What does it mean when a matrix is to the (-1/2) power?

I'm reading a machine learning paper that uses a form of matrix normalization called symmetric divisive; given a matrix A and a diagonal matrix D derived from A, we define $$N=D^{-1/2}AD^{-1/2}$$ I am not sure what that exponent means, am I supposed to invert and then take the square root of the matrix? Of its values? A little lost here.

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For diagonal matrices, it's especially easy. Just apply the exponent to each of the entries on the diagonal. You can verify that squaring the result and multiplying by $D$ gives the identity matrix. – MPW Apr 2 '14 at 23:24
So long as the entries along the diagonal are positive since you're taking the power $-1/2$ – user139388 Apr 2 '14 at 23:25
Thanks that makes sense. Does one of you want to write it in an answer so that I can accept it? – Andrew Apr 2 '14 at 23:32
In general, a $n$-th root of some matrix $A$ is a matrix $B$ with $B^n = A$. Note that there can be zero, one or more than one than one such $B$, therefore writing this as $A^{\frac{1}{2}}$ is a bit dangerous in the general case. – fgp Apr 2 '14 at 23:44
I feel like such an expression ought to carry an "Abuse Of Notation" warning. (Fractional exponents on matrices is a new one on me...) – RecklessReckoner Apr 2 '14 at 23:58

Just for future reference, here is a general way of finding integer and non-integer powers of a square matrix.

The theorem is

$$If \ \ A\vec{v} = \lambda\vec{v} \ ,\\ A^n\vec{v} = \lambda^n\vec{v}$$

where $\lambda$ is an eigenvalue for the eigenvector $\vec{v}$ of matrix $A$.

Source, Examples and Further Information : http://www.blackmesapress.com/Eigenvalues.htm

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And what if the matrix isn't diagonalizable? E.g. $A=\begin{pmatrix}1&1\\0&1\end{pmatrix}$ doesn't have an eigenbasis. – Henning Makholm Apr 3 '14 at 0:42
@HenningMakholm It has, however, the Jordan form. – Algebraic Pavel Apr 3 '14 at 1:50