# Perpendicular of subset within orthogonal complement

I'm currently studying general projections over inner product spaces, and encounter this following generalization.

If $$\mathbb{W}$$ is a $$k$$ dimensional vector subspace of an inner product space $$\mathbb{V}$$, then for any $$\vec{v}\in\mathbb{V}$$ we have $$\mathrm{perp}_\mathbb{W}(\vec{v})=\vec{v}-\mathrm{proj}_\mathbb{W}(\vec{v})\in\mathbb{W}^\bot$$

My book proves this by making the expanding $$\mathrm{proj}_\mathbb{W}(\vec{v})\in\mathbb{W}$$ as

$$\mathrm{perp}_\mathbb{W}(\vec{v})=\vec{v}-\frac{\langle\vec{v},\vec{v_1}\rangle}{\Vert\vec{v_1}\Vert^2}\vec{v_1}-...-\frac{\langle\vec{v},\vec{v_k}\rangle}{\Vert\vec{v_k}\Vert^2}\vec{v_k}$$

, where $$\{\vec{v_1},...,\vec{v_k}\}$$ is an orthogonal basis that spans $$\mathbb{W}$$, and then using using the Gram-Schmidt Orthogonalization Theorem to deduce that $$\{\vec{v_1},...,\vec{v_k},\mathrm{perp}_\mathbb{W}(\vec{v})\}$$ is an orthogonal basis and hence $$\mathrm{perp}_\mathbb{W}(\vec{v})$$ being in the orthogonal complement of $$\mathbb{W}$$ under $$\langle,\rangle$$.

What I don't understand is the usage of the GSOT to prove orthogonality of $$\vec{v}$$, since $$\{\vec{v_1},...,\vec{v_k}\}$$ was not constructed from a basis $$\{\vec{w_1},...,\vec{w_k},\vec{w_{k+1}}\}$$, yet the proof says that any vector $$\vec{v}\in\mathbb{V}$$ to replace $$\vec{w_{k+1}}$$ would instantly make $$\mathrm{perp}_\mathbb{W}(\vec{v})$$ orthogonal.

• Fixed. Thank you – Anson Pang Oct 21 '19 at 2:16

Well, it's not the whole Gram-Schmidt process used here, only its computation part, which shows that the given formula for $$\mathrm{perp}_W(v)$$ yields indeed a vector orthogonal to each $$v_i$$.
Nevertheless, this can be thought of as being in the middle of the Gram-Schmidt process, where $$v_1,\dots, v_n$$ orthogonal vectors are already chosen (spanning $$W$$), and then we are given a next candidate vector $$v$$ (presumably $$v\notin W$$), and make it orthogonal by applying $$v_{n+1}:=\mathrm{perp}_W(v)$$.