# Linear Algebra Problem - columns of $A$ are linearly independent $\Rightarrow$ $A^TA$ non-singular [duplicate]

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. $\blacksquare$

• It appears that you're assuming that $A$ is square. This is not given... – Michael Burr May 16 '17 at 12:10
• How do you define the determinant of a non-square matrix? – Nigel Overmars May 16 '17 at 12:11
• Oh, I see, didn't see that :(. How should I approach this then? – PhysicsMathsLove May 16 '17 at 12:14
• Rank(A'A)=rank(A)=n when A is a real non-singular matrix of order n. – StubbornAtom May 16 '17 at 12:40

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 $Av$?
Hint: Note that $A^{T}Av =0 \implies Av \in \ker A^{T} \cap \text{Image}(A) =\{0\}$.
• 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. – akech May 16 '17 at 13:49