Let $T$ from $R^3$ $\to$ $R^2$ be a transformation defined as $$T \left(\left(\begin{matrix} x \\ y \\ z\end{matrix}\right)\right) \to \left(\begin{matrix} xy \\ xz \end{matrix}\right)$$ Find a basis for $N(T)$ (null space/ kernel of $T$). *Note the original request is to verify if T is linear trough the dimension theorem, so in order to do it you need to find the dimension (i.e number of basis) of $N(T)$.
My reasoning is, let $v$ be a generic vector $\left(\begin{matrix} x \\ y \\ z\end{matrix}\right)$ in $R^3$ it gets transformed to the $0$ vector by $T$ in two cases $$ v1 = \left(\begin{matrix} 0 \\ r \\ s\end{matrix}\right) r,s \in R$$ This case ($v1$) is the only considered by the book.
$$ v2 = \left(\begin{matrix} h \\ 0 \\ 0\end{matrix}\right) h \in R$$ The book doesn't mention $v2$ so from there the doubt if I am wrong or what. Now following, a basis for $v1$ can be $\lbrace\left(\begin{matrix} 0 \\ 1 \\ 0\end{matrix}\right);\left(\begin{matrix} 0 \\ 0 \\ 1\end{matrix}\right) \rbrace$; Same as the book which concludes that $dim(N(T)) = 2$.
Now a basis for $v2$ would be $\{\left(\begin{matrix} 1 \\ 0 \\ 0\end{matrix}\right)\}$, but why doesn't the book even consider it?
I'm stuck here, finding a basis that works for both $v1$ and $v2$ would bring me to $\lbrace\left(\begin{matrix} 1 \\ 0 \\ 0\end{matrix}\right);\left(\begin{matrix} 0 \\ 1 \\ 0\end{matrix}\right);\left(\begin{matrix} 0 \\ 0 \\ 1\end{matrix}\right) \rbrace$ which is a basis for $R^3$ which indeed spans $\{v1; v2\}$ but spanning all $R^3$ it goes further.
As you can see I don't have clear ideas, can someone give me some hint or tell me where I'm wrong?
Thanks
*Sorry for my english