Find the formula of the linear transformation

Question

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.

Answer 1

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 $f$: $$ f(x,y,z) = \left(-\frac{4}{3}x-\frac{2}{3}y+\frac{4}{3}z,-\frac{2}{3}x-\frac{1}{3}y+\frac{2}{3}z,-\frac{2}{3}x-\frac{1}{3}y+\frac{2}{3}z\right). $$