Is it possible when giving a basis for the subspace $N(A^T)$ that $N(A^T) $ = zero vector?
*Note: N(A^T) is the left nullspace
For example, if
$A=\begin{bmatrix} 1 &2 &-1& 0 \\ 2 & 4 & -1 & -1 \\ \end{bmatrix}$ and
$A^T=\begin{bmatrix} 1 &2\\ 2 & 4 \\ -1 & -1\\ 0&-1 \\ \end{bmatrix}$
Then reduced row echelon form is...
$\begin{bmatrix} 1 &2\\ 0 & 1 \\ 0 & 0\\ 0& 0 \\ \end{bmatrix}$
so would $x_1=-2x_2$ so N(A^T): (-2,1)
or is $x_1=0$ and $x_2=0$ so N(A^T): {0}?
I guess what's tripping me up is that fact that each variable is a pivot, so I feel like $N(A^T): (-2,1)$, but part of me thinks is should be$ N(A^T): {0}$. Which one is right? Can someone explain?
From $\begin{bmatrix} 1 &2\\ 0 & 1 \\ 0 & 0\\ 0& 0 \\ \end{bmatrix}*\begin{bmatrix} x_1\\ x_2 \\\end{bmatrix}=\begin{bmatrix} 0\\ 0\\ 0\\ 0 \\ \end{bmatrix}$ you get
$x_1+2x_2=0$ and $x_2=0$, hence $x_1=x_2=0$