Linear Algebra - Minimize the Norm

Click For Summary

Homework Help Overview

The problem involves finding a vector in the span of two given vectors in R4 that minimizes the distance to a specified point. The context is linear algebra, specifically focusing on concepts of vector projection and orthogonality.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning, Problem interpretation, Assumption checking

Approaches and Questions Raised

  • Participants discuss various methods to find the vector that minimizes the norm, including attempts at calculating projections and exploring orthonormal bases. Questions arise about the correctness of formulas and the orthogonality of vectors.

Discussion Status

Participants are actively engaging with the problem, providing guidance on projection techniques and the need for orthonormal bases. There is ongoing exploration of different approaches, with some participants expressing uncertainty about their calculations and seeking clarification.

Contextual Notes

There are discussions about the dimensionality of the span and the properties of the vectors involved, including the need for orthogonality in the basis used for projections. Some participants express confusion about their calculations and the assumptions made regarding the vectors.

steelphantom
Messages
158
Reaction score
0

Homework Statement


In R4, let U = span((1, 1, 0, 0), (1, 1, 1, 2)). Find u in U such that ||u - (1, 2, 3, 4)|| is as small as possible.

Homework Equations



The Attempt at a Solution


I came up with a vector u = (-.5, -.5, 0, 0) + (2, 2, 2, 4) = (1.5, 1.5, 2, 4). Then u - (1, 2, 3, 4) = (0.5, -0.5, -1, 0). Using the standard dot product as the norm, I get ||(0.5, -0.5, -1, 0)|| = sqrt(.25 + .25 + 1) = sqrt(1.5).

This was probably a bad approach, but I can't think of any other real way to do this. Is this the right answer at least? If not, how could I approach this in a better manner other than just randomly picking vectors? Thanks.
 
Physics news on Phys.org
Try playing around with orthogonal projections onto U.
 
Duh... :rolleyes: Thanks. So now I want to find a unit vector which spans U, but how do I do that? Once I find the unit vector, I can find the orthogonal projection P(u), and then the minimal distance will be ||u - P(u)||.
 
Since U is two dimensional you will not find a single vector spanning U. If you are talking about "the" normal vector to U, then I have to tell you that the orthogonal complement of U is two-dimensional as well, so there will be no unique normal vector (even after normalization.)

The rest is right, ||u-Pu|| will be the distance you are looking for.
 
Ok, I did what you said, and ended up with P(u) = (3/2)(1, 1, 0, 0) + 2(1, 1, 1, 2), which was what I originally had, so I guess I was right. Thanks for your help!
 
I was just writing up my work today, and I found that I must have messed up somewhere. Here's how I went about the problem:

P(u) = <u, e1>e1 + <u, e2>e2
||e1|| = sqrt(12 + 12) = sqrt(2) => e1 = (1/sqrt(2))*(1, 1, 0, 0).
||e2|| = sqrt(1 + 1 + 1 + 4) = sqrt(7) => e2 = (1/sqrt(7))*(1, 1, 1, 2).
P(u) = (1/2)[(1*1) + (1*2) + (0*3) + (0*4)]e1 + (1/7)[(1*1) + (1*2) + (1*3) + (1*4)]e2
= (3/2)*(1, 1, 0, 0) + 2*(1, 1, 1, 2) = (3.5, 3.5, 2, 4)

But when I calculate ||P(u) - (1, 2, 3, 4)|| I get sqrt(10), which is higher than my random guess of sqrt(1.5). Where did I go wrong?
 
the coefficient of e2 in P(u) -- shouldn't it be 10/7 instead of 2 ..?

Oh no I see you made a type, it should read 2*4, so the coefficient is right.

As you may have noticed, (Pu-u) as you calculated is orthogonal to neither of the vectors spanning U, so you made a mistake in calculating Pu.

Why do you consider P(u) = <u, e1>e1 + <u, e2>e2 to be the right formula for the projection?
 
Last edited:
Pere Callahan said:
the coefficient of e2 in P(u) -- shouldn't it be 10/7 instead of 2 ..?

Oh no I see you made a type, it should read 2*4, so the coefficient is right.

As you may have noticed, (Pu-u) as you calculated is orthogonal to neither of the vectors spanning U, so you made a mistake in calculating Pu.

Why do you consider P(u) = <u, e1>e1 + <u, e2>e2 to be the right formula for the projection?

