# calculating characteristic row-vector

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.

-
Stochastic matrices are usually defined so that entries in columns sum to $1$, not entries in rows. – alex.jordan Mar 8 '12 at 4:14
The definition of stochastic matrix that I was given is a matrix with all elements non-negative and each row sums to 1, with the property that for some power of teh matrix all elements are positive – jason nesmith Mar 8 '12 at 4:20
OK then in that case you are right-multiplying matrices to row vectors. It's more common to left-multiply matrices to column vectors. – alex.jordan Mar 8 '12 at 4:34
well if it wanted the characteristic column-vector then the left-multiplying would be used. Can you point me in the right direction as to how to start this ? – jason nesmith Mar 8 '12 at 4:44
The point of having entries in a column sum to $1$ (aka probability vectors for columns) is that if you are left-multiplying matrices to vectors (as is more often the case), then a stochastic matrix times a probability vector still works out to be a probability vector. If you have rows with entries adding to one, then the same ideas work, but now you must right-multiply the matrix by a transposed vector. – alex.jordan Mar 8 '12 at 4:53

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 $\begin{bmatrix}10/31 & 11/31 & 10/31\end{bmatrix}$.

-
why did you add in the last row, I thought they had to be square matrices for this to work. And I had derived the matrix by subtracting the identity matrix but when I reduced it i got: 1 0 -1 0 1 -1 0 0 0 How did you derive the solution from the matrix with the additional column? – jason nesmith Mar 8 '12 at 5:23
The matrix I have represents 3 equations in 3 variables. The variables are the entries $a$, $b$, and $c$ for the vector that you are searching for. For example the first column represents $-.8a+.5b+.25c = 0$. Since your setup involves rows that sum to $1$, we need to column-reduce the matrix, not row-reduce it. It looks like you row-reduced. For either of these procedures, square matrices are not necessary. The last solution again has been found through column-reduction. The whole thing might just be more consistent with what you know if you transposed the matrix and used row-reduction. – alex.jordan Mar 8 '12 at 18:33
Ok doing row reduction with the transpose i get: 1 0 -1 0 1 -11/10 0 0 0 – jason nesmith Mar 8 '12 at 21:55
@jason I think you made an arithmetic mistake. The $-1$ that you have should be $-9/8$. Also, here is how you can enter matrices in $\mathrm{\LaTeX}$, which is how math is entered on this site: $\begin{bmatrix} 1&0&1 \\ 0&1&-11/10 \\ 0&0&0 \end{bmatrix}$ produces $\begin{bmatrix} 1&0&1 \\ 0&1&-11/10 \\ 0&0&0 \end{bmatrix}$. – alex.jordan Mar 9 '12 at 0:42
My mistake! I was the one with the arithmetic mistake, not you. I'm editing my answer accordingly. – alex.jordan Mar 9 '12 at 0:49