Finding the dual of linear map

Question

Given $F: \mathbb{R}^2\to \mathbb{R}^3$ by $F(x,y)=(y,x,x+y)$, I have calculated its dual map $F':\mathbb{R}^3\to \mathbb{R}^2$ as $F'(x,y,z)=(x+y+z,x+z)$ by using its matrix representation w.r.t the standard dual basis. Is there any way to deduce it directly (without the matrix representation) using the definition of algebraic adjoint/dual as in the following wikipedia article:

https://en.wikipedia.org/wiki/Transpose_of_a_linear_map

Answer 1

If $\langle \cdot,\cdot \rangle$ is the dot product, then \begin{align} \langle F(x,y), (u,v,w) \rangle &= \langle (y,x,x+y), (u,v,w) \rangle \\ &= yu+xv+(x+y)w \\ &= x(v+w)+y(u+w) \\ &= \langle (x,y), (v+w,u+w) \rangle. \end{align} Thus, $F’$ is the map defined by $$ F’(u,v,w) = (v+w,u+w). $$

Answer 2

Well let's check: \begin{align*} (a,b,c)\cdot F(x,y) = (a,b,c)\cdot (y,x,x+y) = ay+bx+c(x+y) = x(b+c)+y(a+c) \end{align*} while \begin{align*} F'(a,b,c)\cdot (x,y) = (a+b+c,a+c)\cdot (x,y) = x(a+b+c)+y(a+c) \end{align*} which is not the same. Since there is an $ax$ too much I'd say the correct answer is $F'(a,b,c) = (b+c,a+c)$ if I didn't make any mistake in my calculations