Finding eigenvectors through triangularization I have an exam tomorrow and am working through notes. We derived the following stochastic matrix:
$$P= \begin{bmatrix}
 0.8 & 0.5 & 0 & 0\\
0.2 & 0.5 & 0 & 0 \\
0 & 0 & 0.7 & 0.5 \\
0 & 0 & 0.3 & 0.5  
\end{bmatrix}$$
We are looking to find its eigenvalues. He stated in class that we can triangularize $A-\lambda I$ to easily find the $\lambda_i$, but I'm not seeing it. 
$$\det \begin{bmatrix}
 0.8-\lambda & 0.5 & 0 & 0\\
0.2 & 0.5-\lambda & 0 & 0 \\
0 & 0 & 0.7-\lambda & 0.5 \\
0 & 0 & 0.3 & 0.5-\lambda  
\end{bmatrix} = 0$$
I recognize that writing all the steps out in full matrix form is kind of a pain in latex, so please don't feel the need to do that. 
 A: Note that $x = [1,1,1,1]$ is a left eigenvector of $A$ so that $xA = 1x$. The transpose of a matrix preserves eigenvalues and so $1$ is an eigenvalue of $A$ it should be obvious that a similar result holds for any matrix whose columns sum to unity. 
Using Abel's comment, considering the individual block matrices and note that $1$ is an eigenvalue of both
$$A=\begin{bmatrix}
0.8 & 0.5\\
0.2& 0.5
\end{bmatrix} \text{ and } B=\begin{bmatrix}
0.7&0.5\\
0.3& 0.5
\end{bmatrix}.$$
Since 
$$\operatorname{tr}(A) = 0.8 + 0.5 = \lambda_{1} + \lambda_{2} = 1 + \lambda_{2} = 1.3$$
and similarly,
$$\operatorname{tr}(B) = 0.7 + 0.5 = \eta_{1} + \eta_{2} = 1 + \rho_{2} = 1.2$$
(the trace equals the sum of the eigenvalues). It follows the four eigenvalues of $P$ are $1,1,0.2,0.3$.
Consider the matrix $A$ and $\lambda_{2} = 0.3$. Then
$$A-0.3I = \begin{bmatrix}
0.5&0.5\\
0.2&0.2
\end{bmatrix} \rightarrow \begin{bmatrix}
1&1\\
0&0
\end{bmatrix}$$
So one eigenvector of $A$ is $v_{2} = [-1,1]^{T}$ and so an eigenvector of $P$ corresponding to eigenvalue $1$ is $[-1,1,0,0]^{T}$ and repeat for the rest. 
