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Good evening,

I was wondering if there is such a valid operation as raising a matrix to the power of a matrix, e.g. vaguely, if $M$ is a matrix, is $$ M^M $$ valid, or is there at least something similar? Would it be the components of the matrix raised to each component of the matrix it's raised to, resulting in again, another matrix?

Thanks,

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up vote 12 down vote accepted

It is possible to define the exponential $$\exp(M) = \sum_{n \ge 0}^{\infty} \frac{M^n}{n!}$$

of any matrix using power series. Similarly, it is possible to define the logarithm $$\log(I + M) = \sum_{n \ge 1} \frac{(-1)^{n-1} M^n}{n}$$

when the above series converges (this is guaranteed for example if the largest singular value of $M$ is less than $1$). We can therefore define $$M^N = \exp(N \log M)$$

by imitating the identity $a^b = e^{b \log a}$ for, say, positive reals, but this won't have good properties unless $N$ and $M$ commute, I think. It's better to consider the exponential and logarithm separately.

As I have discussed elsewhere on math.SE, the fact that the ordinary exponential takes two inputs which are the same type is misleading. Most (but not all) "exponential-type" operations in mathematics take two inputs which are different types.

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That series defines $\log(I+M)$ –  Thomas Andrews Jun 29 '12 at 1:08
    
@Thomas: whoops. Fixed. –  Qiaochu Yuan Jun 29 '12 at 1:16
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Actually $\log(M)$ can be defined whenever $M$ is invertible, but there are many possible values (there is a choice of branch of the logarithm for each eigenvalue of $M$). –  Robert Israel Jun 29 '12 at 1:45
    
@Robert: that is not so clear to me if $M$ has nontrivial Jordan blocks. Do you want $M$ to be diagonalizable? –  Qiaochu Yuan Jun 29 '12 at 1:54
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See e.g. Rudin, "Functional Analysis", sections 10.21 to 10.30. –  Robert Israel Jun 29 '12 at 21:26
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