Consider the dual space of $$\mathbb{R}$$ as a $$\mathbb{R}$$-vector space. By definition it consists of all linear maps $$f: \mathbb{R} \to \mathbb{R}$$ such that

\begin{align*} f(x+y) &= f(x) + f(y) \\ f(\lambda x) &= \lambda f(x) \end{align*}

for all real $$x,y , \lambda$$. We call elements of the dual space linear functionals.

Given an linear functional $$f$$, it’s clear that $$f(x) = f(1)\times x$$ for all real $$u$$. Similarly for all real numbers $$c$$ we have $$f(x) = cx$$ to be an linear functional. As such the dual space of $$\mathbb{R}$$ is $$\{f: \mathbb{R} \to \mathbb{R} \text{ s.t. } f(x) := cx \;\vert\; c \in \mathbb{R}\}$$.

That’s simple enough. How about the dual space of $$\mathbb{R}^2$$?

In essence, we seek all linear maps $$f:\mathbb{R}^2 \to \mathbb{R}$$ such that

\begin{align*} f((u_1, u_2)+(v_1, v_2)) &= f((u_1, u_2)) + f((v_1,v_2)) \\ f(\lambda (u_1, u_2)) &= \lambda f((u_1,u_2)) \end{align*}

for all real $$u_1, u_2, v_1, v_2, \lambda$$.

Fundamentally we want to make use of the theory we’ve built up for the dual space of $$\mathbb{R}$$. Given an linear functional $$f$$, let’s try to see what happens if we set $$u_2$$ and $$v_2$$ to be zero.

\begin{align*} f((u_1, 0)+(v_1, 0)) &= f((u_1, 0)) + f((v_1,0)) \\ f(\lambda (u_1, 0)) &= \lambda f((u_1,0)) \end{align*}

If we then define a new function $$g_1: \mathbb{R} \to \mathbb{R}$$ by $$g_1(x) := f(x, 0)$$, we observe that

\begin{align*} g_1(u_1 + v_1) &= g_1(u_1) + g_1(v_1) \\ g_1(\lambda u_1) &= \lambda g_1(u_1) \end{align*}

for all real $$u_1, v_1, \lambda$$.

This is exactly what it means for $$g_1$$ to be an linear functional of $$\mathbb{R}$$! As such $$g_1(x) = c_1 x$$ for some real $$c_1$$. Similarly we could define $$g_2(y) = f(0,y)$$ and we find out that $$g_2(y) = c_2 y$$ for some real $$c_2$$. Combining the two, we have

\begin{align*} f(u_1, 0) = g_1(u_1) = c_1 u_1 \\ f(0, u_2) = g_2(u_2) = c_2 u_2\end{align*}

As such $$f(u_1, u_2) = c_1 u_1 + c_2 u_2$$ by additivity. Similarly for any real numbers $$c_1$$ and $$c_2$$ we can see that $$f(u_1, u_2) = c_1 u_1 + c_2 u_2$$ is an linear functional of $$\mathbb{R}^2$$. We conclude that the dual space of $$\mathbb{R}^2$$ is $$\{f: \mathbb{R}^2 \to \mathbb{R} \text{ s.t. } f(u_1, u_2) = c_1 u_1 + c_2 u_2 \;\vert\; c_1, c_2\in \mathbb{R}\}$$.

## Dual Maps

Suppose we have a linear transformation $$T: \mathbb{R}^2 \to \mathbb{R}^2$$. We know it could be represented by a 2 by 2 matrix

$T\bigl(\begin{bmatrix} u_1 \\ u_2 \end{bmatrix}\bigr) = \begin{bmatrix} a & b \\ c & d \end{bmatrix} \begin{bmatrix} u_1 \\ u_2 \end{bmatrix} = \begin{bmatrix} v_1 \\ v_2 \end{bmatrix}$

In other words we could write

\begin{align*} au_1 + bu_2 &= v_1 \\ cu_1 + du_2 &= v_2 \end{align*}

but $$g_1(u_1, u_2) := au_1 + bu_2$$ and $$g_2(u_1, u_2) := cu_1 + du_2$$ are precisely linear functionals of $$\mathbb{R}^2$$! If we interepted $$h_1(v_1, v_2) = v_1$$ and $$h_2(v_1, v_2) = v_2$$ as linear functionals, then $$T$$ could be interepted as a map that maps linear functionals $$h_1$$ to $$g_1$$ and $$h_2$$ to $$g_2$$. This is precisely how the dual map of $$T$$ is defined.

## Further Remarks

Hopefully this makes dual spaces easier to grasp, in particular understanding why the dual basis is a natural and useful concept.

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