Proving Orthogonality of Vectors Using Linear Algebra Techniques

Click For Summary

Homework Help Overview

The discussion revolves around proving the orthogonality of two vector expressions, specifically B|A|+A|B| and A|B|-B|A|, using linear algebra techniques. The participants explore the definitions of vector magnitudes and the properties of the dot product.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the requirement for the dot product to equal zero for orthogonality and question the meaning of |A| and |B|. There are attempts to clarify whether these represent vector magnitudes or determinants of matrices. Some participants suggest evaluating the dot product directly, while others express uncertainty about the steps involved.

Discussion Status

The conversation is ongoing, with some participants providing guidance on the properties of the dot product. There is a mix of interpretations regarding the calculations, and while one participant believes they have arrived at a conclusion, others seek further clarification on the simplifications involved.

Contextual Notes

Participants are navigating the definitions of vector magnitudes and their implications for the problem, with some confusion about the notation used. There is an emphasis on understanding the properties of the dot product without delving into specific calculations.

Yosty22
Messages
182
Reaction score
4

Homework Statement



Show that B|A|+A|B| and A|B|-B|A| are orthogonal.

Homework Equations



N/A

The Attempt at a Solution



I'm not too sure exactly how to start this. I do know that for two things to be orthogonal, the dot product has to be equal to 0, but I'm not sure how to evaluate this at all.
 
Physics news on Phys.org
What specifically is stopping you from taking the dot product of the two?
 
I guess my main issue right now is knowing what exactly is meant by |A| and |B|? Does this mean I can create an arbitrary matrix A and matrix B and replace |B| and |A| with the determinant of the two matrices? So if I have an arbitrary 2x2 matrix |a1 a2| and |b1 b2| I can calculate the determinant, then do the dot product?
|a3 a4| |b3 b4|
 
I would think that |A| denotes the length of vector A, not the determinant of a matrix A. In any case, you don't need to calculate that part out. You just need to know that |A| and |B| are scalars.
 
Okay, I understand they are scalars, but I'm not quite sure where to go from here. Do I need to do something with the Commutator?
 
No. Why would there be a commutator involved? I think you're confusing matrix multiplication with the dot product. They're not the same operation.

The result you're being asked to show is true for any vector space and appropriate dot product. You really don't need to know anything about the specifics of how to calculate the dot product. You just need to know properties of the dot product in general.
 
Ahh, I see. What if I did this:

<B|A|+A|B|> dot <A|B|-B|A|>

If you solve the dot product there, you get B|A|*A|B|-A|B|*B|A|. You have the same thing on either side of the subtraction sign, so it has to be 0. If the dot product is 0, then the vectors are orthogonal. Would that work?
 
What happened to the other two terms?
 
Oh whoops, I had them written down on the paper, I just forgot to type it. So in total, you have:

(B|A|)*(A|B|) - (B|A|)*(B|A|) + (A|B|)*(A|B|) - (A|B|)*(B|A|).

From this, everything cancels down to 0, so the vectors have to be orthogonal because the dot product is zero. Right?
 
  • #10
Yup, you got it.
 
  • Like
Likes   Reactions: 1 person
  • #11
Yosty22 said:
Oh whoops, I had them written down on the paper, I just forgot to type it. So in total, you have:

(B|A|)*(A|B|) - (B|A|)*(B|A|) + (A|B|)*(A|B|) - (A|B|)*(B|A|).

From this, everything cancels down to 0, so the vectors have to be orthogonal because the dot product is zero. Right?

vela said:
Yup, you got it.

I don't quite agree. I would want to see a bit more simplification before I agree that he's got it. Like why the middle two terms cancel. I'm not convinced he understands yet.
 
  • #12
Fair enough. :smile:
 

Similar threads

  • · Replies 5 ·
Replies
5
Views
2K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 7 ·
Replies
7
Views
1K
  • · Replies 1 ·
Replies
1
Views
1K
  • · Replies 4 ·
Replies
4
Views
2K
Replies
15
Views
3K
  • · Replies 25 ·
Replies
25
Views
4K
  • · Replies 24 ·
Replies
24
Views
4K
  • · Replies 13 ·
Replies
13
Views
6K
  • · Replies 11 ·
Replies
11
Views
3K