Proof of The Orthogonal Complement of a Linear Subspace is a Linear Subspace

Conditions are numbered as in Linear Subspace, and the conditions on an inner product as in Complex Inner Product Space. Let $w\in W$ be arbitrary.

Condition 1. By claim 3 of Elementary Identities in a Vector Space we have $0\cdot 0_{V}=0_{V}$, so condition 3 of the inner product gives $\langle w,0_{V}\rangle=\langle w,0\cdot 0_{V}\rangle=0\,\langle w,0_{V}\rangle=0$. Hence $0_{V}\in W^{\perp}$.

Condition 2. If $x,y\in W^{\perp}$, then condition 2 of the inner product gives $\langle w,x+y\rangle=\langle w,x\rangle+\langle w,y\rangle=0+0=0$, so $x+y\in W^{\perp}$.

Condition 3. If $\lambda$ is a complex number and $x\in W^{\perp}$, then condition 3 of the inner product gives $\langle w,\lambda x\rangle=\lambda\langle w,x\rangle=\lambda\cdot 0=0$, so $\lambda x\in W^{\perp}$.

As $w\in W$ was arbitrary in each case, $W^{\perp}$ satisfies all three conditions and is a linear subspace of $V$.