Show that Null$(A) \cap $ Image$(A^T)=\{0\}$. 
Let $A$ be an $m \times n$ matrix with real entries.  
$(1)$ Show that Null$(A) \cap $ Image$(A^T)=\{0\}$.  
$(2)$ If for two suitable matrices $B$ and $C$, we have $AA^TB=AA^TC$, then show that $A^TB=A^TC$.

For the first case we know that Image$(A^T)=$Null$(A)^ \perp$ and Null$(A)^ \perp \cap $Null($A)=\{0\}$. Then from here we can show, but I want some another proof . Like, want to bring contradiction by considering $a \neq 0$ in the intersection.
And for the second one I am totally stuck off. I don't  realize that this can be true for any suitable $B$ and $C$. But its given to prove. Please help me to solve these.
 A: Regarding (1):
I think it is basically the same proof, but maybe it helps you: Assume $x$ lies in $\operatorname{Null}(A) \, \cap \, \operatorname{Image}(A^t)$, i.e. $Ax = 0$ and there exists a vector $y$ such that $A^ty = x$.
This yields $$0 = y^t Ax = y^t A A^t y = (A^ty)^t (A^ty) = \langle A^t y, A^t y \rangle = \langle x, x \rangle$$
where $\langle \, , \, \rangle$ denotes the standard scalar product.
If $x \not= 0$, then $\langle \, , \, \rangle$ is not positive definite. But every scalar product is positive definite, so this is a contradiction. Therefore, our vector $x$ must be $0$.
Regarding (2):


*

*First of all, if we choose any matrix $B$ and let $C = B$, then $AA^tB = AA^tC$ is always true, so this equality can definitely occur.

*If $B$ and $C$ are arbitrary matrices which satisfy $AA^t B = AA^t C$, then $0 = AA^tB - AA^tC = A(A^t B-A^t C)$. Let $D = A^t B- A^t C$ and $v_1, \dots, v_m$ its columns. Then $v_1,\dots,v_m$ lie in $\operatorname{Image}(A^t)$, as $D = A^t B - A^t C = A^t(B-C)$, so they are the images of the columns of $B-C$ under $A^t$. Furthermore, our vectors $v_1,\dots,v_m$ also lie in $\operatorname{Null}(A)$ because they are the columns of $D$ and $AD = 0$. Do you now know why $v_1,\dots, v_m$ are all $0$ and why $A^tB = A^t C$ must be true then?
