For $B$ positive semi-definite matrix find $z\ge0$ with $Bz\ge0$

Question

Question. Given a positive semi-definite matrix $B$ does there exist a non-zero vector $z$ with all components non-negative such that all components of $Bz$ are non-negative?

Here are some details (which you probably do not need, if you understand what is asked above).

I posted an answer to the following question check for a compact set , and in my answer I used a claim which I believe ought to be true, but I do not know.

Suppose that $B$ is an $m\times m$ positive semi-definite matrix. If $y,z$ are (column) vectors with $m$ components each, define $z\ge y$ if $z_j\ge y_j$ for all $1\le j\le m$, and define $z\neq y$ if $z_j\neq y_j$ for at least one $1\le j\le m$.

Question. Does there exist $z\ge0$ with $z\neq0$ such that $Bz\ge0$ ? (That is, does there exist a non-zero vector $z$ with all components non-negative such that all components of $Bz$ are non-negative? Here $B$ is a positive semi-definite matrix, and $0$ is the zero vector with $m$-many components.)

If $Bz=0$ for some non-zero $z\ge0$ then we are done. If $Bz$ is non-zero whenever $z\ge0$ is non-zero, but if the angle between $z$ and $Bz$ is zero, for some such $z$, then again we are done. If the angle is always non-zero, then we may define a vector field on a suitable sub-space of surface and use a combing, or a fixed-point theorem I would think, except I didn't think hard enough how to do this. Something like, consider only $z\ge0$ with $||z||=1$ and assign to such $z$ the non-zero vector $\frac{Bz}{||Bz||}-z$ (or perhaps, better, assign $Bz-z$ or $\frac{Bz-z}{||Bz-z||}$.)

Or, perhaps it is just a matter of solving a certain system of equations, using the properties of the matrix $B$. (I do not know if $B$ is required to be symmetric in the linked question, it shouldn't matter, I think we could assume it if it helps.)

I am sure the answer to this question ought to be known, and I would be grateful for any details or a reference. Thank you!

I tend to believe the answer would come using methods of linear algebra and/or convex optimization, but I did come up with a topological (not purely topological, as it involves the standard metric) question which could be thought of as a generalization of the present one.
Do the closed unit disk $D$ and $f(D)$ intersect, if $||f(x)-x||\le2$ for all $x\in D$?

Answer 1

Let me show that $\{x: \ x\ge 0, \ Bx\ge0\}$ is unbounded.

Consider the linear programming problem $$ \min - e^Tx $$ subject to $$ Bx -z =0 $$ and $x\ge0$, $z\ge0$, and $e$ is the vector of all ones of suitable size.

The dual problem of the above problem is: $$ \max 0 $$ subject to $$ B^Ty \le -e , \ -y\le 0. $$ This set has no feasible point: $y\ne0$ is not allowed. And for $y\ge0$, $y^TBy\ge0>-e^Ty=\|y\|_1$.

By strong duality, the primal problem cannot have a solution. Since the feasible set of the primal problem is non-empty, the primal problem has to be unbounded. Hence, there is $x\ne0$, $x\ge0$, $Bx\ge0$.

The matrix $B$ needs not to be symmetric.

Answer 2

Another try using Farkas lemma: Either there is $x$ such that $$x \ge 0, \ Bx \ge 0, \ e^Tx>0$$ or there is $y_1,y_2$ such that $$ y_1 + B^Ty_2 =-e , \ y_1 \ge0, y_2 \ge0. $$ (Apply the lemma in the notation of the wiki page with $A=\pmatrix{ I & B^T}$ and $x,y$ exchanged). Here, I use $e$ to be the vector of all ones.

Assume the latter system is solvable for $y_1,y_2$. Clearly $(y_1,y_2)\ne 0$. Multiplying the equation by $y_2^T$ yields $$ y_2^Ty_1 + y_2^TBy_2 = -e^Ty_2 = -\|y_2\|_1. $$ The left-hand side is non-negative, the right-hand side is non-positive. Hence both sides have to be zero, so $y_2=0$. Then $y_1=-e$, which is a contradiction to $y_1\ge0$. Hence, the second system is unsolvable, so the first system is solvable: there is $x\ge0$, $x\ne0$, with $Bx\ge0$.