Proof that $\operatorname{ker}(A^{T}A) = \operatorname{ker}(A)$? Is there a proof that can help me understand why this is the case? I can't conceptualize the reason for this in my mind. Thanks.
 A: For
$\vec v \in \ker A \tag{1}$
we have
$A \vec v = 0, \tag{2}$
whence
$A^TA \vec v = 0, \tag{3}$
showing that $\vec v \in \ker A^TA$; thus $\ker A \subset \ker A^TA$.  To go the other way, note that
$A^TA \vec v = 0 \tag{4}$
implies that
$\vec v^T A^TA \vec v = 0, \tag{5}$
or
$(A \vec v)^T(A \vec v) = 0; \tag{6}$
but $(A \vec v)^T(A \vec v)$ is a scalar quantity which if written out in terms of components is
$0 = (A \vec v)^T(A \vec v) = \sum (A \vec v)_i^2, \tag{7}$
where $(A \vec v)_i$ is the $i$-th component of $A\vec v$.  Thus we must have
$(A \vec v)_i = 0 \tag{7}$
for each and every $i$, implying
$A \vec v = 0 \tag{8}$
or
$\vec v \in \ker A, \tag{9}$
as desired.  QED.
Hope this helps.  Cheers,
and as always,
Fiat Lux!!!
A: Hint:  Try using $A^{T}Ax=0\implies x^{T}A^{T}Ax=0\implies (Ax)\cdot(Ax)=0$.
A: $$S:=\text{ker}(A^TA) = \{\textbf{x} \in \mathbb{R}^n : (A^TA)\textbf{x} = 0 \}$$
$$S':=\text{ker}(A) = \{\text{x} \in \mathbb{R}^n: A\textbf{x} =0 \}$$
$(1) \ \ S' \subset S$: This is fairly easy since, $A\textbf{x} = 0 \Rightarrow A^TAx =A^T\cdot0 =0$
$(2)$ Can you show $S \subset S'$?
A: The key is showing that if $A^T Ax = 0$, then $Ax = 0$.  There is a geometric interpretation to this statement.  Consider the vector $Ax$.  Since it is a linear combination of the columns of $A$, it clearly belongs to the column space of $A$.  And since $A^T (Ax) = 0$, $Ax$ belongs to the nullspace (kernel) of $A^T$.  But the column space of $A$ (also known as the row space of $A^T$) and the nullspace (kernel) of $A^T$ are orthogonal complements, and so in particular, the only vector that belongs to both subspaces is $0$.  Therefore, $Ax = 0$.
