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..
$\endgroup$3 Answers
$\begingroup$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.)
$\endgroup$ $\begingroup$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.
$\endgroup$ $\begingroup$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?
$\endgroup$ 2
