Find the formula of the linear transformation Consider the linear transformation $f:R^3\rightarrow R^3$ with    $f(0,1,2)=(2,1,1)$     and     $\ker f=\{(0,2,1),(1,0,1)\}.$ Find:
(i) $f(0,1,0)$ (ii) the general formula of $f$.
Any help, please.

Now, 
$$
(0,1,0)=\frac{1}{3}\big[2(0,2,1)-(0,1,2)\big],
$$
hence
$$f(0,1,0)
=\frac{1}{3}\big[2 \cdot f(0,2,1)-f(0,1,2)\big]
=\frac{1}{3}\big[2 \cdot 0 -(2,1,1)\big]
=\left(-\frac{2}{3},-\frac{1}{3},-\frac{1}{3}\right).
$$
Now,
$$
(0,0,1)
=\frac{1}{2}\big[(0,1,2)-(0,1,0)\big],
$$
hence
$$
f(0,0,1)
=\frac{1}{2}\left[(2,1,1)-(-\frac{2}{3},-\frac{1}{3},-\frac{1}{3})\right]
=\left(\frac{4}{3},\frac{2}{3},\frac{2}{3}\right)
$$
Furthermore,
$$
(1,0,0)
=(1,0,1)-(0,0,1),
$$
hence
$$f(1,0,0)
=f(1,0,1)-f(0,0,1)
=-f(0,0,1)
=\left(-\frac{4}{3},-\frac{2}{3},-\frac{2}{3}\right).
$$
We can get the formula of $f$:
$$
f(x,y,z)
= \left(-\frac{4}{3}x-\frac{2}{3}y+\frac{4}{3}z,-\frac{2}{3}x-\frac{1}{3}y+\frac{2}{3}z,-\frac{2}{3}x-\frac{1}{3}y+\frac{2}{3}z\right).
$$
