# Linear Algebra Vector Proof

1. Oct 11, 2008

### kehler

1. The problem statement, all variables and given/known data
Prove that a vector k is in 'W perp' if and only if k is orthogonal to every vector in the spanning set of W, where W is a subspace of Rn

3. The attempt at a solution
It's so obviously true that I don't know how to prove it! :S

Here's what I did:
Let {w1, w2SUB], .... wm} be a spanning for W.
Let w be a vector in W where
w = c1w1 + c2w2 + ... + cmwm
and all the weights c1, c2,.....,cm are nonzero
Suppose k is in 'W perp' but is not orthogonal to every vector in the spanning set.
Then k.w
= k. (c1w1 + c2w2 + ... + cmwm)
= c1k.w1 + c2k.w2 +... + cmk.wm
= a, where a is a non-zero number
But to be in 'W perp', the dot product of k and any vector in W must be zero
Therefore, k has to be orthogonal to every vector in the spanning set of W to be in 'W perp"

Is this right?? I can't think of any other way to do it :S

Last edited: Oct 11, 2008
2. Oct 11, 2008

### HallsofIvy

Staff Emeritus
Your definition of 'Wperp' is the space of all vector orthogonal to every member of W, right? And you are given that a vector is orthogonal to every member of a spanning set?

Do a direct solution, not indirect. If k.wi= 0, what is k.w?

3. Oct 11, 2008

### kehler

Yep 'W perp' (W perpendicular) is the set of all vectors orthogonal to W.
Oops I realised I made some typo errors. The 'x's in the last two lines were meant to be 'k's. I've edited it now.
What do you mean by a direct solution?
And what's 'wi'? :S

4. Oct 11, 2008

### gabbagabbahey

He means that your basis set is $\{ \vec{w}_i \}= \{\vec{w}_1,\vec{w}_2, \ldots \vec{w}_n \}$ and if $\vec{k}$ is perpendicular to each of your spanning vectors, $\vec{k} \cdot \vec{w}_i=0 \quad \forall i \in [1,n]$. What does that make $\vec{k} \cdot \vec{w}$?

And if $\vec{k}$ is perpendicular to every vector in W, then $\vec{k} \cdot \vec{w}=0$. What does that make $\vec{k} \cdot \vec{w}_i=0$ for each i?

Last edited: Oct 11, 2008
5. Oct 11, 2008

### kehler

If k is perpendicular to each of the spanning vectors, then k.w = 0
and if k is perpendicular to every vector in W, then k.wi = 0

Is that all I have to write? :S

6. Oct 11, 2008

### gabbagabbahey

Well, you should explicitly show that each of those claims is true; but other than that, yes; since any vector for which k.w is zero, is in W_perp.

7. Oct 11, 2008

### kehler

Oh ok cool :). Thanks guys.