Invariant probability vector as a left eigenvector

Question

What is the probability in the long run that the chain is in state 1? Solve this by directly computing the invariant probability vector as a left eigenvector.

\begin{bmatrix} .4 &.2 &.4 \\ .6 &0& .4 \\ .2 &.5& .3 \end{bmatrix}

Please note that this is a 3x3 probability matrix. Specifically I need help with the linear algebra aspect of this. Thanks!

Answer 1

Hint: The invariant probability vector is a left-eigenvector associated with the maximal positive eigenvalue, which in this case is $\lambda = 1$.

Your answer should be:

$\frac{25}{66} \approx 0.38$