How Do Subspaces and Orthogonality Relate in Linear Algebra?

[kdieffen]

Ok so I've been working on this problem and I'm really having some struggles grasping it. Here it is:

Let W be some subspace of Rn, let WW consist of those vectors in Rn that are orthognoal to all vectors in W.

1) Show that WW is a subspace of Rn?

So for this part I'm thinking that because WW is a linear combination of W (maybe) then therefore it forms a subspace of Rn

2) If {v1, v2,...vt} is a basis for W, show that a vector X in Rn lies in WW if and only if x is orthogonal to each of the vectors v1, v2,...vt?

And for this one I'm really at a lose for where to start. Any help would be appreciated.

Thanks

 

[Bacle]

What have you tried, so far.?

Basically, in order to show that any subset S of a vector space V is a subspace,

is to show that elements of S are closed under scaling and under linear combinations;

in this case, show that if w^ and w'^ are in W^ =(" W Perp" ) , so is their sum,

and for any w^, cw^ (c a scalar) is also in W^. The key here is the properties

of the inner-product.



For 2, one side follows by definition. For the other side ( w^ is perpendicular

to each of v1,..,vt ) , think of writing any x in W using the elements of v1,..,vt,

and, again, use the properties of the inner-product (multilinearity).

Good Luck.

 

[kdieffen]

1) Well I understand the definition of a subspace, i think I am just finding it difficult to fully understand the proof. Would it be correct to say then that:

WW is a subspace of Rn because
i) It contains the zero vector
ii) WW= {v1', v2',...vt'}

Let x, y span (WW)
x=av1' + av2' + avt'
y=bv1' + bv2' + bvt'

Therefore (x+y)=(a1+b1)v1' + (a2+b2)v2' + ...(at + bt)vt' and is closed under vector addition.

iii) Let x span (WW)

Then x= a1v1' + a2v2' + atvt' for some a1, a2,...at
Then kx=ka1v1' + ka2v2' + katvt'

Therefore it is closed under scalar multiplication. Since it satisfies all three, it is a subspace of Rn.

So that's what I have for part 1

And for part 2 I'm still lost lol

2)

 

[Bacle]

"" 1) Well I understand the definition of a subspace, i think I am just finding it difficult to fully understand the proof. Would it be correct to say then that:

WW is a subspace of Rn because
i) It contains the zero vector
ii) WW= {v1', v2',...vt'}

Let x, y span (WW)
x=av1' + av2' + avt'
y=bv1' + bv2' + bvt'

Therefore (x+y)=(a1+b1)v1' + (a2+b2)v2' + ...(at + bt)vt' and is closed under vector addition. ""


I don't see how you have shown it is closed. How do you know that x+y is in W^.?



iii) Let x span (WW)

What do you mean here.?. How do you know that a single vector x spans WW.?.
I think (reading below ) you mean: let x be a vector in WW. Right.?

Then x= a1v1' + a2v2' + atvt' for some a1, a2,...at
Then kx=ka1v1' + ka2v2' + katvt'

Therefore it is closed under scalar multiplication.

Not clear to me. How do you know kx is in WW.? You need to check this, or give a good
argument to that effect.

Since it satisfies all three, it is a subspace of Rn.

So that's what I have for part 1

And for part 2 I'm still lost lol.

Well, assume that a vector x in R^n is orthogonal to each of the basis vectors

{v1,..,vt} of W . You then want to show that x is orthogonal to any vector v in W.

How do you show any two vectors are orthogonal.?.

Assume x is orthogonal to each of v1,..,vt (what does this mean.?) and let w be in W.

What do you need to do to show that x and v are orthogonal.?. How can you write w in

order to show this.?