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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.

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Have you heard of the rank-nullity theorem? – Shitikanth Apr 23 '12 at 2:35

Consider $A$ as a homomorphism from $K^{n}$ to $K^{n}$. If $A$ is not invertible, then we would expect $A$ has rank $<n$. By the rank-nullity theorem you should expect the map $x\rightarrow Ax$ satisfies $\dim Ker+\dim Im=n$. Thus $\dim Im<n$ as well.

In your case $\dim Ker=0$, thus $\dim Im=n$ and since vector space is classified by dimension, the image and the domain are isomorphic. Thus we assert that $A$ is an isomophism, i.e, invertible.

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