# Help using eigenvectors to solve Markov chain

I took Linear Algebra last semester and when learning about Markov Chains in my statistics class, I wanted to use eigenvectors/eigenvalues to find the steady-state vector rather than just using systems of equations like our professor taught us. I seem to be having a bit of trouble, however. Here's an example:

Let's say we have a transition matrix P:

$$P = \begin{bmatrix} 0 & 0.5 & 0.5 \\ 0.5 & 0 & 0.5 \\ 1 & 0 & 0 \\ \end{bmatrix}$$

We know that one of the eigenvalues is 1, since it is a Markov chain. The other eigenvalues can be found via $$det(P - \lambda I)$$:

$$\lambda_1 = 1$$ $$\lambda_2 = -0.5$$

Thus making the eigenvectors:

$$x_1 = \begin{bmatrix} 1 \\ 1 \\ 1 \\ \end{bmatrix}$$ $$x_2 = \begin{bmatrix} -0.5 \\ -0.5 \\ 1 \\ \end{bmatrix}$$

This is where I always get stuck. In another example, I simply normalized $$x_1$$ to get $$x_{ss} = \begin{bmatrix} \frac{1}{3} \\ \frac{1}{3} \\ \frac{1}{3} \\ \end{bmatrix}$$

for the long-term forecast/steady state vector. For this example however, a calculator revealed that the steady state vector was actually $$x_{ss} = \begin{bmatrix} \frac{4}{9} \\ \frac{2}{9} \\ \frac{3}{9} \\ \end{bmatrix}$$

How do I get there from the eigenvectors? Any help is much appreciated.

You’ve found right eigenvectors of $$P$$, but what you really need are the left eigenvectors: you’re trying to solve $$\pi P=\pi$$, not $$P\pi=\pi$$. Use whatever technique you used to compute the eigenvectors of $$P$$ on $$P^T$$ instead.
Incidentally, since the rows of a row-stochastic matrix all sum to $$1$$, the vector consisting of all $$1$$s will always be a right eigenvector.