Find $v>0$ so that $Av>0$

Question

Let \begin{align} A=\frac{1}{h^2}\begin{pmatrix} 1 & 0 & 0 & &\cdots & 0\\ -1 & 2 & -1 & 0 & \cdots & 0 \\ 0 & -1 & 2 & -1 & \cdots & 0 \\ 0 & 0 & \ddots & \ddots & \ddots & 0 \\ 0 & \cdots & 0 & -1 & 2 & -1 \\ 0 & \cdots & 0 & 0 & -1 & 1 \end{pmatrix} \in \mathbb R^{(n+1)\times (n+1)} \end{align} with $h=\frac{1}{n}$. I am looking for a $v\in \mathbb R^{n+1}$, $v_i>0$ so that $(Av)_i>0$. for all $i=1,\dots,n+1$. I know that there has to be such $v$ because $A$ is inverse positive, but i havent found a vector. Thanks for help.

Answer 1

The vector $v\in\Bbb R^{n+1}$ defined by $v_i=\sqrt i$ (for $i=1,\dots,n+1$) is a solution.