# Is there any way to use matrix decomposition for finding $A^n$?

If I want to take the power of matrix $$A$$ with e.g 3, $$A^3$$ or with power of $$-\frac {1}{2}$$, e.g $$A^{-\frac {1}{2}}$$ etc.

Is there an easy way to solve $$A^n$$, where $$n\in R$$ and $$A \in R^{nxn}$$ by using QR, LU, SVD, EIG or other decompositions?

• Robert's answer also works with Jordan form, $A = S J S^{-1}.$ the detail is that the diagonal part of $J$ commutes with the nilpotent part (the off-diagonal $1$s) – Will Jagy Feb 10 at 2:19

If you can diagonalize the matrix as $$A = S D S^{-1}$$, then $$A^n = S D^n S^{-1}$$, where for $$D^n$$ you just need to take the $$n$$'th power of the diagonal elements of $$D$$.
Caution: if $$n$$ is not an integer, the $$n$$'th power is multivalued over the complex numbers. This will even affect real matrices which have non-real eigenvalues. Also, $$n$$ is not an integer and the eigenvalues are not all distinct, there may be other $$n$$'th powers not of this form.
In case $$A$$ is diagonizable with a Diagonal matrix $$D$$ (having the eigenvalues on its diagonal and a matrix H, such that $$A=H^{-1}DH$$, then $$A^n = H^{-1}D^nH$$.