Let $V=\mathbb{R}^3$, and define $f_1,f_2,f_3 \in V^{*}$ as follows: $f_1(x,y,z)=x-2y$, $f_2(x,y,z)=x+y+z$, $f_x(x,y,z)=y-3z$.Prove that ${f_1,f_2,f_3}$ is a basis for $V^{*}$, and then find a basis for V for which it is the dual basis.
My attempt: Let $\beta={x_1,x_2,x_3}$ for V. To show that ${f_1,f_2,f_3}$ is a basis for $V^{*}$, it suffices to show that ${f_1,f_2,f_3}$ is linearly independent and spans $V^{*}$.
to prove that ${f_1,f_2,f_3}$ spans $V^{*}$, take any $g \in V^{*}$, and show that $g=c_1 f_1+c_2 f_2+ c_3 f_3$ for some $c_1,c_2,c_3 \in f$.
Call $h=c_1 f_1+c_2 f_2+ c_3 f_3$. we want to show h=g, so it is enough to show that $g(x_i)=h(x_i)$ for any i.
$h(x_i)=(c_1 f_1+c_2 f_2+ c_3 f_3)(x_i)=g(x_i)f_i (x_i)=g(x_i)$. So $g=h \in span {f_1,f_2,f_3}$
to prove linearly independent, it suffices to show that $c_1 f_1+c_2 f_2+ c_3 f_3=0$ and that $c_1=_2=c_3=0$.
$g(x_1)=c_1 (f_1(x_1)+ c_2 f_2(x_1)=c_3 f_3(x_1)=c_1+0+0=0$, so $c_1=0$
The same goes for $x_2$, and $x_3$, and so we have prove linearly independent.
Can someone check my work?