Intersect of U and U perpendicular; Orthogonality

[SeannyBoi71]

Homework Statement

Let U be a subspace of ℝⁿ. Show that if u[itex]\in[/itex]U[itex]\bigcap[/itex]U^{[itex]\bot[/itex]}, then u=0.

Homework Equations

The Attempt at a Solution

I know that U^{[itex]\bot[/itex]} will be orthogonal to U, so any vector u in U dotted with any vector in U^{[itex]\bot[/itex]} will equal 0. But that does not necessarily mean that u = 0 so I must prove something else. I don't know where to include the intersect into all of this, we have never used that kind of thing before.

 

[shaon0]

Homework Statement

Let U be a subspace of ℝⁿ. Show that if u[itex]\in[/itex]U[itex]\bigcap[/itex]U^{[itex]\bot[/itex]}, then u=0.

Homework Equations

The Attempt at a Solution

I know that U^{[itex]\bot[/itex]} will be orthogonal to U, so any vector u in U dotted with any vector in U^{[itex]\bot[/itex]} will equal 0. But that does not necessarily mean that u = 0 so I must prove something else. I don't know where to include the intersect into all of this, we have never used that kind of thing before.

Well, you've almost got it. Think about u^2 and remember that u is in Rⁿ.

 

[SeannyBoi71]

u²? u is a vector, so I'm guessing you're insisting I think about u dot u... I'm a little confused how that helps me here

 

[Deveno]

well one way to look at u.u is to think of "the left u" as being in U, and the "right u" as being in U^⊥ (which we can do since u is in the intersection).

 

[SeannyBoi71]

Well if I do it that way then yes I will get the dot product is 0, because those two are orthogonal. I don't see how that makes u = 0, though. I'm missing something here

 

[Deveno]

review the properties a dot product has to have...(positive definite is the property you're after)

 

[shaon0]

u²? u is a vector, so I'm guessing you're insisting I think about u dot u... I'm a little confused how that helps me here

Yes, but also cancel out other cases using the orthogonal relation. Intuitively, if I rotate my u vector to make it orthogonal. What are the point(s) which are shared by both these vectors?

 

[SeannyBoi71]

review the properties a dot product has to have...(positive definite is the property you're after)

Just looked through my textbook and lecture notes and could not find any property like that, do you think you could explain it?

And now I know that u = 0 because it will be the only vector that is in U and U^{[itex]\bot[/itex]} because, well obviously, they are at a 90° angle. I just don't understand how I'm supposed to show it.

Actually come to think of it, can I use the formula [tex] cosθ = x[itex]\cdot[/itex]y/||x|| ||y|| [/tex] and plug in θ=pi/2, showing that u[itex]\cdot[/itex]w = 0 , so that u must be equal to the 0 vector? This answer seems a bit trivial...

 

[shaon0]

Just looked through my textbook and lecture notes and could not find any property like that, do you think you could explain it?

And now I know that u = 0 because it will be the only vector that is in U and U^{[itex]\bot[/itex]} because, well obviously, they are at a 90° angle. I just don't understand how I'm supposed to show it.

Actually come to think of it, can I use the formula [tex] cosθ = x[itex]\cdot[/itex]y/||x|| ||y|| [/tex] and plug in θ=pi/2, showing that u[itex]\cdot[/itex]w = 0 , so that u must be equal to the 0 vector? This answer seems a bit trivial...

Deveno was probably alluding to the orthogonal inner product space. <x,y>=x.y. An inner product space must never be negative. Also, your method above is correct. Either, x OR y are the 0 vector. But in our example, that is u and u^⊥ satisfy, u=0 and u^⊥=0.
If we write out u=[u₁,...,u_n] and use the fact that u.u^⊥=0 and solve for each u_i

 

[Deveno]

Just looked through my textbook and lecture notes and could not find any property like that, do you think you could explain it?

And now I know that u = 0 because it will be the only vector that is in U and U^{[itex]\bot[/itex]} because, well obviously, they are at a 90° angle. I just don't understand how I'm supposed to show it.

Actually come to think of it, can I use the formula [tex] cosθ = x[itex]\cdot[/itex]y/||x|| ||y|| [/tex] and plug in θ=pi/2, showing that u[itex]\cdot[/itex]w = 0 , so that u must be equal to the 0 vector? This answer seems a bit trivial...

if we are talking about the standard dot product in Rⁿ, then

if u = (u₁,u₂,...,u_n)

u.u = (u₁)² + (u₂)²+...+(u_n)²

now, if u.u = 0, we have:

0 u.u = (u₁)² + (u₂)²+...+(u_n)²

one hopes that you know that the square of a real number is non-negative.

the sum of two (or more) non-negative numbers is also ______?

therefore...

*********

you CANNOT conclude that just because u.w = 0 that one of u or w must be 0. for example, (0,-1).(1,0) = 0(1) + (-1)(0) = 0, but neither of these vectors is the 0-vector.

and, geometrically, saying that the 0 vector is at "a right angle" to something, just doesn't make any sense. how do you choose θ?

 

[SeannyBoi71]

and, geometrically, saying that the 0 vector is at "a right angle" to something, just doesn't make any sense. how do you choose θ?

Didn't mean 0 vector is at a right angle. What I meant is that if two vectors are orthogonal (perpendicular), the only point they are going to have in common is 0 i.e. (0,0,0) in R³. And I chose θ to be pi/2 because that is a 90° angle and therefore the two vectors would be orthogonal... Is this logic not correct? I assumed this is what shaon0 was getting at. The thing is that I don't know how to "show" this, like the question asks. I can put it into words but not into equations.