Existance of an eigenvector $x$ such that $Ax\le x$ could anyone help me with the following problem?
$A$ is positive semidefinite with all entries $\ge 0$, with spectral radius $\rho(A)<1$, then does there exists $x>0$ with the property $Ax\le x$?
Thanks for helping
 A: $\textbf{ Your claim is true for symmetric matrices.}$
Suppose the eigenvectors of $A$ are $v_1, \ldots, v_n$ with eigenvalues $0 \leq \lambda_1, \ldots, \lambda_n < 1$. Then we have $A ( \sum \mu_i v_i ) = \sum \mu_i \lambda_i v_i$. 
Consider the subset $C = \{ (\mu_1, \ldots, \mu_n) ~|~ \sum \mu_i v_i \geq 0\}$. Note that this is cut out by $n$ closed half spaces, each defined by a hyperplane whose normal is one of the rows of the matrix $P$ whose columns are the eigenvectors $v_i$. It is easy to see that $C$ is a convex cone.
Now $Ax \leq x$ is equivalent to $\sum \mu_i (1 - \lambda_i) v_i \geq 0$. 
We can restate the claim as follows: we want to find $(\mu_1, \ldots, \mu_n) \in C$ such that $(\mu_1 (1-\lambda_1), \ldots, \mu_n (1 -\lambda_n) \in C$. If we define $((1-\lambda_1), \ldots, (1-\lambda_n))\cdot C$ to be the cone obtained by scaling every element of $C$ entrywise by the indicated scalar, then this claim is the same as saying that $((1-\lambda_1), \ldots, (1-\lambda_n))\cdot C \cap C \neq \emptyset$
Now if we don't restrict our PSD to be nonsymmetric, then our eigenvectors aren't orthogonal, so we don't necessarily have a solution to this. For example our convex cone could live in $\mathbb{R}^2$ cut out by the lines $y = x$ and $y = 2x$. Then if $\lambda_1 = 0$ and $\lambda_2 = .75$ we see $C \cap (1,\frac{1}{4})\cdot C = \emptyset$
However, if we restrict our PSD to be symmetric (which I assume you meant), we get orthogonal vectors, and hence our cone is just the image of the positive orthant under some rotation. If our rotated orthant contains one of the axes, WLOG assume $(1,0,0,\ldots, 0) \in C$. Then we have $(\mu_1 (1-\lambda_1), \ldots, \mu_n (1 -\lambda_n)) = ((1-\lambda_1), 0 , 0 , \ldots, 0) \in C$, as desired.
Otherwise, note that our cone $C$ is the convex cone generated by $\sigma_i v_i$ for some choice of sign $\sigma_i$ for each eigenvector, since its the intersection of a bunch of half spaces whose normals are the rows of our matrix, and our matrix is symmetric. In order for $((1-\lambda_1), \ldots, (1-\lambda_n))\cdot C \cap C = \emptyset$, we would need the all the images of $\sigma_i v_i$ under scaling by $(1-\lambda_i)$ on each coordinate to lie one the other side of some given defining hyperplane of $C$. However, $\sigma_i v_i$ and its image lie on the same side of the hyperplane that $v_i$ is normal too since the hyperplane contains the whole octant the normal is in, and the scaling by positive scalars doesnt take $\sigma_i v_i$ out of its octant.
