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While I was studying for my exam, I found this question about range and nullspaces of a matrix, which seemed easy at the first glance, but I couldn't find a way to prove it. Could you please take a look at it?
Let $A\in R^{mxn}$
Prove that:
i) If $R(A) = R^m \implies det(AA^T)\neq0.$
ii)If $N(A)=\{0\} \implies det(A^TA)\neq0$
For (ii), note that $$ \det(A'A)=0\implies\exists x\neq 0:A'Ax=0\implies 0=x'A'Ax=|Ax|^2\implies 0\neq x\in N(A). $$ For (i), we use (ii) along with the observation that $R(A)=\mathbb{R}^m$ implies $N(A')=\{0\}$. Indeed, suppose there is some $x\neq 0$, $x\in\mathbb{R^m}$ such that $A'x=0\in\mathbb{R}^n$. Claim that $A$ cannot map to $x$ (which means the surjectivity of $A$ will be violated). Suppose otherwise: $$ \exists\beta\in\mathbb{R}^n: A\beta=x\implies 0=A'x=A'A\beta\implies 0=\beta'A'A\beta=|A\beta|^2=|x|^2 $$ which would imply $x=0$. But this would be a contradiction because we have supposed $x\neq 0$.
