91Ó°ÊÓ

The dual space V* is the set of all linear functionals on V, which are linear maps from V to its base field F (usually the Real numbers R or Complex numbers C). Mathematically, we can express it as: \[ V^{*} = \{ f: V \rightarrow F | f \text{ is linear} \}\]

For each \(b \in \mathcal{B}\), we define a map \(b^{\prime} \in V^{*}\) such that \(b^{\prime}(c) = 1\) if \(c = b\), and 0 otherwise, for all \(c \in \mathcal{B}\). To determine if the given set \(\{b^{\prime} \mid b \in \mathcal{B}\}\) forms a basis for V*, we need to show that this set is linearly independent and spans V*. First, let's show that the set is linearly independent: Assume that we have a linear combination of the elements in the set that equates to the zero functional, i.e., \[\sum_{b \in \mathcal{B}} k_b b^{\prime} = 0\] where \(k_b \in F\). Now, let's apply this functional to any basis element \(c \in \mathcal{B}\): \[\sum_{b \in \mathcal{B}} k_b b^{\prime}(c) = 0\] Due to the definition of \(b^{\prime}\), only one term in the sum will be non-zero, which is the term corresponding to \(c = b\). So, we have \(k_c = 0\), which implies that all coefficients \(k_b\) are zero. Thus, the set \(\{b^{\prime} \mid b \in \mathcal{B}\}\) is linearly independent. Next, showing that the set spans V* is more difficult due to the infinite dimension of V. In fact, it turns out that the given set does not span V*. To see why, consider a functional \(f \in V^{*}\), such that for any finite subset \(S \subseteq \mathcal{B}\), we have: \[f \left( \sum_{b \in S} b \right) = 1\] Clearly, this functional cannot be a linear combination of elements from the set \(\{b^{\prime} \mid b \in \mathcal{B}\}\) (if it were, it would evaluate to 0 for any element b in B that is not in S). Therefore, the given set does not span V*, and thus cannot form a basis for V*.

We have found that the given set of functionals \(\{b^{\prime} \mid b \in \mathcal{B}\}\) is linearly independent but does not span the dual space V*. This implies that for infinite-dimensional vector spaces, there may exist functionals in V* not formed by a linear combination of the basis element functionals. Consequently, the concept of a dual basis (i.e., a basis for V* formed from the linear functionals of the basis elements of V) is not sufficient to describe the entire dual space V* for infinite-dimensional vector spaces.