Given $V$ a inner product space and $W_1$, $W_2$ subspaces of $V$

Show $(W_1 + W_2 )^\perp = W_1^\perp \cap W_2^\perp$ and $W_1^\perp + W_2^\perp ⊆ (W_1 \cap W_2 )^\perp$.


1 Answer 1


1) We prove that $(W_1+W_2)^\perp=W_1^\perp\cap W_2^\perp$.

Any $v\in (W_1+W_2)^\perp$ is such that $\langle v,w_1+w_2\rangle=0$ for all $w_1\in W_1$ and $w_2\in W_2$. In particular $v\in W_1^\perp$ (choosing $w_2=0$) and similarly $v\in W_2^\perp$ (choosing $w_1=0$). We conclude that $v\in W_1^\perp\cap W_2^\perp$.

The converse is just as easy. Any $v\in W_1^\perp\cap W_2^\perp$ satisfies $\langle v,w_i\rangle=0$ for all $w_i\in W_i$ and $i=1,2$. This implies $\langle v,w_1+w_2\rangle=\langle v,w_1\rangle+\langle v,w_2\rangle=0$ for all $w_1+w_2\in W_1+W_2$, that is $v\in (W_1+W_2)^\perp$.

2) As well, we show $W_1^\perp + W_2^\perp ⊆ (W_1 \cap W_2 )^\perp$.

For if, let $v\in W_1^\perp + W_2^\perp$ be the sum of $v_1\in W_1^\perp$ and $v_2\in W_2^\perp$ and let $w\in W_1\cap W_2$.

One computes $\langle v,w\rangle=\langle v_1+v_2,w\rangle=\underbrace{\langle v_1,w\rangle}_{=0}+\underbrace{\langle v_2,w\rangle}_{=0 }=0$. We conclude $v\in (W_1 \cap W_2 )^\perp$.


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