Elementary Linear Algebra 7th Edition

Since for any $w\in W$, we have $$\langle 0,w\rangle =0$$ then $0\in W^\perp$ and $W$ in nonempty set. Now, let $v_1,v_2\in W^\perp$, then for any $w\in W$, we get \begin{aligned} \langle v_1+v_2, w\rangle &=\langle v_1,w\rangle +\langle v_2,w\rangle=0 \end{aligned} hence, $v_1+v_2\in W^\perp$ which means that $W^\perp$ is closed under addition. For any $v\in W^\perp$, $w\in W$ and $k\in R$, we have $$\langle kv, w\rangle =k\langle v, w\rangle=0.$$ Hence, $W^\perp$ is closed scalar multiplication. Consequently, $W^\perp$ is a subspace of $W$.