Finding the nth power of a matrix via change of basis

Question

So I have a matrix $$\mathcal A=\begin{bmatrix}0&1\\1&1\end{bmatrix}$$ with corresponding eigenvectors $$\vec v_1=\begin{bmatrix}2\\1+\sqrt{5}\end{bmatrix}\ \ \ \ \ \ \ \ \ \ \vec v_2=\begin{bmatrix}2\\1-\sqrt{5}\end{bmatrix}$$To diagonalise this matrix I changed the basis via $$\begin{bmatrix}2&2\\1+\sqrt{5}&1-\sqrt{5}\end{bmatrix}^{-1}\begin{bmatrix}0&1\\1&1\end{bmatrix}\begin{bmatrix}2&2\\1+\sqrt{5}&1-\sqrt{5}\end{bmatrix}$$And got the new diagonalised matrix to be $$\begin{bmatrix}\frac{1+\sqrt{5}}{2}&0\\0&\frac{1-\sqrt{5}}{2}\end{bmatrix}$$I think that the $n^{th}$ power of this matrix is $$\begin{bmatrix}\bigg(\frac{1+\sqrt{5}}{2}\bigg)^n&0\\0&\bigg(\frac{1-\sqrt{5}}{2}\bigg)^n \end{bmatrix}$$However I'm unsure where to go from here. I thought of doing $$\begin{bmatrix}1&0\\0&1\end{bmatrix}^{-1}\begin{bmatrix}\bigg(\frac{1+\sqrt{5}}{2}\bigg)^n&0\\0&\bigg(\frac{1-\sqrt{5}}{2}\bigg)^n \end{bmatrix}\begin{bmatrix}1&0\\0&1\end{bmatrix}$$However this clearly has no effect. Any help is appreciated, thanks :)

Answer 1

No. The final step should be$$\begin{bmatrix}2&2\\1+\sqrt5&1-\sqrt5\end{bmatrix}\begin{bmatrix}\left(\frac{1+\sqrt5}2\right)^n&0\\0&\left(\frac{1-\sqrt5}2\right)^n\end{bmatrix}\begin{bmatrix}2&2\\1+\sqrt5&1-\sqrt5\end{bmatrix}^{-1}.$$