This exercise appears in Section 6.B "Orthonormal Bases" in Linear Algebra Done Right by Sheldon Axler. Inner product spaces, norms, orthogonality, and orthonormal bases have been introduced.
The unofficial solution manual I am using presents a somewhat involved proof of this exercise that involves induction and the Gram-Schmidt procedure, which I will not reproduce here. However, I arrived at an alternative proof that is much simpler in my opinion. I would like to check if my proof is correct.
$(V, \langle \cdot, \cdot \rangle)$ is an inner product space over the field $\mathbb{F}$, which stands for either $\mathbb{R}$ or $\mathbb{C}$. (Axler, pp. 4, 167)
- Suppose $v_1, \dotsc, v_m$ is a linearly independent list in $V$. Show that there exists $w \in V$ such that $\langle w, v_j \rangle > 0$ for all $j \in \{1, \dotsc, m\}$. (Axler, p. 191)
Proof. Let $U = \operatorname{span}(v_1, \dotsc, v_m)$. Consider the linear map $\phi : U \to \mathbb{F}^m$ defined by $$ \phi(u) = (\langle u, v_1 \rangle, \dotsc, \langle u, v_m \rangle). $$ We show that $\phi$ is injective. Suppose that $u \in U$ and $\phi(u) = 0$. Then, $$ \langle u, v_1 \rangle = \dotsb = \langle u, v_m \rangle = 0. $$ Since $u \in \operatorname{span}(v_1, \dotsc, v_m)$, we have $\langle u, u \rangle = 0$, so $u = 0$. Hence, $\phi$ is injective. But $\dim U = \dim \mathbb{F}^m = m$, so $\phi$ is surjective. Choose $w \in U$ such that $$ \phi(w) = (1, \dotsc, 1). $$ Therefore, $\langle w, v_j \rangle = 1 > 0$ for all $j \in \{1, \dotsc, m\}$.