Suppose $v_1, \dots v_m$ is a linearly independent list in $V$. Show that there exists $w \in V$ such that $\langle w, v_j \rangle > 0$ for all $j \in {1, \dots ,m}$.
I understand this question is saying given a linearly independent list, there is $w \in V$ such that the vector $w$ is not orthogonal to any $v$ in that linearly independent set. I'm also confused as to why it is significant that the inner product be greater than zero and instead of just $\neq 0$. Can someone give me a hint on how to do this problem?
I know that $\langle v, v \rangle >0$ for all $v$ not equal to zero, and since $v_1, \dots v_m$ is linearly independent, then none of the $v_j$ will be zero, but it is impossible to have w equal to all $v_j$?
Hint: suppose that $w_{n-1}$ is such that $\langle w, v_j \rangle > 0$ for $j = 1,\dots,n-1$. Let $S_{n-1}$ be the span of $v_1,\dots,v_{n-1}$. Suppose furthermore (without loss of generality) that $w \in S_{n-1}$.
Let $v^\perp$ be the component of $v_n$ perpendicular to $S_{n-1}$. Let $w_{n} = w_{n-1} + a v^\perp$ for some constant $a>0$. We may select $a>0$ so that $w_n$ satisfies $\langle w, v_j \rangle > 0$ for $j = 1,\dots,n$. Note that $w_n \in S_n$, the span of $v_1,\dots,v_n$.
In particular, for $i \leq n-1$, we have $$ \langle w_n, v_i \rangle = \langle w_{n-1}, v_i \rangle + a\langle v^\perp, v_i \rangle = \langle w_{n-1}, v_i \rangle $$ and then $$ \langle w_n, v_n\rangle = \langle w_{n-1}, v_n \rangle + a\langle v^\perp, v_n \rangle = \langle w_{n-1}, v_n \rangle + a\|v^\perp\|^2 \\ \geq a \|v_n\|^2-\|w_{n-1}\| \|v_n\| = (a \|v_n\|-\|w_{n-1}\|) \|v_n\| $$ So, in particular, it suffices to set $a > \frac{\|w_{n-1}\|}{\|v_n\|}$.