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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 $0$.

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  • $\begingroup$ you are confusing trace and ranks of a matrix for the statement as you put it is completely false. $\endgroup$ – dezdichado Sep 5 '18 at 4:41
  • $\begingroup$ Trace is only defined for square matrices. $\endgroup$ – xbh Sep 5 '18 at 4:48
  • $\begingroup$ Thanks, I corrected $\endgroup$ – Paprika Sep 5 '18 at 4:57
  • $\begingroup$ By the rank-nullity theorem the null space of $A$ is $\{0\}$. $\endgroup$ – amsmath Sep 5 '18 at 7:14
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Hint: $Ax = 0$ is equivalent to $||Ax|| = 0$.

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