6

If I have a stochastic matrix $X$- the sum of each row is equal to $1$ and all elements are non-negative.

Given this property, how can I show that:

$x'X=x'$ , $x\geq 0$

Has a non-zero solution?

I'm assuming this has something to do with proving a feasible dual, but not may be wrong..

Martin Sleziak
  • 56,060
GBa
  • 1,076

3 Answers3

7

Here's the linear programming approach (with considerable help from another user, Fanfan). Consider the LP

$$\min 0^T y$$ subject to $$(X - I)y \geq 1,$$ where $0$ and $1$ are, respectively, vectors containing all 0's and all 1's.

By Fanfan's answer to my question this LP is infeasible. Thus its dual, $$\max 1^T x$$ subject to $$x^T (X-I) = 0^T,$$ $$x \geq 0,$$ is either infeasible or unbounded. But $x = 0$ is a solution, and so this dual problem is feasible. Thus it must be unbounded. But that means there must be a nonzero solution $x_1$ in its feasible region. Thus we have $x_1^T X = x_1^T$ with $x_1 \geq 0$, $x_1 \neq 0$.

(Fanfan's answer to my question also includes another answer to your question - one that uses Farkas's Lemma rather than LP duality. It ends up being quite similar to my answer here, as, of course, Farkas's Lemma and LP duality are basically equivalent.)

Community
  • 1
Mike Spivey
  • 56,818
3

If your matrix $X$ is $n\times n$, then $X$ is the transition matrix of a Markov chain with state space $\lbrace 1 , 2 , \dots , n \rbrace$. The vector $x$ that you are looking for is a stationary measure for the Markov chain.

The existence of such a non-zero $x$ can be proven using probability theory, but a purely linear approach falls under Perron-Frobenius theory.

For instance, the averages ${1\over n}\sum_{k=1}^n X^k$ of the powers of $X$ converge to a matrix $M$ (this statement needs proof!). Then, it is easy to see that any row of $M$ can serve as $x'$ and solves $x'X=x'$. Note that the row sums of $M$ all equal 1, so $x\neq 0$.

As you suggest, maybe there is also an approach via linear programming.

1

Here is a hint for you:

Take $v = (1, 1, 1, \dots, 1)'$, the vector with only ones. By definition of your Matrix, you have $Xv = v$. Thus, $v$ is an eigenvector with eigenvalue 1.

Now change the first basis vector of your vector space to $v$ and keep the other unchanged. In this new basis, what will $X$ looks like?

Djaian
  • 582
  • So you're saying to replace Row 1 of A with v? Then, when looking at x'X we'll have sum of all x as the resulting first row, I'm still now sure how this fits in.. – GBa Nov 02 '10 at 17:11
  • @Mark: Keep the same notation please. The matrix is $X$, not $A$. I don't suggest to replace the first row with vector $v$. A square matrix is always given in a vector space relatively to a basis. Each column represent the result of the linear map expressed as a combination of the basis vector. Since $Xv = v$, this would imply that the first column of the matrix would be $(1, 0, 0, \dots, 0)$ if you take $v$ as the first vector of the basis. – Djaian Nov 02 '10 at 22:38