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 \mathcal{V} is an n-dimensional vector space, if \{x_{1}, \ldots, x_{n}\} is a basis in \mathcal{V} , and if \{\alpha_{1}, \ldots, \alpha_{n}\} is any set of n scalars, then there is one and only one linear functional y on \mathcal{V} such that [x_{i}, y]=\alpha_{i} for i=1, \ldots, n .
Proof. Every x in \mathcal{V} may be written in the form x=\xi_{1} x_{1}+\cdots+\xi_{n} x_{n} in one and only one way; if y is any linear functional, then [x, y]=\xi_{1}[x_{1}, y]+\cdots+\xi_{n}[x_{n}, y]. From this relation the uniqueness of y is clear; if [x_{i}, y]=\alpha_{i} , then the value of [x, y] is determined, for every x , by [x, y]=\sum_{i} \xi_{i} \alpha_{i} . The argument can also be turned around; if we define y by [x, y]=\xi_{1} \alpha_{1}+\cdots+\xi_{n} \alpha_{n}, then y is indeed a linear functional, and \left[x_{i}, y\right]=\alpha_{i} . ◻
Theorem 2. If \mathcal{V} is an n -dimensional vector space and if \mathcal{X}=\{x_{1}, \ldots, x_{n}\} is a basis in \mathcal{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}]=\delta_{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 \mathcal{X} .
Proof. It follows from Theorem 1 that, for each j=1, \ldots, n , a unique y_{j} in \mathcal{V}^{\prime} can be found so that [x_{i}, y_{j}]=\delta_{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 \alpha_{1} y_{1}+ \cdots+\alpha_{n} y_{n}=0 , in other words, if [x, \alpha_{1} y_{1}+\cdots+\alpha_{n} y_{n}]=\alpha_{1}[x, y_{1}]+\cdots+\alpha_{n}[x, y_{n}]=0 for all x , then we should have, for x=x_{i} ,
0=\sum_{j} \alpha_{j}[x_{i}, y_{j}]=\sum_{j} \alpha_{j} \delta_{i j}=\alpha_{i}.
In the second place, every y in \mathcal{V}^{\prime} is a linear combination of y_{1}, \ldots, y_{n} . To prove this, write [x_{i}, y]=\alpha_{i} ; then, for x=\sum_{i} \xi_{i} x_{i} , we have [x, y]=\xi_{1} \alpha_{1}+\cdots+\xi_{n} \alpha_{n}.
On the other hand [x, y_{j}]=\sum_{i} \xi_{i}[x_{i}, y_{j}]=\xi_{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=\alpha_{1} y_{1}+\cdots+\alpha_{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 \mathcal{V} , then there exists a linear functional y on \mathcal{V} such that [u, y]\neq[v, y] ; or equivalently, to any non-zero vector x in \mathcal{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 \mathcal{X}=\{x_{1}, \ldots, x_{n}\} be any basis in \mathcal{V} , and let \mathcal{X}^{\prime}=\{y_{1}, \ldots, y_{n}\} be the dual basis in \mathcal{V}^{\prime} . If x=\sum_{i} \xi_{i} x_{i} , then (as above) [x, y_{j}]=\xi_{j} . Hence if [x, y]=0 for all y , and, in particular, if [x, y_{j}]=0 for j=1, \ldots, n , then x=0 . ◻
