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I want to solve $\mathbf{A}^x = \mathbf{B}$ where $\mathbf{A}$ and $\mathbf{B}$ are both $n$-by-$n$ matrices and $x$ is real. I see that in general there may be no solutions, or multiple solutions.

I was trying to find the period of a discrete-time linear dynamical system, but now I'm interested in the equation itself.

By "solve" I mean, given the two matrices, how can I find $x$ using the linear algebra primitives that MATLAB has, for example. Or maybe there's a name for this equation.

edit maybe less generally $\mathbf{A}^x = \lambda \mathbf{I}$

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You have to be rather lucky to find an $x$ for "random" $A$ and $B$. Just think about the degrees of freedom... –  Fabian Jul 7 '12 at 17:18
What kind of values do $A$ and $B$ take? Do you know them exactly or only to finite precision? –  Qiaochu Yuan Jul 7 '12 at 18:54

1 Answer 1

It is not clear what you mean by $A^x$ for $x$ irrational or even for $x$ not an integer. This is an issue already when $n = 1$; for example, what do you mean by $(-1)^{1/2}$? ( Which square root of $-1$?) What do you mean by $(-1)^{\pi}$?

One fix in the case $n = 1$ is to restrict to the case that $A, B$ are positive reals. The corresponding restriction for matrices is to restrict to the case that $A, B$ are positive-definite Hermitian matrices. In that case the spectral theorem guarantees that $A$ has an orthonormal basis of eigenvectors $e_1, ... e_n$ with positive real eigenvalues $\lambda_1, ... \lambda_n$, so you can define $A^x$ by requiring that $$A^x e_i = \lambda_i^x e_i$$

for all $i$. A necessary condition for the existence of a solution to $A^x = B$ is that $B$ also has eigenvectors $e_1, ... e_n$ (since this is true of $A^x$), and a necessary and sufficient condition is that its eigenvalues $\mu_1, ... \mu_n$ with respect to those eigenvectors satisfies $$\lambda_i^x = \mu_i$$

for all $i$.

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