Let \(A = \begin{bmatrix} 1 & -1 & -1 &3\\ 2 & -2 & 0 & 4\end{bmatrix}\).
Do the vectors
\(\begin{bmatrix} -2 \\ 0 \\ 1 \\ 1\end{bmatrix}\) and
\(\begin{bmatrix} 1 \\ 0 \\ -1/2 \\ -1/2\end{bmatrix}\)
span the nullspace of \(A\)?
The answer is “No”.
Applying the elementary row operation
\(R_2 \leftarrow R_2 -2 R_1\) to \(A\) gives
\(\begin{bmatrix} 1 & -1 & -1 &3\\ 0 & 0 & 2 & -2\end{bmatrix}\).
Hence, there are going to be two non-pivot columns and so the nullspace
is spanned by two linearly independent vectors. However,
the two given vectors are scalar multiple of each other and so they
cannot span \(N(A)\).