I also have a separate question that asks to prove that if a system $AX = b$ has infinitely many solutions, then the null space does not consist only of the zero vector.
I am thinking they're asking the same thing, as I know linearly dependent rows imply at least one row of zeros in the $RREF$ and imply that the matrix is non-invertible (same as infinite solutions).
However, I'm not sure on how to proceed with the proof.
Any help and hints are appreciated.
Thanks!
If a matrix has linear independent rows it has full rank. Using the rank theorem we conclude that $Ax=0$ only has the trivial solution. What can we conclude if $A$ does NOT have full rank?
this is false, consider the transformation $f(x,y)\rightarrow (x,y,0)$. Clearly the kernel is just $(0,0)$.
the matrix when we take the canonical basis is clearly:
$\begin{pmatrix} 1 & 0 \\ 0 & 1 \\ 0 & 0 \\ \end{pmatrix}$.
The result it true for square matrices though, because liner independence of the rows would imply that the rank is smaller than the dimension of the domain.