Check of a problem about nullspace

[Zero2Infinity]

Homework Statement

Let $V\subset \mathbb{R}^3$ be the subspace generated by $\{(1,1,0),(0,2,0)\}$ and $W=\{(x,y,z)\in\mathbb{R}^3|x-y=0\}$. Find a matrix $A$ associated to a linear map $f:\mathbb{R}^3\rightarrow\mathbb{R}^3$ through the standard basis such that its nullspace is $V$.

Homework Equations

Nullspace definition

The Attempt at a Solution

Using the nullspace definition I get that $f(0,2,0)=f(1,1,0)=(0,0,0)$. Thus,
\begin{equation} f(0,2,0) =(0,0,0)\end{equation}
\begin{equation}f(1,1,0) = (0,0,0) \end{equation}
\begin{equation}f(0,0,1)=(1,1,1) \end{equation}
Since the matrix has to be written wrt the standard basis of $\mathbb{R}^3$, which is $(1,0,0),(0,1,0),(0,0,1)$, I infer that
\begin{equation} f(1,0,0)=f(1,1,0)-\frac{1}{2}f(0,2,0)=(0,0,0)-\frac{1}{2}(0,0,0)=(0,0,0) \end{equation}
\begin{equation}f(0,1,0)=\frac{1}{2}f(0,2,0)=\frac{1}{2}(0,0,0)=(0,0,0)\end{equation}
while I know from the text of the problem that $f(0,0,1)=(1,1,1)$. In conclusion,
\begin{equation} A=\begin{pmatrix} 0&0&1\\0&0&1\\0&0&1\end{pmatrix}\end{equation}
If I look for a basis of the kernel I effectively get the two vectors $(1,1,0),(0,2,0)$, so it should be ok. Is it correct?

 

[andrewkirk]

Let $V\subset \mathbb{R}^3$ be the subspace generated by $\{(1,1,0),(0,2,0)\}$ and $W=\{(x,y,z)\in\mathbb{R}^3|x-y=0\}$.

Does this mean

1. Let $V\subset \mathbb{R}^3$ be the subspace generated by $\{(1,1,0),(0,2,0)\}$ and let $W=\{(x,y,z)\in\mathbb{R}^3|x-y=0\}$; or does it mean
2. Let $V\subset \mathbb{R}^3$ be the subspace generated by $\{(1,1,0),(0,2,0)\}\cup\{(x,y,z)\in\mathbb{R}^3|x-y=0\}$?

If it means the second one then $V=\mathbb R^3$, so $f$ must be the zero map. If it means the first one then $W$ is not used in the problem, which raises the question of why it has been defined.