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

We can simply put our vectors in the matrix and do row operations in the following way - we are trying to get a basis in the right part:
$$\left(
\begin{array}{ccc|ccc}
0 & 1 & 1 &-1 & 1 & 1\\
1 & 0 & 1 & 1 &-1 & 1\\
1 & 1 & 0 & 1 &-1 & 0
\end{array}
\right)\sim
\left(
\begin{array}{ccc|ccc}
0 & 1 & 1 &-1 & 1 & 1\\
0 &-1 & 1 & 0 & 0 & 1\\
1 & 1 & 0 & 1 &-1 & 0
\end{array}
\right)\sim
\left(
\begin{array}{ccc|ccc}
1 & 2 & 1 & 0 & 0 & 1\\
0 &-1 & 1 & 0 & 0 & 1\\
1 & 1 & 0 & 1 &-1 & 0
\end{array}
\right)\sim
\left(
\begin{array}{ccc|ccc}
1 & 1 & 0 & 1 &-1 & 0\\
0 &-1 & 1 & 0 & 0 & 1\\
1 & 2 & 1 & 0 & 0 & 1
\end{array}
\right)\sim
\left(
\begin{array}{ccc|ccc}
1 & 1 & 0 & 1 &-1 & 0\\
0 &-1 & 1 & 0 & 0 & 1\\
1 & 3 & 0 & 0 & 0 & 0
\end{array}
\right)$$
What have we found out by doing this? 
Notice that at the beginning we have $a|b$, where $T(a)=b$ in each row. This property is not changed using row operations.
So we see that $T(1,3,0)=(0,0,0)$. This means that $(1,3,0)\in\operatorname{Ker} T$.
We have also found out that the image is generated by the vectors $(1,-1,0)$ and $(0,0,1)$. (If we look only on the right part, then we have tried to row reduce the matrix consisting of images of basis vectors.) Since these vectors are linearly independent, we get that $\dim\operatorname{Im} T =2$. By rank-nullity theorem we know that $\dim\operatorname{Ker} T=1$.
