The following is problem 13 from chapter 3F of the third edition of Axler's Linear Algebra Done Right
Define $T:\mathbb{R}^3\to\mathbb{R}^2$ by
$$T(x,y,z)=(4x+5y+6z, 7x+8y+9z)\tag{1}$$
Suppose $\phi_1,\phi_2$ denotes the dual basis of the standard basis of $\mathbb{R}^2$ and $\psi_1, \psi_2, \psi_3$ denotes the dual basis of the standard basis of $\mathbb{R}^3$.
(a) Describe the linear functionals $T'(\phi_1)$ and $T'(\phi_2)$.
(b) Write $T'(\phi_1)$ and $T'(\phi_2)$ as linear combinations of $\psi_1,\psi_2,\psi_3$.
Someone else has asked about their solution to this problem before, though without any answers.
Here is my attempt at a solution
a)
By definition of dual basis we have
$$\phi_j v_i=\begin{cases} 1,\ \text{if}\ i=j \\ 0,\ \text{if}\ i\neq j \end{cases},\ i=1,2\tag{2}$$
$$\psi_j w_i=\begin{cases} 1,\ \text{if}\ i=j \\ 0,\ \text{if}\ i\neq j \end{cases},\ i=1,2,3\tag{3}$$
Then, by definition of dual map we have
$$T'\phi_i = \phi_i\circ T \in L(\mathbb{R}^3,\mathbb{R})$$
Every vector in $\mathbb{R}^3$ that $T$ maps to $(1,0)\in\mathbb{R}^2$, $T'\phi_1$ maps to 1 and it maps all other vectors to 0.
Similarly, every vector in $\mathbb{R}^3$ that $T$ maps to $(0,1)\in\mathbb{R}^2$, $T'\phi_2$ maps to 1 and it maps all other vectors to 0.
Finding such vectors means solving
$$T(x,y,z)=(4x+5y+6z,7x+8y+9z)=(1,0)\tag{4}$$ $$T(x,y,z)=(4x+5y+6z,7x+8y+9z)=(0,1)\tag{5}$$
I'll omit the calculations but the solutions to (4) are $(-4c_1,\frac{7}{3}c_2,0)$ and the solutions to (5) are $(-5c_1,-\frac{4}{3}c_2,0)$.
Thus,
$$T'\phi_1v = \begin{cases} 1,\ \text{if}\ v= (-4c_1,\frac{7}{3}c_2,0) \\ 0,\ \text{otherwise} \end{cases}\tag{6}$$
$$T'\phi_2v = \begin{cases} 1,\ \text{if}\ v= (-5c_1,-\frac{4}{3}c_2,0) \\ 0,\ \text{otherwise} \end{cases}\tag{7}$$
b)
Since
$$T\psi_1=(4,7)\tag{8}$$ $$T\psi_2=(5,8)\tag{9}$$ $$T\psi_3=(6,9)\tag{10}$$
then the matrix of $T$ relative to the standard bases of $\mathbb{R}^2$ and $\mathbb{R}^3$ is
$$\begin{bmatrix} 4 & 5 & 6\\ 7 & 8 & 9 \end{bmatrix}\tag{11}$$
and the matrix of $T'$ relative to the dual bases of $\mathbb{R}^3$ and $\mathbb{R}^2$ is the transpose of (11) which is
$$\begin{bmatrix} 4 & 7 \\ 5 & 8 \\ 6 & 9 \end{bmatrix}\tag{12}$$
Thus,
$$T'\phi_1 = 4\psi_1 + 5\psi_2+6\psi_3\tag{13}$$
$$T'\phi_2 = 7\psi_1 + 8\psi_2+9\psi_3\tag{13}$$