$\ker(A^TA) = \ker(A)$ I need help to understand the red part of the part of the proof below:
$\operatorname{rank}(X) = D$. One can show that then $D = \operatorname{rank}(X) = \operatorname{rank}(X^TX)$ follows which means that $X^TX \in \mathbf{R}^{D \times D}$ has full rank and is invertible.
Proof: To show: $\operatorname{rank}(A) = \operatorname{rank}(A^TA)$ for any matrix $A \in \mathbf{R}^{N \times D}$. This is equivalent to showing that the null space of both matrices are the same (remember the Rank–nullity theorem stating that for $A \in \mathbf{R}^{NxD}$, we have that $\operatorname{rank}(A) + nul(A) = D$). So we have to show that $nul(A) = nul(A^TA)$ which is equivalent to:
\begin{equation}
A^T A x = 0 \Longleftrightarrow A x = 0
\end{equation}
"$\impliedby$" Assuming that $Ax=0$, we obtain that $A^T A x = A 0 = 0$.
"$\implies$"  Assuming $A^T A x = 0$
$\implies x^T A^T A x = 0$
$\implies \color{red}{(Ax)^T (Ax) = 0 \implies Ax = 0}$.
Note that this last step only holds for real matrices A. We have shown that the null spaces are identical and therefore, the rank of both matrices is the same which completes the proof.
 A: $$A^T (Ax) = 0$$Multiply by $x^T$, you get
$$x^TA^T(Ax) = (Ax)^T(Ax) = \Vert Ax \Vert^2 = 0$$
If $\Vert Ax \Vert^2 = 0$ then $Ax = 0$ due to the norm property. So, if we let $y = Ax$ and expand the norm we get, 
$$\Vert y \Vert^2 = \sum y_i^2  = 0$$
Sum of positive numbers is equal to $0$, when each $y_i = 0$.
A: We have 
$$
Ax=\begin{pmatrix}
 A_{11} & A_{12} & \ldots & A_{1n} \\
 A_{21} & A_{22} & \ldots & A_{2n} \\
 \vdots & \vdots &  \ddots & \vdots \\
A_{n1} & A_{n2} & \ldots & A_{nn} \\
\end{pmatrix}
\begin{pmatrix}x_1\\x_2\\\vdots \\ x_n\end{pmatrix}
=
\begin{pmatrix}\sum_{k=1}^{n}A_{1k}x_k\\\sum_{k=1}^{n}A_{2k}x_k\\\vdots \\ \sum_{k=1}^{n}A_{nk}x_k\end{pmatrix}
$$
On the other hand, we also have
$$
x^TA^T=\begin{pmatrix}x_1,&x_2,&\cdots, & x_n\end{pmatrix}
\begin{pmatrix}
 A_{11} & A_{21} & \ldots & A_{n1} \\
 A_{12} & A_{22} & \ldots & A_{n2} \\
 \vdots & \vdots &  \ddots & \vdots \\
A_{1n} & A_{2n} & \ldots & A_{nn} \\
\end{pmatrix}
=
\begin{pmatrix}\sum_{k=1}^{n}x_kA_{1k}, &\sum_{k=1}^{n}x_kA_{2k}&\cdots & \sum_{k=1}^{n}x_kA_{nk}\end{pmatrix}
$$
It is now easy to see that
$$
(Ax)^T=x^TA^T.
$$
Continuing we have to
$$
(Ax)^T(Ax)= (\sum_{k=1}^{n}x_1A_{nk})^2+ (\sum_{k=1}^{n}x_2A_{nk})^2+\ldots + (\sum_{k=1}^{n}x_1A_{nk})^2=0
$$
if, only if, 
$$
 (\sum_{k=1}^{n}x_1A_{nk})^2=0\quad \mbox{ for all } k=1,2,\ldots,n
$$
A: Firstly, let $y=Ax$ so as to reinforce that it's merely a vector. Since both $A$ and $x$ were arbitrary so is $y$ because the identity matrix is a matrix. By substitution we have that $y^Ty=0$, but this is merely the dot product so the only vector that can satisfy this equation is $0$.
