When is the multiplication of rectangular matrices an invertible matrix? Let $W\in\mathbb{R}^{N\times n }$ with $N>n$. Under which conditions is the matrix $S=W^T W$ invertible?
 A: For $A\in\mathbb{R}^{m\times n}$ the matrix $A^\intercal A$ is called the Gramian of $A$. This is an extremely important construction!
Fact 1. $A^\intercal A$ is an $n\times n$ symmetric matrix
Proof. Exercise $\Box$
Fact 2. $\operatorname{rank}(A^\intercal A)=\operatorname{rank}(A)$
Proof. Note that $\boldsymbol{v}\in\operatorname{Null}(A)$ implies $A^\intercal A\boldsymbol{v}=A^\intercal\boldsymbol{O}=\boldsymbol{O}$ so $\operatorname{Null}(A)\subset\operatorname{Null}(A^\intercal A)$.
Next, suppose $\boldsymbol{v}\in\operatorname{Null}(A^\intercal A)$. Then
$$
\lVert A\boldsymbol{v}\rVert^2
= \langle A\boldsymbol{v}, A\boldsymbol{v}\rangle
= \langle\boldsymbol{v}, A^\intercal A\boldsymbol{v}\rangle
= \langle\boldsymbol{v}, \boldsymbol{O}\rangle
= 0
$$
It follows that $A\boldsymbol{v}=\boldsymbol{O}$, so $\operatorname{Null}(A^\intercal A)\subset\operatorname{Null}(A)$.
Our work above then demonstrates that $\operatorname{Null}(A)=\operatorname{Null}(A^\intercal A)$. Thus
$$\operatorname{rank}(A^\intercal A)=n-\dim\operatorname{Null}(A^\intercal A)=n-\operatorname{Null}(A)=\operatorname{rank}(A)\qquad\Box$$
This now gives a satisfactory answer to your question:
Fact 3. $A^\intercal A$ is invertible if and only if $A$ has full column rank.
A: $S$ will be invertible if $W$ is injective.  Maybe not the best proof...
Suppose not ($W$ injective, $S$ not bijective).  Then there exists $x\neq0$ s.t. $Sx=0$ and $\langle Sx,y\rangle=\langle Wx,Wy\rangle=0$ for all $y$, in particular for $y=x$.  Hence $\|Wx\|=0$, a contradiction.
