Coordinates of the linear function in dual space.

Question

Assume we have dual basis {$\phi_1,\phi_2, \phi_3 $} in $V^*$ of basis {$v_1, v_2, v_3$} in $\mathbb{R}^3$. How can i find coordinates of the function $f(x) = x_1 - x_2 + 4x_3$ in basis {$\phi_1, \phi_2, \phi_3$}.

Answer 1

Write $f=b_1\phi_1+b_2\phi_2+b_3\phi_3$. Since $\{\phi_1,\phi_2,\phi_3\}$ is the dual basis of $\mathcal{B}=\{v_1,v_2,v_3\}$, the following holds $$ \phi_i(v_j)= \begin{cases} 1 \text{ if } i=j\\ 0 \text{ otherwise} \end{cases} $$

Then go to compute $f(v_1)$. $v_1$ has coordinates $(1,0,0)$ with respect to $\mathcal{B}$, hence $f(v_1)=1$. On the other hand, $f(v_1)=b_1$, hence $b_1=1$. A similar reasoning applied to $v_2$ and $v_3$ tells you that $b_2=-1$ and $b_3=4$.

Hence, $$f=\phi_1-\phi_2+4\phi_3$$