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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!

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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.

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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?

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