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 $\operatorname{Sp}\{(1,1,1)\} = \ker (T)$?  
 A: The matrix associated to $T$ with respect to the basis $\mathscr{B}=\{(0,1,1),(1,0,1),(1,1,0)\}$ on the domain and the canonical basis on the codomain is
$$
A=\begin{bmatrix}
-1 & 1 & 1 \\
1 & -1 & -1 \\
1 & 1 & 0
\end{bmatrix}
$$
The matrix associated to $T$ with respect to the canonical basis on both the domain and the codomain is
$$
B=AS^{-1}
$$
where
$$
S=\begin{bmatrix}
0 & 1 & 1 \\
1 & 0 & 1 \\
1 & 1 & 0
\end{bmatrix}
$$
and
$$
S^{-1}=\begin{bmatrix}
-1/2 & 1/2 & 1/2 \\
1/2 & -1/2 & 1/2 \\
1/2 & 1/2 & -1/2
\end{bmatrix}
$$
so that
$$
B=\begin{bmatrix}
3/2 & -1/2 & -1/2 \\
-3/2 & 1/2 & 1/2 \\
0 & 0 & 1
\end{bmatrix}
$$
Can you compute the null space of $B$? That is, the set of vectors $v$ such that $Bv=0$, which is the kernel of $T$.
A: 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.
A: 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 $(1,-1,1)\in \ker T$?
A: 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 $\dim\operatorname{Ker} T=1$.
A: Hint.
Choose the basis $B = \{(0,1,1),\;(1,0,1),\;(1,1,0)\}$. 
What does the matrix representation $[T]^B_E$ of the linear map $T$ with respect to the bases $B$ and $E$ (the standard basis) look like?
Recall that the kernel of a matrix is invariant under basis transformation.
A: Assuming you mean the vector $\;(111)\;$ wrt the usual, canonical basis, first write it as a linear combination of the given basis:
$$\begin{pmatrix}\;\;1\\-1\\\;1\end{pmatrix}=a\begin{pmatrix}0\\1\\1\end{pmatrix}+b\begin{pmatrix}1\\0\\1\end{pmatrix}+c\begin{pmatrix}1\\1\\0\end{pmatrix}\iff$$
$$\begin{align*}b+c=1\\{}\\a+c=-1\\{}\\a+b=1\end{align*}\;\implies\;\;b-c=2\implies b=\frac32\;,\;\;a=c=-\frac12$$
Thus we get
$$T\begin{pmatrix}1\\-1\\1\end{pmatrix}=\frac12\left(-T\begin{pmatrix}0\\1\\1\end{pmatrix}+3T\begin{pmatrix}1\\0\\1\end{pmatrix}-T\begin{pmatrix}1\\1\\0\end{pmatrix}\right)=\ldots$$
If the above is zero you've found one non-zero element in the kernel and thus a basis for it, otherwise you need to find another vector that actually is in the kernel.
