I'm reading Brémaud's Markov Chains, Gibbs Fields, Monte Carlo Simulation and Queues 2nd edition and I'm not following example 4.3.4 in page 159. Here's what's in it:
Let $A$ be a square matrix of dimension $r$ with $r$ distinct eigenvalues $\lambda_1 , \dots , \lambda_r$, and let $u_1 , \dots , u_r$ be the associated left eigenvectors and
$v_1 , \dots , v_r$ be the associated right eigenvectors. Then both the systems $u_1 , \dots , u_r$ and $v_1 , \dots , v_r$ are linearly independent. Moreover, $u^T_i v_j =0$ if $i \neq j$,and we can choose them in a way that $u^T_i v_i =1$ for each $1\leq i \leq r$, in which case we have the spectral decomposition: \begin{equation} A^n = \sum_{i=1}^r \lambda^n_i v_i u^T_i. \end{equation}
Example 4.3.4: The transition matrix on $E=\{1,2\}$, $$ P=\begin{bmatrix} 1-\alpha & \alpha\\ \beta & 1-\beta \end{bmatrix} $$ where $\alpha,\beta \in (0,1)$ has for characteristic polynomial $(1-\alpha-\lambda)(1-\beta-\lambda)-\alpha\beta$, which admits the roots $\lambda_1 =1$ and $\lambda_2 = 1-\alpha-\beta$. The right eigenvector associated with $\lambda_1 =1$ is $v=\mathbf{1}$, with all entries equal to $1$. Also, the stationary distribution $$ \pi^T = \left(\frac{\beta}{\alpha + \beta} , \frac{\alpha}{\alpha + \beta}\right) $$ is the left eigenvector corresponding to the eigenvalue $1$. In this example, the spectral decomposition takes the form: $$ P^n =\frac{1}{\alpha + \beta} \begin{bmatrix} 1-\alpha & \alpha \\ \beta & 1-\beta \end{bmatrix} + \frac{(1-\alpha-\beta)^n}{\alpha + \beta} \begin{bmatrix} \alpha & -\alpha \\ 1-\beta & -\beta \end{bmatrix} $$ and therefore, since $|1-\alpha-\beta|<1$, $$ \lim_{n \to \infty} P^n = \frac{1}{\alpha + \beta} \begin{bmatrix} 1-\alpha & \alpha \\ \beta & 1-\beta \end{bmatrix}. $$ In particular, the result of convergence to steady state, $$ \lim_{n \to \infty} {P}^n = \mathbf{1} \pi^T $$ was recovered for this special case in a purely algebraic way.
If I understood correctly, by the last line of the example we should have: $$ \lim_{n \to \infty} {P}^n = \mathbf{1} \pi^T = \frac{1}{\alpha + \beta}\begin{bmatrix} 1 \\ 1 \end{bmatrix}\begin{bmatrix} \beta & \alpha \end{bmatrix}=\frac{1}{\alpha + \beta}\begin{bmatrix} \beta & \alpha\\ \beta & \alpha\end{bmatrix}, $$ but this obviously doesn't coincide with the expression in the 3rd to last line: $$ \lim_{n \to \infty} P^n = \frac{1}{\alpha + \beta} \begin{bmatrix} 1-\alpha & \alpha \\ \beta & 1-\beta \end{bmatrix}. $$
Also, I've done the calculations myself and the spectral decomposition I got is: $$ P^n = \frac{1}{\alpha+\beta}\begin{pmatrix}{\beta} & {\alpha}\\{\beta} & {\alpha} \end{pmatrix} +\frac{(1-\alpha-\beta)^n}{\alpha+\beta}\begin{pmatrix}{\alpha} & -{\alpha}\\-{\beta} & {\beta}\end{pmatrix} $$ which, taking limit when $n \to \infty$, coincides with $\mathbf{1} \pi^T$. Are the spectral decomposition and the equation in the 3rd to last line of the example wrong?