Dual bases

One more word before embarking on the proofs of the important theorems. The concept of dual space was defined without any reference to coordinate systems; a glance at the following proofs will show a superabundance of coordinate systems. We wish to point out that this phenomenon is inevitable; we shall be establishing results concerning dimension, and dimension is the one concept (so far) whose very definition is given in terms of a basis.

Theorem 1. If $𝒱$ is an n-dimensional vector space, if ${x_{1}, \dots, x_{n}}$ is a basis in $𝒱$ , and if ${α_{1}, \dots, α_{n}}$ is any set of $n$ scalars, then there is one and only one linear functional $y$ on $𝒱$ such that $[x_{i}, y] = α_{i}$ for $i = 1, \dots, n$ .

Proof. Every $x$ in $𝒱$ may be written in the form $x = ξ_{1} x_{1} + \dots + ξ_{n} x_{n}$ in one and only one way; if $y$ is any linear functional, then $[x, y] = ξ_{1} [x_{1}, y] + \dots + ξ_{n} [x_{n}, y] .$ From this relation the uniqueness of $y$ is clear; if $[x_{i}, y] = α_{i}$ , then the value of $[x, y]$ is determined, for every $x$ , by $[x, y] = \sum_{i} ξ_{i} α_{i}$ . The argument can also be turned around; if we define $y$ by $[x, y] = ξ_{1} α_{1} + \dots + ξ_{n} α_{n},$ then $y$ is indeed a linear functional, and $[x_{i}, y] = α_{i}$ . ◻

Theorem 2. If $𝒱$ is an $n$ -dimensional vector space and if $𝒳 = {x_{1}, \dots, x_{n}}$ is a basis in $𝒱$ , then there is a uniquely determined basis \mathcal{X}^{\prime} in \mathcal{V}^{\prime} , \mathcal{X}^{\prime}=\{y_{1}, \ldots, y_{n}\} , with the property that $[x_{i}, y_{j}] = δ_{i j}$ . Consequently the dual space of an $n$ -dimensional space is $n$ -dimensional.

The basis \mathcal{X}^{\prime} is called the dual basis of $𝒳$ .

Proof. It follows from Theorem 1 that, for each $j = 1, \dots, n$ , a unique $y_{j}$ in \mathcal{V}^{\prime} can be found so that $[x_{i}, y_{j}] = δ_{i j}$ ; we have only to prove that the set \mathcal{X}^{\prime}=\{y_{1}, \ldots, y_{n}\} is a basis in \mathcal{V}^{\prime} .

In the first place, \mathcal{X}^{\prime} is a linearly independent set, for if we had $α_{1} y_{1} +$ $\dots + α_{n} y_{n} = 0$ , in other words, if $[x, α_{1} y_{1} + \dots + α_{n} y_{n}] = α_{1} [x, y_{1}] + \dots + α_{n} [x, y_{n}] = 0$ for all $x$ , then we should have, for $x = x_{i}$ ,

$0 = \sum_{j} α_{j} [x_{i}, y_{j}] = \sum_{j} α_{j} δ_{i j} = α_{i} .$

In the second place, every $y$ in \mathcal{V}^{\prime} is a linear combination of $y_{1}, \dots, y_{n}$ . To prove this, write $[x_{i}, y] = α_{i}$ ; then, for $x = \sum_{i} ξ_{i} x_{i}$ , we have $[x, y] = ξ_{1} α_{1} + \dots + ξ_{n} α_{n} .$

On the other hand $[x, y_{j}] = \sum_{i} ξ_{i} [x_{i}, y_{j}] = ξ_{j}$

so that, substituting in the preceding equation, we get

\begin{align} {[x, y] } & =\alpha_{1}[x, y_{1}]+\cdots+\alpha_{n}[x, y_{n}] \\ & =[x, \alpha_{1} y_{1}+\cdots+\alpha_{n} y_{n}] \end{align}

Consequently $y = α_{1} y_{1} + \dots + α_{n} y_{n}$ , and the proof of the theorem is complete. ◻

We shall need also the following easy consequence of Theorem 2.

Theorem 3. If $u$ and $v$ are any two different vectors of the $n$ -dimensional vector space $𝒱$ , then there exists a linear functional $y$ on $𝒱$ such that $[u, y] \neq [v, y]$ ; or equivalently, to any non-zero vector $x$ in $𝒱$ there corresponds a $y$ in \mathcal{V}^\prime such that $[x, y] \neq 0$ .

Proof. That the two statements in the theorem are indeed equivalent is seen by considering $x = u - v$ . We shall, accordingly, prove the latter statement only.

Let $𝒳 = {x_{1}, \dots, x_{n}}$ be any basis in $𝒱$ , and let \mathcal{X}^{\prime}=\{y_{1}, \ldots, y_{n}\} be the dual basis in \mathcal{V}^{\prime} . If $x = \sum_{i} ξ_{i} x_{i}$ , then (as above) $[x, y_{j}] = ξ_{j}$ . Hence if $[x, y] = 0$ for all $y$ , and, in particular, if $[x, y_{j}] = 0$ for $j = 1, \dots, n$ , then $x = 0$ . ◻

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