This is my initial trail:

Since $Nul(A) = 0$, by the invertible matrix theorem the matrix A is full rank and thus invertible. This implies $det(A)\neq0$, and likewise $det(A^T)\neq0$.

This implies that $det(A^T A ) = det(A^T)\, det(A) \neq 0$, and thus $A^T A$ is invertible.

However, on second thought this implies that A is itself square, which that is not given. How do I prove/explain this if we don't know if A is square or not?


1 Answer 1


If $A^T A v = 0$, then $v^T A^T Av = 0$, so $\|Av\| = 0$, so $Av =0$, so $v = 0$.

Therefore, $\operatorname{Nul}(A^T A) = 0$.

  • $\begingroup$ Makes sense. Thanks! $\endgroup$
    – eirik-ff
    Commented May 28, 2018 at 9:15
  • $\begingroup$ The main thing to add is that $A^TA$ is square, which is why we can then conclude it is invertible. I believe non-square matrices can have $0$ null space, but of course are not invertible. $\endgroup$
    – tsojtc
    Commented May 28, 2018 at 10:01

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