Let $E$ be subset of a vector space $V$. Let $B =\{w_1,\dots,w_k\}$ be a basis for $E$. Prove: $E^\perp =\{y\in V | y\perp w_i, 1\leq i \leq k \}$
Is my proof correct?
Define two sets:
(a) $E_1 = \{x\in V| \langle x,e \rangle=0 \forall e\in E\}$
(b) $E_2 =\{y\in V ~| ~y\perp w_i, 1\leq i \leq k \}$
I need to show that those two sets are equal. First, I show that $E_1\subseteq E_2$
$v\in E_1 \implies \langle v,e\rangle=0 \forall e$
since $w_1,\dots,w_k\in E$, then by definition $\langle v,w_i\rangle=0 \forall i$. Hence $v\in E_2$
Now, we need to show $E_2\subseteq E_1$:
$v\in E_2 \implies \langle v,w_i\rangle=0 \forall i$ hence, $$ c_1\langle v,w_1\rangle +\dots +c_k\langle v,w_k \rangle =0\\ \langle v,c_1 w_1+\dots c_k w_k\rangle =0\\ \langle v,e\rangle=0 \forall e\in E \implies v\in E_1 \implies E_1=E_2 $$