Non-singularity of A^T*A

  • #1
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Main Question or Discussion Point

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!
 

Answers and Replies

  • #2
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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).
 
  • #3
AlephZero
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Prove it by contradition. If the columns of X are linearly dependent, there is a non-zero vector ##y## such that ##X^TXy = 0##.
 
  • #4
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Thank you. I suspected it was true, but couldn't prove it to myself.
 
  • #5
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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".
 
  • #6
AlephZero
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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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