Is there a Linear Transformation from $T : \Bbb R^5 \rightarrow \Bbb R^4 $ so $$\operatorname{Ker}T = \{( x,y,z,t,w) \in \Bbb R^5 \; | \; x = 2y, \text{ and, } z = 2t = 3w\}$$ if so find an example of this transformation. It's enough to define it on base $\Bbb R^5 $. If not, explain why.
Well I tried to solve it and I'm not sure that what I've done was correct... I said that $x=2y$ , $z=3w$ and $t=\frac{3w}{2}$ so it should look like this
$$\operatorname{Ker}T = \left\{\left( 2y,y,3w,\frac{3w}{2},w \right) \right\}$$ so I got $x$ and $w$ out in order to get 2 vectors... $$\operatorname{Ker}T = \left\{y( 2,1,0,0,0)+w \left( 0,0,3,\frac{3}{2},1 \right) \right\}$$ I changed the second vector so it will be nicer and get $y$ and $w$ out so..
$$\operatorname{Ker}T = \operatorname{Sp}\{( 2,1,0,0,0),(0,0,6,3,2)\}$$
Hopefully so far I'm correct? since they are kernal i can say that:
$T(2,1,0,0,0)=(0,0,0,0)$ and $T(0,0,6,3,2)=(0,0,0,0)$
from here I defined 3 more vectors from (which I made)
$$T(0,1,0,0,0)=(0,1,0,0), T(0,0,0,1,0)=(0,0,1,0) \text{ and } T(0,0,0,0,1)=(0,0,0,1)$$
I wrote those matrix and used row reduction in order to see that it's not linearly dependent and because of that it is a base (english is not my main language so sorry if I didn't write it correctly)
$$\begin{bmatrix} 2 & 1 & 0 & 0 & 0 \\ 0 & 0 & 6 & 3 & 2 \\0 & 1 & 0 & 0 & 0 \\0 & 0 & 0 & 1 &0 \\0 & 0 & 0 & 0 &1\end{bmatrix};$$
and we get
$$\begin{bmatrix} 1 & 0 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 & 0 \\0 & 0 & 1 & 0 & 0 \\0 & 0 & 0 & 1 &0 \\0 & 0 & 0 & 0 &1\end{bmatrix};$$
so it must be a base therefore since $T(v_i) =w_i$ for each $i=1,2...n$. and it is linear transformation... Is it correct or am I wrong? Thank you very much!