Exercise 11.2 (The Dual Basis)

Proof. To prove that $\left({𝜀}^1,\dots ,{𝜀}^n\right)$ is a base for $V^{\text{∗}}$, we verify that it is linearly independent and that it spans $V^{\text{∗}}$.

Linear independence. Suppose for the sake of contradiction that $\left({𝜀}^1,\dots ,{𝜀}^n\right)$ are linearly dependent, i.e., we can find scalars $a_1,\dots ,a_n$, not all zero, such that

Note that the $0$ on the right denotes the function that sends every element of $V$ to $0\in ℝ$. The above equality above is an equality of maps, which should hold on any $v$ we evaluate either side on. In particular, it should hold for all $E_j$ for $j=1,\dots ,n$, i.e.,

In other words, $a_j=0$ for all $j=1,\dots ,n$ - a contradiction.

Span. Now we show that $\left({𝜀}^1,\dots ,{𝜀}^n\right)$ spans $V^{\text{∗}}$. Let $\omega \in V^{\text{∗}}$. For each $j=1,\dots ,n$, let $w_j$ denote the scalar $\omega (w_j)$. We claim that

By Exercise B.13, to verify that two functions, LHS and RHS, are equal, it suffices to demonstrate that they agree on the basis elements. Indeed, for each $1\le j\le n$,

by the definition of the $w_j$ and the ${𝜀}^j$. Thus, $\omega$ and $w_1{𝜀}^1+\dots w_n{𝜀}^n$ agree on the basis, so we conclude that they are equal as elements of $V^{\text{∗}}$. Hence $\left({𝜀}^1,\dots ,{𝜀}^n\right)$ spans $V^{\text{∗}}$. □