Suppose $\{v_1,v_2,...v_k\}$ spans $\mathbb{R}^n$, and that $w\in \mathbb{R}^n$ is orthogonal to $v_i$ for every $i=1,2,...,k$. Prove that $w=0$
My proof:
Let $A=\{v_1,v_2,...v_k\}$
$A$ spans $\mathbb{R}^n$. Then,
$\forall x\in \mathbb{R}^n$ , $x= \Sigma_{i=1}^ka_iv_i$
Given $w$ is orthogonal to $v_i$, then, $w*v_i=0$
Consider $x\in \mathbb{R}^n$,
$w*x=w*\Sigma_{i=1}^ka_iv_i=\Sigma_{i=1}^ka_i(wv_i)=\Sigma_{i=1}^ka_i(0)=0$
Then, $\forall x\in \mathbb{R}^n,w*x=0$
Since $w\in \mathbb{R}^n$, we can take $x=w$
Then, $w*w=0$
Then, $w=0$
Does this make sense?
Your proof is fine.
Just minor things:
Perhaps add some quantifiers:
$$\forall x \in \mathbb{R}^n, \exists a_i \in \mathbb{R}, i \in \{ 1, \ldots k\}, x = \sum_{i=1}^k a_iv_i$$
More common use of notations for inner product:$w^Tx$ or $\langle w, x \rangle$ or $w\cdot x$.