Given the matrix for $a, b \in (0,1)$
$$ \begin{pmatrix} 1-a & a \\ b & 1-b \end{pmatrix}$$
The eigenvalues are $\lambda_1 = 1$ and $\lambda_2 = -a -b +1$.
We want to find the eigenvectors based on that so we solve
$$ (\begin{pmatrix} 1-a & a \\ b & 1-b \end{pmatrix} - I) x = \begin{pmatrix} -a & a \\ b & -b \end{pmatrix}x = 0$$
which gives $-a x_1 + a x_2 = 0 \Rightarrow x_1 = x_2$, and $bx_1 - bx_2 = 0$ which doesn't get me anywhere ...
Same for the other eigenvalue
$$ \begin{pmatrix} b & a \\ b & a \end{pmatrix}x = 0$$
This matrix is even linear dependent. What am I missing?
1) For $\lambda=1$, you have $x_1=x_2$, so the eigenvalues are the nonzero multiples of $\begin{bmatrix}1\\1\end{bmatrix}$
2) For $\lambda=1-a-b$, you have $bx_1+ax_2=0$, so $bx_1=-ax_2$ and
the eigenvalues are the nonzero multiples of $\begin{bmatrix}a\\-b\end{bmatrix}$