1
$\begingroup$

Here is the question:

enter image description here

I set up the probability matrix as: \begin{bmatrix} 0 & 1/3 & 1/3 & 1/3 \\ 0 & 1/2 & 1/2 & 0 \\ 1 & 0 & 0 & 0 \\ 1/2 & 0 & 1/2 & 0 \\ \end{bmatrix}

Then, $1$ is an eigenvalue. I calculated the eigenvector for $1$ as $(1,1,1,1)$. So is the answer $(1/4,1/4,1/4,1/4)$ ?

$\endgroup$
  • 5
    $\begingroup$ Instead of looking for an eigenvector of $P$, you should be looking for an eigenvector of $P^T$ (that is, a left-eigenvector of $P$). You should calculate this eigenvector to be $(6,4,5,2)$. $\endgroup$ – Omnomnomnom Aug 13 at 18:55

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.