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 \(x=0\) . ◻