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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?

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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  • $\begingroup$ Many thanks. I should have realised that. $\endgroup$ – Jeromy Anglim Mar 25 '13 at 4:45
  • $\begingroup$ @JeromyAnglim: no problem - I am blind to most things until I work them! Regards $\endgroup$ – Amzoti Mar 25 '13 at 4:48
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    $\begingroup$ Clear to me! ;-) $\endgroup$ – Namaste Apr 13 '13 at 0:27
  • $\begingroup$ Same here...and it can be embarrassing when I miss something in answers here on math.se! So it it does keep oneself humble. $\endgroup$ – Namaste Apr 13 '13 at 2:26
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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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    $\begingroup$ 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. ) $\endgroup$ – Nikita Evseev Mar 25 '13 at 5:06
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    $\begingroup$ 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. $\endgroup$ – Glen O Mar 25 '13 at 5:31
  • $\begingroup$ Yes, even $\sqrt{1}$ is not unique. $\endgroup$ – Nikita Evseev Mar 25 '13 at 6:27
  • $\begingroup$ 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. $\endgroup$ – Jeromy Anglim Mar 25 '13 at 6:57
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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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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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    $\begingroup$ 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 $\begin{bmatrix} 0 & 1 \\ 0 & 0 \end{bmatrix}$. $\endgroup$ – Ian Feb 13 '16 at 23:48
  • $\begingroup$ Also, the square root of a rectangular matrix doesn't make any sense anyway. $\endgroup$ – Ian Feb 14 '16 at 2:51

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