$\exists t\in \mathbb{R}, t>0$ such that $v_t = x - ty \ge 0$

The following fact is something I have seen being used in proofs of the Perron-Frobenius theorem:

Let $x,y \in \mathbb{R}^n, x>0, y\neq 0$. Then $\exists t\in \mathbb{R}, t>0$ such that $v_t = x - ty \ge 0$ where $v_t$ has atleast one entry equal to $0$.

I do not manage to prove this fact on my own. Any tips?

Clarification: If $x = (x_1, x_2, ..., x_n)$, by $x>0$ I mean that $\forall i, x_i > 0$. $x\geq 0$ is defined likewise. When I say $x$ has at least one entry equal to 0, I mean $\exists i$, such that $x_i = 0$.

Solution 1:

Since $t>0$ the components with $y_i \le 0$ are "no problem". We now look at the differences $x_i-ty_i$ for all the components $i$ such that $y_i >0$. Those are only finitely many components. Choose $t_i$ such that $x_i -t_iy_i=0$ which is possible by setting $t_i = \frac{x_i}{y_i}$. Then we have $x_i-ty_i >0$ for all $t < t_i$. If we then take $t_0= \min{\{t_i: i \text{ such that } y_i>0\}}$ We have $x-ty \ge 0$ and there is one component $=0$ - namely the one with minimal $t_i$.

However, as MXXZ pointed out, this all only works if there is a component $y_i>0$. If $y \le 0$ the components of $x-ty$ are all bigger or equal to the components of $x$ and we don't necessarily have a component $=0$.