Find $T(x,y,z)$ for $(x,y,z)$ in $\mathbb R^3$ I have a question with the following facts:
$T: \mathbb R^3\to\mathbb R^3$ is a linear transformation represented by the basis $B=((1,0,0),(1,1,0),(1,1,1))$, by the matrix $[T]_{B}$, where $[T]_{B}$ is:
$\begin{bmatrix}-1 & 2 & 0\\-1 & -2 & 0 \\ 0 & 0 &0\end{bmatrix}$
We also know that $(2,2,2)\in\ker T$.
My objective:
To determine $T(x,y,z)$ for each $(x,y,z)\in\mathbb R^3$
I know I have to use the standard basis but I don't know why, or how.
Any help is appriciated.
 A: let $a=(1,0,0),b=(1,1,0),c=(1,1,1)$
Hence $T(a)=-a-b=(-2,-1,0),T(b)=2a-2b=(0,-2,0),T(c)=0$ [This is concluded from the matrix]
Thus $(0,1,0)=b-a\Rightarrow T(0,1,0)=T(b)-T(a)=(2,-1,0)$
And $(0,0,1)=c-b\Rightarrow T(0,0,1)=T(c)-T(b)=(0,2,0)$
So we have that $T(x,y,z)=xT(1,0,0)+yT(0,1,0)+zT(0,0,1)=(-2x+2y,-x-y+2z,0)$
A: In the standard basis:
$$T((1,0,0))=(-2,-1,0)$$
$$T((1,1,0))=(0,-2,0)$$
$$T((1,1,1))=(0,0,0)$$
$$\therefore T((0,1,0))=(0,-2,0)-(-2,-1,0)=(2,-1,0)$$
$$\therefore T((0,0,1))=(0,0,0)-(2,-1,0)-(-2,-1,0)=(0,2,0)$$
$$\therefore T=\begin{bmatrix}-2 & 2 & 0\\-1 & -1 & 2 \\ 0 & 0 &0\end{bmatrix}$$
Also $\ker T=\langle(1,1,1)\rangle$ as you suggested.
A: Actually, I think we don't need to know that $(2, 2, 2) \in \mathrm{ker} T$. Let $S$ be the standard basis of $\Bbb R^3$. Then: $$ [I]_{B, S} = \left[ \begin{array}{c c c} 1 & 1 & 1 \\ 0 & 1 & 1 \\ 0 & 0 & 1 \end{array} \right]$$
We invert this matrix, and get: $$[I]_{S,B} = \left[ \begin{array}{c c c} 1 & -1 & 0 \\ 0 & 1 & -1 \\ 0 & 0 & 1 \end{array} \right] $$
Finally, you can use that: $$ [T]_S = [I]_{B,S} \cdot [T]_B \cdot [I]_{S,B} $$
Having $[T]_S$, multiply it by the vector $(x,y,z)^t$ (as column) to obtain $T(x,y,z)$ in the standard basis. Also, verify that $[T]_S \cdot (2,2, 2)^t = (0,0, 0)^t$.
(I have to go to the doctor now, if by the time I'm back you haven't managed to solve it yet, I'll edit and elaborate more my answer)
