If W is a subspace of V, show orthogonal complement of W is also a subspace of V $W$ is a subspace of the vector space $V$. Show that $W^{\perp}$ is also a subspace of $V$.
 A: You need to show three things:


*

*$W^\perp$ is non-empty.

*$W^\perp$ is closed under scalar multiplication; that is, if ${\bf
    v}\in W^\perp$, then  $\alpha{\bf v}\in W^\perp$ for all scalars
$\alpha$.

*$W^\perp$ is closed under vector addition; that is, if ${\bf
        v_1}\in W^\perp$ and  ${\bf v_2}\in W^\perp$, then  ${\bf v_1}+{\bf
        v}_2\in W^\perp$.

Recall that $\bf v$ is in $W^\perp$ if and only if ${\bf v}\cdot {\bf w}=0$ for all ${\bf w}\in W$.

Towards showing 1) holds, note (and verify) that  the zero vector is in $W^\perp$. 
Towards showing  3) holds, suppose that ${\bf v}_1$ and ${\bf v}_2$ are both  in $W^\perp$.   We have to show that the vector ${\bf v}_1+{\bf v}_2$ is in $W^\perp$; so we need to verify that $({\bf v}_1+{\bf v}_2)\cdot {\bf w}=0$ for all ${\bf w}\in W$.  Towards this end, use the fact that $({\bf v}_1+{\bf v}_2)\cdot {\bf w}=  {\bf v}_1\cdot {\bf w}+{\bf v}_2\cdot {\bf w} $.
I'll leave the verification that 2) holds, and the rest of the verification that 3) holds for you.
