Why is the matrix invertible if its null space is zero? I'm trying to understand the proof of the following statement:

If $A$ has independent columns then $A^TA$ is invertible.

Please keep in mind that these structured are based on real field.
The proof starts with following equation:
$A^TAx=0$
Then it is stated, that for $A^TA$ to be invertible, $x$ must be the zero vector.
In other words, for $A^TA$ to be invertible, it's null space must be the zero vector.

I found this answer, but I couldn't completely understand it. The author of the answer defined variables: $A$ being a matrix containing columns of $a_n$, $x$ being a vector containing columns of $(x_n)^T$, and $Ax$ being equal to $0$.
Then it's obvious that $x_na_n=0$, but author states that unless $x_n=0$, the columns will not be linearly independent.
I couldn't understand this. If $A$ has linearly independent columns, then how will it change at all when being multiplied by $x$? Maybe they mean $Ax$, but still even if $x$ is zero vector, isn't $Ax$ just a simple linear combination of $A$ and $x$? (And if that is true then it's definitely not linearly independent).
 A: As stated in the comments, this does not hold in general, but holds if $A$ is a real matrix.
Claim 1: $ker(A) = ker(A^TA)$
$\subseteq$ is clear. To prove the other, let $v \in ker(A^TA)$. Then $0 = \langle v, A^TAv \rangle = ||Av||^2 \iff Av = 0 \iff v \in ker(A)$.
Claim 2: matrix $A$ is invertible iff injective (as a linear map given bases in domain and codomain). Note that for a linear map, or a matrix, $A$, it is injective iff $ker(A) = \{0\}$, i.e. trivial.
So since $ker(A)$ is trivial as A has independent column vectors, $A^TA$ is invertible by claims 1 and 2.
A: 
Then it's obvious that $x_na_n=0$, but author states that unless $x_n=0$, the columns will not be linearly independent.

That's not what the author says. 
He explains that if $a_1,\dots,a_n$ are the columns of $A$ (thus each $a_i$ is a vector in $\Bbb R^m$ - where $m$ is the number of rows of $A$) and if $x = \begin{bmatrix}x_1 & \dots & x_n\end{bmatrix}^T$ (thus each $x_i$ is a scalar), then $\displaystyle Ax = \sum_{i=1}^n x_ia_i$ is a linear combination of the vectors $a_1,\dots,a_n$. Thus, if at least one of the $x_i$ is non-zero, the definition of linear dependence tells us that $a_1,\dots,a_n$ are linearly dependent. 
But if the only $x$ giving $Ax=0$ is the zero vector, then the columns $a_1,\dots,a_n$ are linearly independent. 
