Is {C, D} Independent? Proving Matrix Independence in 3 Simple Steps

Click For Summary

Homework Help Overview

The discussion revolves around proving the independence of two matrices, C and D, within the context of linear algebra. Participants explore the implications of matrix relations and the conditions under which matrices can be considered independent.

Discussion Character

  • Exploratory, Assumption checking, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the implications of assuming C and D are not independent, questioning what relations would hold true and how transposing these relations affects the analysis. There is exploration of the conditions under which coefficients a and b can be non-zero, and the validity of equating coefficients in the context of matrix equations.

Discussion Status

The discussion has progressed through various interpretations and mathematical manipulations. Participants have offered guidance on how to approach the problem, particularly regarding the implications of non-zero coefficients and the use of properties of matrices. There is an ongoing exploration of the conditions necessary to conclude independence.

Contextual Notes

Participants note that the matrices C and D are n x n and nonzero, which is a critical aspect of the problem being discussed. There is also a recognition of the initial misunderstanding regarding the independence of the matrices.

jumbogala
Messages
414
Reaction score
4
Edit: There was a mistake in the question, see below for right question.
 
Last edited:
Physics news on Phys.org
If {C,D} was not a set of independent matrices then what sort of relation would C and D satisfy? What does taking the transpose of both sides of said relation tell you?
 
If {C,D} were not a set of independent matrices, then I would be able to find some numbers a and b not both zero, such that aC + bD = 0.

Transpose of this... aCT + bDT = 0?
 
Right, now use the relations given in the question to rewrite that last equation.
 
Okay, so that means

aC + b(-D) = 0.

Now that means that aC + b(-D) = aC + bD

and can be rewritten aC -b(D) = aC + bD... which would mean that a = a, and -b = b. But the only way -b = b can be true is if b = 0. So therefore, b is zero and my initial condition is false, and they are independent! right?
 
oh wait, but a can still be something other than zero... my condition is that they are not BOTH zero. so my initial condition is not necessarily false =/
 
Sort of. You can't exactly just equate coefficients if that's what you did. The last equation you wrote implies that bD = 0 by just rearranging. You can't cancel the D here either, but you can use one property given in the question which you haven't used yet. :)
 
Wait, why can't I cancel the D? If bD = 0 and D is not zero, then isn't b zero?

Am I supposed to use the fact that they are n x n?
 
I just realized the question is not supposed to say that C and D are independent matrices, just that they are n x n and nonzero, if that makes a difference.
 
  • #10
Sure it makes a difference. You've got aC+bD=0 and aC-bD=0. To show C and D are independent, you want to show that a and b are both zero. Suppose a is nonzero?
 
  • #11
If it's still true that bD is zero, then if a is nonzero, you'd have aC is not zero.

Then aC + bD = 0 --> aC + 0 = 0 but if aC is not zero this is obviously not true.
 
  • #12
Add the two equations. Who cares what bD is? If a is nonzero then you get 2aC=0. a is nonzero, C is nonzero. Possible, or not?
 
  • #13
No, not possible. Therefore a must be zero.

But I still need to show that b is zero, right?

Now that I know a is zero can I just plug it into the equation aC + bD = 0?
 
  • #14
Sure you can. Or you could subtract the two equations and conclude 2bD=0, D being nonzero means b must be zero. Same conclusion either way.
 
  • #15
Great, so then a = b = 0, thus {C, D} is independent. And we're done!

Thanks to both of you for your help :)
 

Similar threads

  • · Replies 5 ·
Replies
5
Views
2K
Replies
1
Views
2K
  • · Replies 2 ·
Replies
2
Views
2K
Replies
6
Views
2K
Replies
8
Views
3K
Replies
2
Views
2K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 7 ·
Replies
7
Views
3K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 3 ·
Replies
3
Views
3K