Are Non-Singular Column Vectors of X Linearly Independent?

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Discussion Overview

The discussion revolves around the relationship between the non-singularity of the matrix \(X^T X\) and the linear independence of the column vectors of matrix \(X\). Participants explore whether the non-singularity of \(X^T X\) implies that the columns of \(X\) are linearly independent, examining both theoretical proofs and counterexamples.

Discussion Character

  • Debate/contested
  • Mathematical reasoning

Main Points Raised

  • One participant questions if \(X\) is a square matrix and suggests using the determinant properties to analyze the relationship between \(X\) and \(X^T X\).
  • Another participant proposes proving the statement by contradiction, indicating that if the columns of \(X\) are linearly dependent, there exists a non-zero vector \(y\) such that \(X^T X y = 0\).
  • A later reply emphasizes the validity of the contradiction proof and suggests that the original poster should have been able to prove the converse if they understood the proof provided.
  • One participant comments on the terminology used, suggesting that "contraposition" might be a more appropriate term than "contradiction" in this context.
  • Another participant reflects on the importance of understanding concepts over remembering specific terminology.

Areas of Agreement / Disagreement

Participants express differing views on the implications of non-singularity and linear independence, with no consensus reached on the validity of the original claim. The discussion remains unresolved regarding the proof of the converse statement.

Contextual Notes

Some assumptions about the nature of matrix \(X\) (e.g., whether it is square or not) are not fully explored, and the discussion does not clarify the conditions under which the proposed proofs hold.

IniquiTrance
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Is it true to say that if [itex]X^T X[/itex] is non-singular, then the column vectors of X must be linearly independent? I know how to prove that if the columns of X are linearly independent, then [itex]X^T X[/itex] is non-singular. Just not sure about the other way around. Thanks!
 
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Is X a square matrix? If so use
[tex]det(X^TX) = det(X)^2=0[/tex]

If not X is not square, but is real, then QR decomposition should reduce the problem to that of square matrices (something simpler may suffice, but this is the simplest approach I can think of right now).
 
Prove it by contradition. If the columns of X are linearly dependent, there is a non-zero vector ##y## such that ##X^TXy = 0##.
 
Thank you. I suspected it was true, but couldn't prove it to myself.
 
AlephZero said:
Prove it by contradition. If the columns of X are linearly dependent, there is a non-zero vector ##y## such that ##X^TXy = 0##.

That's a very nice proof, especially because you established necessary and sufficient conditions. The OP claimed that he could prove the converse, but, if he knew this proof he should have had no trouble.

Just one remark, instead of "contradiction", maybe you should have used "contraposition".
 
Dickfore said:
Just one remark, instead of "contradiction", maybe you should have used "contraposition".

Well, I used to know what "contrapositive" meant when I was a student, but these days I find understanding the concepts is more useful than remembering their names.
 

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