Complexity of computing matrix powers

Question

I am interested in calculating the $n$'th power of a $n\times n$ matrix $A$. Suppose we have an algorithm for matrix multiplication which runs in $\mathcal{O}(M(n))$ time. Then, one can easily calculate $A^n$ in $\mathcal{O}(M(n)\log(n))$ time. Is it possible to solve this problem in lesser time complexity?

Matrix entries can, in general, be from a semiring but you can assume additional structure if it helps.

Note: I understand that in general computing $A^m$ in $o(M(n)\log(m))$ time would give a $o(\log m)$ algorithm for exponentiation. But, a number of interesting problems reduce to the special case of matrix exponentiation where m=$\mathcal O(n)$, and I was not able to prove the same about this simpler problem.

Answer 1

If the matrix is diagonalizable then taking the $n$th power can be done in time $$O(D(n)+ n\log n)$$ where $D(n)$ is the time to diagonalize $A$.

Just to complete the details, if $A=P^{-1}DP$ with a diagonal $D$, then $$A^n = (P^{-1}DP)^n = P^{-1}D^nP$$

and $D^n$ can be computed by just taking each element of the diagonal (each eigenvalue of $A$) to the $n$th power.

Answer 2

One good way out is Singular Value Decomposition SVD. Given an $n\times n$ real matrix $A$ of full rank , SVD splits it apart as $A=U\Sigma U^T$ where $\Sigma$ is a diagonal matrix, in time $O(n^3)$. By the properties of SVD, $A^m = U \Sigma^m U^T$, so only the diagonal matrix need be exponentiated, and this can be done in $O(n\log m)$ time. Performing the final multiplication $U \times \Sigma^m \times U^T$ takes $O(n^{2.3727})$, so we have altogether $O(n^3 + n \log m)$ operations.

Update after comment The point is that once the SVD is found, any power takes only $O(n^{2.3727}+n \log m)$ time to compute by your own C-W algorithm. But this isn't your question.If there were really an $o(M(n)\log(m))$ algorithm, it would convert immediately to an $o(\log n)$ algorithm for integers. I suspect that one such doesn't exist.