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I'm proving that $A^tA$ will be positive definite iff $A$ is invertible. I guess that there are ways to show this with determinants, eigenvectors. But I've just gone with

  1. Positive definites must have trivial null space by definition
  2. $A^tA$ and A share their null space (demonstrated)
  3. An invertible matrix can only have a trivial null space (if $Ax = 0$; $A^{-1} A x = A^{-1} 0$; $x = 0$)

As far as I can reason, this only demonstrates that if $A$ is invertible then $A^tA$ is positive definite. I'd like to demonstrate that if the null space of $A$ is trivial then it MUST be invertible, but am not sure how to do this. I imagine there's something straightforward but my brain is pretty fried.

  • 0
    Have you heard of the [rank-nullity theorem](http://en.wikipedia.org/wiki/Rank%E2%80%93nullity_theorem)?2012-04-23
  • 0
    See http://math.stackexchange.com/a/172897/105752016-05-01

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