I need to find a linear transformation $\ T: \mathbb R^4 \rightarrow \mathbb R^4 $ in which $\ (\ker T)^\perp = Sp\{(1,2,0,4),(-1,0,1,0) \} $ and $\ T(1,0,1,1) = (1,2,1,1) $
So to find KerT $\ (x,y,z,t) \cdot (1,2,0,4) = 0 \\ (x,y,z,t) \cdot (-1,0,1,0) = 0 \\ \ker T= Sp \{(-2,1,-2,0),(-4,0,-4,1)\} $
I'm not sure how can I pick a basis for ImT ? How do I come up with such transformation?
$(-2,1,-2,0),(-4,0,-4,1),(1,0,1,1)$ are linearly independent. You just have to find another vector that is linearly independent of these and a lucky guess is $(1,0,0,0)$. All these fact are very easy to verify if you first concentrate on the second coordinates. Now you can map $(-2,1,-2,0),(-4,0,-4,1)$ to $0$, $(1,0,1,1)$ to $(1,2,1,1)$ and $(1,0,0,0)$ to itself.