Answer
See below
Work Step by Step
Let $A=\begin{bmatrix}
a_{11} & a_{12} & ... & a_{1n}\\
a_{21} & a_{22} & ... & a_{2n}\\
. & . & ... & .\\
a_{m1} & a_{m2} & ... & a_{mn}
\end{bmatrix}$ be an m × n matrix.
then we have $(a_{11},a_{21},...,a_{m1}),(a_{12},a_{22},...a_{m2})$ is linearly independent.
Let $A^TA$ be $n\times n$ matrix.
Since $A^TA$ can be linearly independent, then rank of $A^TA$ is $n$.
$A_i$ is the i-th column of $A$, but $A_1x_1+A_2x_2+...=A_nx_n=0$ is a linear independence.
Then we see $Ax=0\\ \rightarrow A^TAx=0\\
\rightarrow x_1=x_2=...x_n=0$ where $x=(x_1,...x_n)^T$
It shows that the columns of $A$ are linearly independent if and only if $A^T A$ is invertible
