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 $\operatorname{Sp}\{(1,1,1)\} = \ker (T)$?  

Let the basis for the domain be $B=\{v_1,v_2,v_3\}=\{(0,1,1),\;(1,0,1),\;(1,1,0)\}$. Let $w_1,w_2,w_3$ be the respective images of $v_i's$ under $T$.
A simple observation shows that: the set $\{w_1, w_2\}$ is linearly independent (as they are not multiples of each other) whereas $\{w_1,w_2,w_3\}$ is a dependent set because $w_1-w_2+2w_3=0$. This means the dimension of the range space is exactly $2$, hence the kernel will be of dimension $1$ (by the rank-nullity theorem).
In fact we can now get the basis vector for the kernel as well:
Since $w_1-w_2+2w_3=0$, this means $T(v_1-v_2+2v_3)=0$. Thus the vector $v_1-v_2+2v_3=(1, 3, 0)$ forms the basis of the kernel.
