Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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.

share|cite|improve this question
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
up vote 0 down vote accepted

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 :

share|cite|improve this answer
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

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.