Kernel of linear transformation in $\Bbb R^3$ Let  $T: \Bbb R^3 \rightarrow \Bbb R^3$ be a linear transformation satisfying
\begin{align*}
T(0,1,1) =& (-1,1,1) \\
T(1,0,1) =& (1,-1,1) \\
T(1,1,0) =& (1,-1,0) .
\end{align*}
Is it necessary true that $\ker(T) = \operatorname{Sp}\{(1,-1,1)\}$ ?
Well, I tried to say that we know that $\operatorname{Im}(T) = \operatorname{Sp}\{T(0,1,1),\,T(1,0,1),\,T(1,1,1)\}$
So, $\operatorname{Im}(T) = \operatorname{Sp}\{(-1,1,1),\,(1,-1,1),\,(1,-1,0)\}$ which means $\operatorname{Sp}\{(1,-1,1)\} \in \operatorname{Im}(T)$
and also $(1,1,1)$ is linearly independent by $2$ other vectors which are in $\operatorname{Im}(T)$.
Now, how can I prove that $\operatorname{Sp}\{(1,1,1)\}$ not inside $\ker(T)$? or maybe $\operatorname{Sp}\{(1,1,1)\} \in \ker (T)$  which makes it $\operatorname{Sp}\{(1,1,1)\} = \ker (T)$?  

You know that ${\rm im}\; f=\langle (-1,1,1),(1,-1,1),(1,-1,0)\rangle$. This is the same as
 $$\langle (-1,1,1)\color{red}{+(1,-1,0)},(1,-1,1),(1,-1,0)\rangle=\langle (0,0,1),(1,-1,1),(1,-1,0)\rangle$$ which in turn is the same as $$\langle (0,0,1),(1,-1,1)\color{blue}{-(0,0,1)},(1,-1,0)\rangle=\langle (0,0,1),(1,-1,0),(1,-1,0)\rangle$$
Thus the dimension of the image is $2$. By the rank-nullity theorem, the dimension of the kernel is $1$. It follows that if you find $v\in\ker T$ nonzero, $\langle v\rangle =\ker T$. Can you check whether $(1,-1,1)\in \ker T$?
