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I am trying to go in this direction only. Is this proof correct? ($A$ is an $m \times n$ matrix).

I am considering the contrapositive of the statement:

Suppose $A^TA$ is singular. Then $\det (A^TA) =0$. $\det(A^TA) = \det(A^T) \det(A)$ and since $\det (A^T) = \det(A)$, we must have $det(A) = 0$. In other words, the columns of $A$ are linearly dependent. $\blacksquare$

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    $\begingroup$ It appears that you're assuming that $A$ is square. This is not given... $\endgroup$ – Michael Burr May 16 '17 at 12:10
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    $\begingroup$ How do you define the determinant of a non-square matrix? $\endgroup$ – Nigel Overmars May 16 '17 at 12:11
  • $\begingroup$ Oh, I see, didn't see that :(. How should I approach this then? $\endgroup$ – PhysicsMathsLove May 16 '17 at 12:14
  • $\begingroup$ Rank(A'A)=rank(A)=n when A is a real non-singular matrix of order n. $\endgroup$ – StubbornAtom May 16 '17 at 12:40
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Notice that $$ A^TA v\cdot v = Av \cdot Av = \Vert Av \Vert^2 $$ So if $A^TA v= 0$, what does this tell you about $Av$?

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  • $\begingroup$ got it - cheers! $\endgroup$ – PhysicsMathsLove May 16 '17 at 13:36
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Hint: Note that $A^{T}Av =0 \implies Av \in \ker A^{T} \cap \text{Image}(A) =\{0\}$.

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  • $\begingroup$ This does not provide an answer to the question. To critique or request clarification from an author, leave a comment below their post. - From Review $\endgroup$ – M. Winter May 16 '17 at 12:58
  • $\begingroup$ A partially correct answer with details left to the OP should not be discouraged. In fact, if the OP understands why the null space of $A^{T}$ is an orthogonal complement of the range of $A$, then this hint gives a complete answer along the line of using contrapositive argument just as the OP attempted. $\endgroup$ – akech May 16 '17 at 13:49
  • $\begingroup$ Actually I am not sure what is the correct behavior, but for me the answer section is for answers. A hint, especially of this length, can be given as a comment. Maybe someone familiar with the SE mechanics can clarify this. However, this might not be the correct review response and I am sorry, but it was my choice because of a lack of a better option in the menu. $\endgroup$ – M. Winter May 16 '17 at 13:53

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