Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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.

share|improve this question
    
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

1 Answer 1

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

share|improve this answer
    
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

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.