Find Basis for Uperp from x1, x2, x3: Orthogonal Bases

[theRukus]

Homework Statement

Let U be a subspace of R⁴ and let S={x₁, x₂, x₃} be an orthogonal basis of U.

Given x₁, x₂, x₃, find a basis for Uperp (the subspace containing all vectors orthogonal to all vectors in U).

I am actually given three vectors x₁, x₂, x₃, but I am looking more to understand the concept.

Homework Equations

The Attempt at a Solution

Since all vectors in Uperp are orthogonal all vectors in U, I am led to believe that all vectors in a basis of Uperp would be orthogonal to the vectors in the basis of U... But the vectors I am given are all orthogonal to one another, so I am somewhat confused.

Any insight is appreciated. Thank you.

 

[ystael]

Since all vectors in Uperp are orthogonal all vectors in U, I am led to believe that all vectors in a basis of Uperp would be orthogonal to the vectors in the basis of U... But the vectors I am given are all orthogonal to one another, so I am somewhat confused.

Given that \dim U = 3 and you are working in \mathbb{R}^4, what do you expect the dimension of U^\perp to be? (Just because x_1, x_2, x_3 are all mutually orthogonal, that doesn't necessarily mean you can't have another vector orthogonal to all of them.)

 

[theRukus]

Yes, given. However, wouldn't the basis for U^\perp have to consist of 3 new vectors, each orthogonal to x_1, x_2, x_3?

 

[micromass]

No, the basis of U^\bot will not consist of three vectors. It will consist of one vector, which is orthogonal to the basis in U.

 

[theRukus]

Thank you

 

[theRukus]

Alright. I have been trying to solve in the following manner:

Pick a random vector x.

Solve for z = x - proj_x1(x) - proj_x2(x) - proj_x3(x)

That is, z equals the random vector x minus x's projections on the three orthogonal vectors.

This should leave me with one vector z orthogonal to all three original vectors, correct?

I have tried this with three random vectors x, and for some reason every z is orthogonal to x₁ and x₂, but not x₃. I have been redoing my calculations repeatedly, but the same thing occurs. Is this possible, or have I some error?

 

[micromass]

Yes you should get to the correct answer this way (provided that the vector x is linearly independent from x_1,x_2,x_3.
If you can't find your mistake, then you can always post your work here. So we can see what went wrong...

 

[theRukus]

I tried again, instead using the formula

z = x - proj_x1(x) - proj_x3(x) - proj_x2(x)

Using this method, z is orthogonal to x₁ and x₃, but not x₂.. What givesss...?

 

[theRukus]

The latest effort:

GIVEN: x1 = [1 0 -1 -1] x2 = [2 1 1 1] x3 = [-1 3 -1 0]

Let x = [1 2 3 4] (I have also tried with x = [1 1 1 1] and x = [2 1 2 1])
z = x - proj_x1(x) - proj_x3(x) - proj_x2(x)

= [1 2 3 4] - (-6/3)[1 0 -1 -1] - (2/11)[-1 3 -1 0] - (11/7)[2 1 1 1]

= [1 2 3 4] - [-2 0 2 2] - [-2/11 6/11 -2/11 0] - [22/7 11/7 11/7 11/7]

= [3 2 1 2] - [-14/77 42/77 -14/77 0] - [242/77 121/77 121/77 121/77]

=[231/77 154/77 77/77 154/77] - [-256/77 -79/77 -135/77 -121/77]

= [487 233 212 275]This vector is orthogonal to x₁ and x₃ but not x₂ (dot product test)

 

[micromass]

[3 2 1 2] - [-14/77 42/77 -14/77 0] - [242/77 121/77 121/77 121/77]

=[231/77 154/77 77/77 154/77] - [-256/77 -79/77 -135/77 -121/77]

It seems that you made some sign errors in this step...

 

[theRukus]

I don't mean to question your authority, but I don't see any sign errors. Are you sure they're there?

 

[micromass]

Well, for me we have

[3 2 1 2] - [-14/77 42/77 -14/77 0] - [242/77 121/77 121/77 121/77]
= [3 2 1 2] - ([-14/77 42/77 -14/77 0] + [242/77 121/77 121/77 121/77])
= [231/77 154/77 77/77 154/77]-[228/77 163/77 107/77 121/77]

This is a bit different then your solution:
[231/77 154/77 77/77 154/77] - [-256/77 -79/77 -135/77 -121/77]

 

[theRukus]

I'm at a loss to understand how the two are different. I'm comparing it to simple arithmetic:

-3 - 2 - 1 = - 3 - (2 + 1)
LS = -6, RS = -6
...

So how is this any different?

I can see that you are right; my homework problem is solved.. But now I am completely confused.

 

[micromass]

Well, for the first coordinate, you have

-(-14/77)-242/77=14/77-242/77=-228/77

thesame with the other coordinates...