If $v_1, \dots, v_m$ are linearly independent, then there is $w$ such that $\langle w, v_j \rangle > 0$ for all $j$ 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 $v_j$?
 A: Consider a linear functional $\phi$ from $span(v_1,…,v_m)$ to scalar field $\mathbb F$  (I assume $\mathbb R$ or $\mathbb C$) given by formula $$\phi(a_1v_1+\cdots+a_mv_m)=a_1+\cdots+a_m.$$ Clearly $\phi(v_j)=1$ for $j \in \{1,…,m\}$. By the Riesz Representation Theorem there is a vector $w \in span(v_1,…,v_m) \subset V$ such that $\phi(v)=\langle v,w \rangle$ for every $v \in span(v_1,…,v_m)$. Thus $\langle v_j,w \rangle = \phi(v_j) = 1 >0$ for all $j \in \{1,…,m\}$.
A: 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 $a > \frac{\|w_{n-1}\|}{\|v_n\|}$.
A: 
If v1,…,vm are linearly independent, then there is w such that ⟨w,vj⟩ > 0 for all j
