Let $A$ be a matrix with $Nul(A) = 0$. Explain why the matrix $A^T A$ is invertible.

Question

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.

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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?

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