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Let $V=(\Bbb R^3)$ and define $f_1, f_2, f_3 \in V^*$ as follows:

$$f_1(x, y, z) = x − 2y,f_2(x, y, z) = x + y + z, f_3(x, y, z) = y − 3z.$$

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 work for the part (a):

Suppose $c_1f_1+c_2f_2+c_3f_3=0$, the zero transformation, for some scalars $c_1,c_2,c_3$. Let $v=(x,y,z)$$\in(\Bbb R^3)$. Then $(c_1f_1+c_2f_2+c_3f_3)(v)=0$. So

\begin{align} 0 & =c_1f_1(v)+c_2f_2(v)+c_3f_3(v) \\[10pt] & =c_1(x-2y)+c_2(x+y+z)+c_3(y-3z) \\[10pt] & =(c_1+c_2)x+(-2c_1+c_2+c_3)y+(c_2-3c_3)z \end{align}

Note that all equations hold for all $(x,y,z)\in(\Bbb R^3)$. Thus
\begin{cases} c_1+c_2&=0 \\ -2c_1+c_2+c_3&=0 \\c_2-3c_3&=0 \end{cases}
It is not difficult to show that $(c_1,c_2,c_3)=(0,0,0)$ is the only solution of the equation. Thus $\{f_1,f_2,f_3\}$ is linearly independent. Since dim$V^*$= dim$V$= $3$, we conclude that $\{f_1,f_2,f_3\}$ is a basis for $V^*$.

Now I am stuck with the second part (b). Please anyone help me out.

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The first dual basis vector should be $(a, b, c) \in \mathbb{R}^3$ such that $f_1(a, b, c) = 1$, and $f_2(a, b, c) = f_3(a, b, c) = 0$. Write this system out to get

$$\begin{matrix} a & -&2b & && = & 1 \\ a & +&b & +&c & = & 0 \\ & &b & -&3c & = & 0 \end{matrix} $$

And solve. Now do the same for the second and third dual basis vectors.

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You must find $\{v_1,v_2,v_3\}$, $v_j \in \mathbb{R}^3$ such that $f_i(v_j) = \delta_{ij}$.

For example, $f_1(v_1) = x_1 - 2y_1 = 1$, $f_2(v_1) = x_1 + y_1 +z_1 = 0$, $f_3(v_1) = y_1 - 3z_1 = 0$. Solve this system, you get $v_1$.

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