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Let $A\in \mathbb{R}^{m\times n}, m\ge n$ and $\operatorname{rank}(A) = n$. Prove that $A^tA$ is nonsingular

I found this answer https://math.stackexchange.com/a/1840814/166180

which gives some hints. It says that $A$ and $A^tA$ have the same null space, that is, if $x$ is such that $A^tAx = 0$, then

$$A^tAx = 0 \implies Ax = 0$$

To say that $A^tA$ is nonsingular (invertible) is the same as saying that the null space of $A^tA$ is $0$, which is the same, by above, as saying that the null space of $A$ is $0$.

So by hypothesis on the exercise we have that $rank(A)=n$ and $m\ge n$. What should daying that the rank of $A$ is $n$ mean? I think that since the trace is less than $m$, the system $Ax=0$ has multiple solutions (just a guess, I don't know how to formalize this), so the null space of $A$ wouldn't be $0$.

Bernard
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Paprika
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Hint: $Ax = 0$ is equivalent to $||Ax|| = 0$.

Seewoo Lee
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  • sorry but i need another hint. how does $| A x | \ne 0$ imply then that $| A^* A x | \ne 0$ for $x \ne 0$? How do I know $A x$ isn't in the kernel of $A^*$? thanks so much for any help. – william_grisaitis Mar 07 '23 at 22:07
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    @william_grisaitis ${\lVert Ax \rVert} \neq 0$ implies $0 \neq {\lVert Ax \rVert}^2 = (Ax)^(Ax) = x^A^* Ax$. Then, since $x^A^ Ax \neq 0$ that implies $A^* Ax \neq 0$. – clay Sep 18 '23 at 05:20