How to find the exponent $x$ of a matrix in the following "eigenvalue" equation
$$\mathbf{A}^x \mathbf{v} = a \mathbf{v} ?$$
If it was a scalar eqution, then one could take logarithms on both sides to solve for $x$. I have heard of logarithm of a matrix, but as far as I know, there is no such thing as a logarithm of a vector and this is a vector equation.
Add 1 If I work along the line of the eigenvalue problem, then one gets a characteristic polynomial
$$|\mathbf{A}^x - a\mathbf{I}| = (\lambda_1^x - a)(\lambda_2^x - a)...(\lambda_n^x - a)$$
which suggests that if I know the eigenvalues of $\mathbf{A}$ then I can just solve $\lambda_1^x = a$ for $x$ (And hope that all other parts of the decomposition, i.e. $\lambda_j^x = a$, also result in the save value? Or will I have $n$ solutions for $x$, counting the multiplicity of the eigenvalues?)
We will assume that all $\lambda_k$ are $>0$, which is a condition under which one can take a non-integer power of a matrix by different methods (see (https://en.wikipedia.org/wiki/Matrix_function)).
I think that all is the consequence of what you have written :
$$|\mathbf{A}^x - a\mathbf{I}| = (\lambda_1^x - a)(\lambda_2^x - a)...(\lambda_n^x - a)=0$$
if and only if, for some $k$,
$$\lambda_k^x=a \ \ \ \iff \ \ \ x=\dfrac{\ln a}{\ln \lambda_k} $$
under the condition that it makes sense... in particular $a$ has to be positive.
This will leave you in general with real values of $x$ and exceptionaly with integer values.