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I am fairly new to matrices, especially stochastic matrices. In an effort to become more comfortable with them I am doing working out some problems. One of them that is giving me a hard time is to calculate the characteristic row-vector associated with eigenvalue 1 of the following matrix:

$\begin{bmatrix} .2 & .6 & .2\\ .5 & 0 & .5\\ .25& .5 &.25 \end{bmatrix}$

I am not quite sure how to start this problem out. I have tried reducing the matrix down as well as the transpose of the matrix but neither has proved fruitful so far.

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    Here's a nice tutorial [in PDF](http://www.scss.tcd.ie/Rozenn.Dahyot/CS1BA1/SolutionEigen.pdf) on how to compute *right* eigenvalues and eigenvectors. In your case, you're computing *left* eigenvector corresponding to left eigenvalue $\lambda = 1.$2012-03-08

1 Answers 1

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If $A$ is your matrix and $\vec{v}^t$ is a row vector such that $\vec{v}^tA=\vec{v}^t$ then $\vec{v}^t(A-I)=\vec{0}^t$ We can solve for $\vec{v}^t$ by column reducing the corresponding augmented matrix $\begin{bmatrix} -.8 & .6 & .2\\ .5 & -1 & .5\\ .25 & .5 & -.75\\ 0 & 0 & 0 \end{bmatrix}\to \begin{bmatrix} 1 & 0 & 0\\ 0 & 1 & 0\\ -1 & -11/10 & 0\\ 0 & 0 & 0 \end{bmatrix}$

EDIT: Formerly there was an arithmetic error in this reduction.

So the solution space is the span of $\begin{bmatrix}1,11/10,1\end{bmatrix}$.If you would like to rescale so that entries sum to $1$, then you can add a column to the matrix above representing the condition that the entries sum to $1$:

$\begin{bmatrix} 1 & 0 & 0&1\\ 0 & 1 & 0&1\\ -1 & -11/10 & 0&1\\ 0 & 0 & 0 &1 \end{bmatrix}\to \begin{bmatrix} 1 & 0 & 0 & 0\\ 0 & 1 & 0 & 0\\ 0 & 0 & 1 & 0\\ 10/31 & 11/31 & 10/31 & 0 \end{bmatrix}$

So the solution with entries summing to one is $\begin{bmatrix}10/31 & 11/31 & 10/31\end{bmatrix}$.

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    My mistake! I was the one with the arithmetic mistake, not you. I'm editing my answer accordingly.2012-03-09