Finding the $\lim_{n\to\infty}(A^{n})$ of a complex matrix Let $$ A = .5\begin{pmatrix} 1+\alpha & -1+\alpha\\ -1+\alpha & 1+\alpha\end{pmatrix}. $$ where $\alpha$ is a complex number.
For which values of alpha does the limit $\lim_{n\to\infty}(A^{n})$ exist?
Determine the limit for the cases where it exists.
I have worked out a similar problem without the complex number alpha. For that problem I diagonalized the matrix A and then $$A^{n}=\left(PDP^{-1}\right)^{n}=PD^{n}P^{-1}$$ and I took the $\lim_{n\to\infty}(A^{n}=\left(PDP^{-1}\right)^{n}=PD^{n}P^{-1})$
I am having difficulty diagonalizing the matrix A. 
 A: This particular $A$ is very easy to diagonalize. You need to notice that
$$
A=I-\frac{1-\alpha}2\,\begin{bmatrix}1&1\\1&1\end{bmatrix}.
$$
You have $$\begin{bmatrix}1/\sqrt2&1/\sqrt2\\1/\sqrt2&-1/\sqrt2\end{bmatrix}\,\begin{bmatrix}1&1\\1&1\end{bmatrix}\,\begin{bmatrix}1/\sqrt2&1/\sqrt2\\1/\sqrt2&-1\sqrt2\end{bmatrix}=\begin{bmatrix}2&0\\0&0\end{bmatrix}.$$
Thus
$$
\begin{bmatrix}1/\sqrt2&1/\sqrt2\\1/\sqrt2&-1/\sqrt2\end{bmatrix}\,A\,\begin{bmatrix}1/\sqrt2&1/\sqrt2\\1/\sqrt2&-1/\sqrt2\end{bmatrix}\,=I-\frac{1-\alpha}2\,\begin{bmatrix}2&0\\0&0\end{bmatrix}=\begin{bmatrix}\alpha&0\\0&1\end{bmatrix}
$$
Thus
$$
A^n=\begin{bmatrix}1/\sqrt2&1/\sqrt2\\1/\sqrt2&-1/\sqrt2\end{bmatrix}\,
\begin{bmatrix}\alpha^n&0\\0&1\end{bmatrix}\,\begin{bmatrix}1/\sqrt2&1/\sqrt2\\1/\sqrt2&-1/\sqrt2\end{bmatrix}\,=\frac12\,\begin{bmatrix}1+\alpha^n&-1+\alpha^n\\-1+\alpha^n&1+\alpha^n\end{bmatrix}.
$$
A: Let $l$ denote the eigenvalues of $A$. We then have
$$(1+a-2l)^2 - (-1+a)^2 = 0 \implies (1+a-2l-1+a)(1+a-2l+1-a) = 0$$
Hence, $l=a$ or $l=1$. Hence, for $A^n$ to exist, we need $\vert l \vert \leq 1 \implies -1 \leq a \leq 1$
