# Finding eigenvectors through triangularization

I have an exam tomorrow and am working through notes. We derived the following stochastic matrix: $$P=\left[ \begin{matrix} 0.8 & 0.5 & 0 & 0\\ 0.2 & 0.5 & 0 & 0 \\ 0 & 0 & 0.7 & 0.5 \\ 0 & 0 & 0.3 & 0.5 \end{matrix} \right]$$

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.

$$\det \left( \left[ \begin{matrix} 0.8-\lambda & 0.5 & 0 & 0\\ 0.2 & 0.5-\lambda & 0 & 0 \\ 0 & 0 & 0.7-\lambda & 0.5 \\ 0 & 0 & 0.3 & 0.5-\lambda \end{matrix} \right] \right)=0$$

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.

• this is a block matrix. find the eigenvalues/vectors of the $2\times 2$ matrices. then compose to find the eigenvalues/vector of the full matrix. – abel Apr 20 '15 at 4:11

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.