1
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ Fixed. Thank you $\endgroup$ – Anson Pang Oct 21 '19 at 2:16
0
$\begingroup$

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 $v_{n+1}:=\mathrm{perp}_W(v)$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.