Finding Basis of Kernel $\Bbb R^4 \to \Bbb R^2$ 
The linear transformation $T:\Bbb R^4→\Bbb R^2$ defined by
  $$T(x,y,z,s)=(2x−2y+z+2s,4y−4x−5s)$$
Give a basis for the kernel of T.
Basis of $\operatorname{Ker}(T)$ is { ... }
Enter your answers as comma separated lists of vectors, 
for example $(1,2,3),(4,5,6)$

So, I have this question. I solved some questions similar to this but they were $\Bbb R^3$ to $\Bbb R^2$ and I could eliminate some letters for example: I found $x(-1,0,1)$ and eliminated $y$ and $z$, but in this one, I couldn't eliminate any letters so I'm stuck. Can someone help me with this?

First, we need to find the elements $(x,y,z,s)$ of the kernel. That is find the elements  $(x,y,z,s)$ such that:
\begin{cases} 2x-2y+z+2s = 0 \\ 4y-4x-5s=0\end{cases}
Solving that, we get:
$$x = t-\frac{5}{4}w\quad ; \quad z=\frac{1}{2}w$$
Where $t,w \in \mathbb{R}$
So the elements in the kernel are:
\begin{align}
(x,y,z,s)&=\left( t-\frac{5}{4}w,t,\frac{1}{2}w,w\right) \\
&=(t,t,0,0)+\left( -\frac{5}{4}w,0,\frac{1}{2}w,w\right) \\
&=t(1,1,0,0)+w\left( -\frac{5}{4},0,\frac{1}{2},1\right)
\end{align}
Then you need check that $(1,1,0,0)$ and $\left( -\frac{5}{4},0,\frac{1}{2},1\right)$ are linearly independent and generate the kernel.