I have a formula that says Pv = v, e1>e1 + ... + <v, em>em, where {e1, ..., em} is an orthonormal basis for U. Apparently {e1, e2} is not an orthonormal basis. Do I just need to find an orthonormal basis and then calculate P(u)?
 
Last edited:
steelphantom said:
Do I just need to find an orthonormal basis and then calculate P(u)?

Sounds like a good idea. :smile:
 
  • #10
Ok, to be orthonormal, <v_i, v_j> = 0 for i != j. So I want <(a, a, 0, 0), (b, b, b, 2b)> = 0. This means sqrt(ab + ab) = sqrt(2ab) = 0. But the only way this can happen is if a and/or b are zero. But if I solve <(a, a, 0, 0), (a, a, 0, 0)> = 1 I get a = 1/sqrt(2) and if I solve <(b, b, b, 2b), (b, b, b, 2b)> = 1 I get a = 1/sqrt(7)>.

This doesn't seem to make sense. What am I doing wrong? Thanks for your continued help.
 
  • #11
steelphantom said:
Ok, to be orthonormal, <v_i, v_j> = 0 for i != j. So I want <(a, a, 0, 0), (b, b, b, 2b)> = 0. This means sqrt(ab + ab) = sqrt(2ab) = 0. But the only way this can happen is if a and/or b are zero. But if I solve <(a, a, 0, 0), (a, a, 0, 0)> = 1 I get a = 1/sqrt(2) and if I solve <(b, b, b, 2b), (b, b, b, 2b)> = 1 I get a = 1/sqrt(7)>.

This doesn't seem to make sense. What am I doing wrong? Thanks for your continued help.

What you were trying to do was taking v_1 to be a multiple of (1,1,0,0) and v_2 a multiple of (1,1,1,2). However, (1,1,0,0) and (1,1,1,2) are not orthogonal, so any multiples of them will not be orthogonal either. That's why the method didn't work.

Try taking v_1 to be equal to (1,1,0,0), that;s fine. Then you need to find another vector in U which is orthogonal to this v_1. A general vector in U is a linear combination of (1,1,0,0) and (1,1,1,2) that is

[tex] v_2 = a (1,1,0,0) + b (1,1,1,2)[/tex]

Try to find a and b such that <v_1, v_2> = 0. Then you can normalize v_1 and v_2 and have an orthonormal basis.

In general you might want to look up Gram-Schmidt orthogonalization which is a method to find an orthogonal basis if you start from any given basis in a finite-dimensional Euclidean vector space. It's not necessary for this problem, though.
 
Last edited:
  • #12
Ok, so I take u1 = (1, 1, 0, 0) and then take u2 = a(1, 1, 0, 0) + b(1, 1, 1, 2) = (a + b, a + b, b, 2b). Then <u1, u2> = a+b + a+b = 2(a+b). So the only way for this to be zero is if a = -b. So I take a = 1, b = -1 to get u2 = (0, 0, -1, -2). Then ||u1|| = sqrt(2) and ||u2|| = sqrt(5), so e1 = (1/sqrt(2))*u1, and e2 = (1/sqrt(5))*u2.

So now we have P(u) = 1/2[(1*1) + (1*2) + (0*3) + (0*4)](1, 1, 0, 0) + 1/5[(0*1) + (0*2) + (-1*3) + (-2*4)](0, 0, -1, -2) = (3/2)*(1, 1, 0, 0) - (11/5)*(0, 0, -1, -2) = (3/2, 3/2, 11/5, 11/5).

Then P(u) - u = (0.5, 0.5, -0.8, 0.4) and ||P(u) - u|| = sqrt((0.25) + (0.25) + (0.64) + (0.16)) = sqrt(1.3).

I think I've FINALLY got this problem solved. Please let me know if I did this right. Thanks a lot for your help, Pere! :smile:
 
  • #13
Seems right. So your first guess was pretty close :smile:
 

Similar threads

Replies
1
Views
2K
  • · Replies 4 ·
Replies
4
Views
3K
  • · Replies 12 ·
Replies
12
Views
3K
  • · Replies 7 ·
Replies
7
Views
1K
  • · Replies 10 ·
Replies
10
Views
2K
  • · Replies 5 ·
Replies
5
Views
2K
  • · Replies 2 ·
Replies
2
Views
1K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 18 ·
Replies
18
Views
3K
  • · Replies 3 ·
Replies
3
Views
1K