Recursive matrices raised to the kth power

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.

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I ain't quite familiar with "combination factors", but if the eigendecomposition of $\mathbf A$ is $\mathbf V\mathbf \Lambda\mathbf V^{-1}$ (that is, $\mathbf A$ has a complete eigenvector set), then the $n$-th power of $\mathbf A$ is $\mathbf V\mathbf \Lambda^n\mathbf V^{-1}$. Thus, multiplying by just the diagonal matrix would be wrong... –  Ｊ. Ｍ. Aug 3 '11 at 8:20
The combination factors that I refer to are just the weights to $\vec{u}$. Yea, raising $A$'s power is just raising the eigenvalues' powers. But Say if I have $A^{k}\vec{u_{0}}=\vec{u_{1}}$ and I want to do it recursively until $u_{k}$, I'm thinking if I could just multiply the diagonal matrix. Maybe I should update my question a little on this part. I didn't express myself well enough on this. –  xenon Aug 3 '11 at 8:47
"I'm thinking if I could just multiply the diagonal matrix. " - Still no (there are factors on both the left and right hand sides, and matrix multiplication isn't commutative). If you have a computing environment handy, you can try it out yourself... –  Ｊ. Ｍ. Aug 3 '11 at 8:50
I have updated my question a little. So, with $\Lambda$, I can say that $A^{k}\vec{u_{0}}=S\Lambda ^{k}\vec{c}$ is not equivalent to $A^{k}\vec{u_{0}}=\Lambda ^{k}S\vec{c}$ because matrix multiplication isn't commutative, am I right? I have tried to work it out by hands. They are different but I just felt weird because the idea "feels" like the same. –  xenon Aug 3 '11 at 8:58
Right.$\text{}$ –  Ｊ. Ｍ. Aug 3 '11 at 8:59

It is true that if $A$ has a complete set of eigenvectors, you can expand $\vec{u_{0}}=\sum c_n \vec{s_n}$ with the $\vec{s_n}$ being eigenvectors of $A$. Then, if $\lambda_n$ is the eigenvalue corresponding to $\vec{s_n}$, you have $\vec{u_1}=A\vec{u_0}=\sum c_n \lambda_n \vec{s_n}$ and $\vec{u_k}=A^k\vec{u_0}=\sum c_n \lambda_n^k \vec{s_n}$. You can show this by using the linearity of the operations.
As J.M. suggests, the matrix multiplication's non-commutative rule applies and therefore $A^{k}\vec{u_{0}}=\Lambda ^{k}\vec{u_{0}}$ is not valid. The thing is $A^{k}\vec{u_{0}}=S\Lambda ^{k}\vec{c}\neq\Lambda ^{k}S\vec{c}$