given a linear map $T: \mathbb{R}^4 \to \mathbb{R}^2$ which exists:
$$T(x, y, z, w) = \left(\begin{array}{cccc} 1 & 1 & 1 & 1\\ 2 & 0 & -1 & 1 \end{array}\right)\cdot\left(\begin{array}{c} x\\ y\\ z\\ w \end{array}\right)$$
Find a basis for $\ker T$.
My attempt: $\left(\begin{array}{cccc} 1 & 1 & 1 & 1\\ 2 & 0 & -1 & 1 \end{array}\right)\cdot\left(\begin{array}{c} x\\ y\\ z\\ w \end{array}\right) = \left(\begin{array}{c} x+y+z+w\\ 2x -z +w \end{array}\right) = \left(\begin{array}{c} 0 \\ 0 \end{array}\right) $
Let $z \triangleq s, \ w \triangleq t$
preforming calculations, I have:
$$ \operatorname{Ker} T = \operatorname{sp} \left\{ \left(\begin{array}{c}1\\-3\\2\\0\end{array}\right),\left(\begin{array}{c}1\\1\\0\\-2\end{array}\right)\right\} $$
Is this correct?
A good thing about this exercise is that you can check yourself.
Denote the vectors you found as $v,w$. If $Tv=Tw=0$ and $v$ and $w$ are independent, then you found a basis for $kerT$.