a) Ta có :
- $f(\alpha x)=f(\alpha x_{1},\alpha x_{2},\alpha x_{3})$
$=(2\alpha x_{1}-6\alpha x_{2}+2\alpha x_{3};\alpha x_{1}-3\alpha x_{2}+\alpha x_{3};3\alpha x_{1}-9\alpha x_{2}+3\alpha x_{3})$
$=\alpha(2x_{1}-6x_{2}+2x_{3};x_{1}-3x_{2}+x_{3};3x_{1}-9x_{2}+3x_{3})$
$=\alpha f(x)$
- $f(x+x')=f(x_{1}+x'_{1},x_{2}+x'_{2},x_{3}+x'_{3})$
$=2(x_{1}+x'_{1})-6(x_{2}+x'_{2})+2(x_{3}+x'_{3}),(x_{1}+x'_{1})-3(x_{2}+x'_{2})+(x_{3}+x'_{3}),3(x_{1}+x'_{1})-9(x_{2}+x'_{2})+3(x_{3}+x'_{3})$
$=(2x_{1}-6x_{2}+2x_{3},x_{1}-3x_{2}+x_{3},3x_{1}-9x_{2}+3x_{3})+(2x'_{1}-6x'_{2}+2x'_{3},x'_{1}-3x'_{2}+x'_{3},3x'_{1}-9x'_{2}+3x'_{3})$
$=f(x)+f(x')$
=> f là 1 ánh xạ tuyến tính
b) $Ker f=\left \{ (x_{1},x_{2},x_{3}):2x_{1}-6x_{2}+2x_{3}=0;x_{1}-3x_{2}+x_{3}=0;3x_{1}-9x_{2}+3x_{3}=0 \right \}$
Xét hệ pt :
$\left\{\begin{matrix} 2x_{1}-6x_{2}+2x_{3}=0\\ x_{1}-3x_{2}+x_{3}=0\\ 3x_{1}-9x_{2}+3x_{3}=0 \end{matrix}\right.$
$\Leftrightarrow \left\{\begin{matrix} x_{1}=3a-b\\ x_{2}=a\\ x_{3}=b \end{matrix}\right.$
=> 1 cơ sở của Ker f là ( 3,1,0) , ( -1,0,1) và dim (Ker f) = 2
$dim(Kerf)+dim(Imf)=3$
$\Rightarrow dim(Imf)=1$
c) Đặt $U=\left \{ u_{1},u_{2},u_{3} \right \}$
$[f(u_{1})]_{U}=\begin{pmatrix} -4\\ -2\\ -6 \end{pmatrix}$ , $[f(u_{2})]_{U}=\begin{pmatrix} 4\\ 2\\ 6 \end{pmatrix}$ , $[f(u_{3})]_{U}=\begin{pmatrix} -4\\ -2\\ -6 \end{pmatrix}$
=> Ma trận của f trong cơ sở U
$A=\begin{pmatrix} -4 &4 &-4 \\ -2 &2 &-2 \\ -6 &6 &-6 \end{pmatrix}$