Let $W_1$ and $W_2$ be subspaces of a finite dimensional inner product space space. Prove that $$(W_1 \cap W_2)^\perp=W_1^\perp + W_2^\perp $$
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One direction is easy : Let $\alpha \neq 0$ and $\alpha \in W_1^\perp + W_2^\perp$, i.e. $\alpha$ can be written as $\alpha = \beta + \gamma$ such that $\beta \in W_1^\perp$ and $\gamma \in W_2^\perp$, hence $(\beta|\eta)=0$ for all $\eta \in W_1 $ and $(\gamma|\delta)=0$ for all $\delta \in W_2 $. Now for all $\eta \in W_1 \cap W_2$ it is clear that $(\alpha | \eta) =0$ . hence $$(W_1 \cap W_2)^\perp\supset W_1^\perp + W_2^\perp $$ For proving the other containment, let $\alpha \neq 0$ and $\alpha \in (W_1 \cap W_2)^\perp$, it means that for all $\beta \in W_1 \cap W_2$, $(\alpha|\beta)=0$. Hence $\alpha \in V \setminus (W_1 \cap W_2) = V \setminus (W_1) \cup V \setminus (W_2)$. Hence $\alpha \in W_1^c$ or $\alpha \in W_2^c$. WLOG suppose $\alpha \in W_1^c$. We also have that $$V=(W_1 ) \oplus (W_1 )^\perp$$therefore $\alpha = \eta + \delta $ where $\eta \in W_1$ and $\delta \in W_1^\perp$ and indeed $\delta \neq 0$. From here I have to somehow show that $\eta=0$, but I am stuck in here...
I already appreciate any help
Some relevant facts are (1) that if $W$ is a closed subspace, then $W^{\bot \bot} = W$, and (2) if $W \subset X$, then $X^\bot \subset W^\bot$, and (3) finite dimensional subspaces are always closed (and so (1) applies).
You have shown one direction, you wish to show that $(W_1 \cap W_2)^\bot\subset W_1^\bot + W_2^\bot$. Because of the above facts, this is equivalent to showing $W_1 \cap W_2 \supset (W_1^\bot + W_2^\bot)^\bot$.
So, suppose that $x \in (W_1^\bot + W_2^\bot)^\bot$. This means that $\langle x,w_1'+w_2'\rangle = 0$ whenever $w_k' \in W_k^\bot$, $k=1,2$.
In particular, we have $\langle x,w_1'\rangle = 0$ for all $w_1' \in W_1^\bot$, and so $x \in W_1^{\bot\bot} = W_1$. Similarly, we have $x \in W_2$. And so, $x \in W_1 \cap W_2$.