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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.

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1 Answer 1

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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.

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  • $\begingroup$ Thank you so so much!! This was the key! $\endgroup$
    – amstrudy
    Mar 12, 2019 at 7:09

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