Suppose $A$ is a $m \times n$ matrix and $x \in K^n$. Does $A^T\cdot Ax=0 \implies Ax=0$? I am trying hard to prove this but it's really difficult, i want to show the intersection of $Ker(A^T)$ and $im(A)$ is $0$ to prove the statement but I just end up with a loop.
Assuming $\Bbb K = \Bbb Q$, $K =\Bbb R$, or $K = \Bbb C$ and $A^T$ denotes the ordinary transpose (in the cases $\Bbb K = \Bbb Q, \Bbb R$) or the "conjugate transpose" or hermitian adjoint (also denoted $A^\dagger$, when $\Bbb K = \Bbb C$), we may say:
$A^T \cdot Ax = 0, \tag 1$
$x^T A^T \cdot Ax = 0, \tag 2$
$\Vert Ax \Vert^2 = (Ax)^T \cdot (Ax) = x^TA^T \cdot Ax = 0, \tag 3$
$Ax = 0. \tag 4$