The question is:
Consider the linear transformation $L: \mathbb{R}^3 \to \mathbb{R}^3 : (x,y,z) \mapsto (2x+y-z,y-2z,-2x-z)$. When $U = \text{Span}\{(0,0,1),(1,1,1)\} \subset \mathbb{R}^3$ is a linear subspace of $(\mathbb{R},\mathbb{R}^3,+)$, what is $\{v \in \mathbb{R}^3 | L(v) \in U\}$?
I've been searching for quite a while now, thinking that I should find $\{X | L_\beta^\beta X \in \text{Span}\{(0,0,1),(1,1,1)\}\}$ ($\beta$ the standard basis of $\mathbb{R}^3$). I found that $L_\beta^\beta = \begin{pmatrix} 2&1&-1\\0&1&-2\\-2&0&-1\end{pmatrix}$, where the columns are the images of the basis vectors of $\mathbb{R}^3$. So I tried to solve:
$$ \begin{pmatrix} 2&1&-1\\0&1&-2\\-2&0&-1\end{pmatrix} X = \lambda_1 \begin{pmatrix} 0\\0\\1\end{pmatrix} + \lambda_2\begin{pmatrix} 1\\1\\1\end{pmatrix}$$
This gave me a set of solutions with one free variable and $\lambda_2$ in it (I found $ V = \text{Span} \{ (\frac{3}{2}\mu,\lambda_2 + 2\mu,\mu) | \mu \in \mathbb{R} \} $), but I have no idea what to do with it now.
The answer is $\text{Span}\{(-1,4,2),(1,0,-2)\}\}$ (not unique of course).
Edit: so looking back at my solution and at the given solution, what I had been doing was not very wrong. I should have taken one solution of $\{X | L_\beta^\beta X = (0,0,1)\}$ and one of $\{X | L_\beta^\beta X = (1,1,1)\}$, and then the span of those two is $\{v \in \mathbb{R}^3 | L(v) \in U\}$, since the transformation is linear.
Best Answer
Well, this leads to the systems of linear equations $(2x+y−z,y−2z,−2x−z) = (0,0,1)$ and $(2x+y−z,y−2z,−2x−z) = (1,1,1)$. Then take the span of these solutions giving the span of $\{(−1,4,2),(1,0,−2)\}$.