For a $n\times n$ matrix $A$ that has independent eigenvectors, I want to raise the power of $A$ recursively like $A^{1}\vec{u_{0}}=\vec{u_{1}}$ and then to find out $\vec{u_{k}}$, I could use $A^{k}\vec{u_{0}}=\vec{u_{k}}$.
And if I expand $A^{k}\vec{u_{0}}=S\Lambda ^{k}\vec{c}$, where $S$ is the eigenspace and $\vec{c}$ is the combination factors of the eigenvectors in $S$ such that $\vec{u_{k}}=c_{1}\vec{s_{1}}+\cdots+c_{n}\vec{s_{n}}$ and $\Lambda $ is the eigenvalues matrix.
Can I say I do it this way instead: $ A^{k}\vec{u_{0}}=\Lambda ^{k}S\vec{c} $ Then... $ A^{k}\vec{u_{0}}=\Lambda ^{k}\vec{u_{0}} $
Would everything still be the same? I have a hard time trying to prove to myself. It doesn't look like they the same since the multiplication sequence for matrices does make a difference. But it's like they have the same meaning as: $ A^{k}\vec{u_{0}}=C_{1}\lambda _{1}^{k}\vec{x_{1}}+C_{2}\lambda _{2}^{k}\vec{x_{2}}+\cdots+C_{n}\lambda _{n}^{k}\vec{x_{n}} $
Thanks for any help.