote accepted

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:

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

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


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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! 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

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 $A$.

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This doesn't do the right thing: $A^{1/2} A^{1/2}$ needn't be $A$. But this trick does work for converting an eigendecomposition into a square root. Note that some matrices have no square root at all, for instance $\begin{bmatrix} 0 & 1 \\ 0 & 0 \end{bmatrix}$. – Ian Feb 13 at 23:48
Also, the square root of a rectangular matrix doesn't make any sense anyway. – Ian Feb 14 at 2:51

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