I have the following matrix: \begin{bmatrix}0&0&0\\0&0&1\\1&2&3\end{bmatrix} I'm tasked with finding the nullspace and the range of it. I can clearly see that the rank is 2. The nullspace is 1 because of some theorem where the nullspace+rank=number of rows or columns.
What I don't understand is why the nullspace is 1 and not 2 in this matrix.
If I rewrite it I get: 1x=-2y-3z=-2s-3t
y=s
z=t
s[-2,1,0]
t[-3,0,1]
Videos such as these: https://www.youtube.com/watch?v=EpcCNOe-oyE imply this is how you find the nullspace.
If $A$ is your matrix, then its null space is the set of those vectors $(x,y,z)$ such that $A.(x,y,z)=(0,0,0)$. But\begin{align}A.(x,y,z)=(0,0,0)&\iff\left\{\begin{array}{l}0=0\\z=0\\x+2y+3z=0\end{array}\right.\\&\iff\left\{\begin{array}{l}x=-2y\\z=0.\end{array}\right.\end{align}So, the null space is $\{(-2y,y,0)\mid y\in\mathbb R\}$, whose dimension is $1$.