# How do you prove that if the rows of a matrix are linearly dependent then the null space does not consist only of the zero vector?

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!

this is false, consider the transformation $f(x,y)\rightarrow (x,y,0)$. Clearly the kernel is just $(0,0)$.
$\begin{pmatrix} 1 & 0 \\ 0 & 1 \\ 0 & 0 \\ \end{pmatrix}$.
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?