Consider the dual space of \(\mathbb{R}\) as a \(\mathbb{R}\)-vector space. By definition it consists of all 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 \(u\). Also for all real numbers \(c\), \(f(x) = c \times x\) is a linear functional. As such the dual space of \(\mathbb{R}\) is \(\{f: \mathbb{R} \to \mathbb{R}, f(x) := cx \;\vert\; c \in \mathbb{R}\}\).

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

In essence, we seek all 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. We would have

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

for all real \(u_1, v_1, \lambda\).

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 \(\{f: \mathbb{R}^2 \to \mathbb{R}, f(u_1, u_2) = c_1 u_1 + c_2 u_2 \;\vert\; c_1, c_2\in \mathbb{R}\}\).

These ideas can be extended to find out the dual spaces of \(\mathbb{R}^n\) for \(n\geq 3\).

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.