# Let W = span \{v_1, v_2,...,v_n\}. Show that if x is orthogonal to each v_j, for 1 \leq j \leq ...

## Question:

Let {eq}W = span \{v_1, v_2,...,v_n\} {/eq}. Show that if {eq}x {/eq} is orthogonal to each {eq}v_j {/eq}, for {eq}1 \leq j \leq p {/eq}, then {eq}x {/eq} is orthogonal to every vector in {eq}W. {/eq}

## Orthogonal Vectors

First we have a set W which is a span of a finite set of vectors. That would imply that any element of W can be expressed as a linear combination of the elements or vectors of the finite set of vectors. Then we have a vector x that is orthogonal to each element in the finite set of vectors. The question asks us to prove or show that this vector x is orthogonal to any element in W. The concepts used to show orthogonality include inner products of vectors from the area of Linear Algebra in Mathematics. The concept of orthogonality in an abstract vector space is akin to the concept of two vectors being perpendicular in a two or three dimensional setting.

Since {eq}x {/eq} is orthogonal to each {eq}v_j \; \colon \; j=1,\cdots,n, {/eq} therefore

{eq}\left\langle x, v_j \right\rangle = 0 \; \forall \; j=1,\cdots,n \qquad (1) {/eq}

Then let

{eq}w \in W = span \left\{ v_1, v_2,\cdots,v_n \right\} \implies w = k_1 v_1 + k_2 v_2 + \cdots + k_n v_n \quad {\rm where} \quad k_1,k_2,\cdots,k_n \in \mathbb{R} \qquad (2) {/eq}

From (2)

{eq}\left\langle x, w \right\rangle = \left\langle x, k_1v_1+k_2v_2+\cdots+k_nv_n \right\rangle = k_1 \left\langle x, v_1 \right\rangle + k_2 \left\langle x, v_2 \right\rangle + \cdots + k_n \left\langle x, v_n \right\rangle \qquad (3) {/eq}

Using the orthogonality property from (1) in (3) leads us to

{eq}\left\langle x, w \right\rangle = \left\langle x, k_1v_1+k_2v_2+\cdots+k_nv_n \right\rangle = k_1 \left\langle x, v_1 \right\rangle + k_2 \left\langle x, v_2 \right\rangle + \cdots + k_n \left\langle x, v_n \right\rangle = k_1 (0)+k_2(0)+\cdots+k_n(0)=0 \qquad (4) {/eq}

From (4) above {eq}x {/eq} is orthogonal to {eq}w. {/eq}

Since {eq}w \in W {/eq} is arbitrary, therefore {eq}x {/eq} is orthogonal to every vector in the set {eq}W. {/eq} Hence the proof. 