Take the 2-minute tour ×
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.

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 $\mathbf{A}^x = \lambda \mathbf{I}$

share|improve this question
1  
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
add comment

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 $i$.

share|improve this answer
add comment

Your Answer

 
discard

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.