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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?

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    $\begingroup$ Your proof looks OK to me! Cheers! $\endgroup$ – Robert Lewis Apr 6 '18 at 1:30
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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 $w\cdot x$.

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