For $ A \in M_{m \times n}(\mathbb{R})$, does $\ker(A)$ relate to $\ker(A^T) $? For $ A \in M_{m \times n}(\mathbb{R})$, does $\ker(A)$ relate to $\ker(A^T) $? And if so, what would be the connection? I wouldn't imagine there being any without additional conditions on $A$ (like symmetry or something), but I'm not sure.  
To provide some context, I'm trying to show that if $Ax = 0_m $ has a unique solution $x \in \mathbb{R^n}$, i.e. $\ker(A) = \{0_n\}$, then the matrix $A^TA$ is invertible. I reasoned that $Ax=0_m$ having a unique solution will not necessarily yield $\ker(A^TA) = \{0_n\}$ as $\ker(A^T)$ might not be $ \{0_m\}$. However, if $\ker(A) = \{0_n\}$ implies $\ker(A^T) = \{0_m\}$ then the solution should follow.
Thanks in advance,
Brandon      
 A: $\text{ker}(A)$ and $\text{ker}(A^T)$ are not related in general. However, $\text{ker}(A)$ and $\text{ker}(A^TA)$ do.
Lemma: $\text{ker}(A)=\text{ker}(A^TA)$.
Proof: It is obvious that $\text{ker}(A)\subset\text{ker}(A^TA)$. 
Conversely let $x\in \text{ker}(A^TA)$, i.e. $A^TAx=0$. Premultiply with $x^T$ to get
$$
x^T\underbrace{A^TAx}_{=0}=(Ax,Ax)=\|Ax\|^2=0\qquad\Rightarrow\quad Ax=0
\qquad\Rightarrow\quad x\in \text{ker}(A).
$$
It gives $\text{ker}(A^TA)\subset\text{ker}(A)$, hence, equality.
A: Suppose $x \in \mathbf{R}^n$ and $A^T A x = 0$. Then
\begin{align*}
0 &= \langle A^T Ax, x \rangle \\
&= \langle Ax, Ax \rangle,
\end{align*}
which implies that $Ax = 0$, which implies that $x = 0$ (because $Ker(A) = \{0\}$). Thus we have shown that $Ker(A^T A) = \{0\}$, which implies that the square matrix $A^TA$ is invertible.
A: The ranks of $A$ and $A^T$ are the same, and equal to the rank of $A^TA$ (this can be seen with singular value decomposition). Thus, if $m \leq n$, and $A$ is $m \times n$ and full rank $m$, then $A^TA$ will be invertible. However, if $m < n$ then $AA^T$ cannot be invertible, it is $n \times n$ and has rank $m$ where $m < n$.
A: You may be also interested in the following, familiar from the theory of Hilbert spaces operators, result: if $A:H \to H$ is some bounded operator on Hilbert space and $A^*$ denotes its adjoint then you have the decomposition
$\ker(A) \oplus \overline{im(A^*)}=H$. If your space is finite dimensional, then $A$ can be viewed as a matrix (with respect to some orthogonal basis) and if all of the entries are real, then $A^*$ becomes the transposition $A^*=A^T$ so the above decomposition reads as $\ker(A) \oplus \overline{im(A^T)}=H$. So in this case $\ker(A)$ relates rather to the image of $A^T$ 
A: A very useful fact to be aware of is that if $A \in M_{m \times n}(\mathbb R)$ then $A$ and $A^TA$ have the same null space:
\begin{align*}
& Ax = 0 \\
\implies & A^T A x = 0 \\
\implies & x^T A^T A x = 0 \\
\implies & \| Ax \|^2 = 0 \\
\implies & Ax = 0.
\end{align*}
Thus, if the null space of $A$ is trivial, then the null space of $A^T A$ is trivial, so $A^T A$ is invertible.
