Let \(V\) be a vector space over the field \(F\). Suppose there are a finite number of vectors \(\alpha_1, \ldots, \alpha_r\) in \(V\) which span \(V\). Prove that \(V\) is finite-dimensional.
Solution
1. Definition of finite-dimensionality:
1.1. A vector space \(V\) over a field \(F\) is finite-dimensional if it possesses a finite basis (a finite linearly independent subset of \(V\) that spans \(V\)).
Hoffman–Kunze definition of finite-dimensional vector spaces.
1.2. If \(V = \{0\}\), the empty set \(\emptyset\) is a basis with \(0\) elements, so \(V\) is finite-dimensional.
\(\operatorname{span}(\emptyset) = \{0\}\).
2. Extract a linearly independent spanning subset from \(S = \{\alpha_1, \dots, \alpha_r\}\):
2.1. Assume \(V \neq \{0\}\), so \(S\) contains at least one non-zero vector.
a spanning set for a non-trivial space must contain a non-zero vector.
2.2. Let \(\mathcal{F} = \{T \subseteq S : T \text{ is linearly independent over } F\}\).
definition of the collection of independent subsets.
2.3. Since any non-zero vector \(\{\alpha_i\} \in \mathcal{F}\), \(\mathcal{F}\) is non-empty.
a single non-zero vector is linearly independent.
2.4. Since \(S\) is a finite set of \(r\) elements, the power set \(\mathcal{P}(S)\) is finite, so \(\mathcal{F}\) is a finite non-empty collection of sets.
subset of a finite set is finite.
2.5. Choose a maximal element \(\mathcal{B} = \{\alpha_{i_1}, \dots, \alpha_{i_k}\} \in \mathcal{F}\) under set inclusion (e.g. an independent subset of maximal cardinality).
every finite non-empty poset has a maximal element.
3. Show that \(\mathcal{B}\) is a basis for \(V\):
3.1. By definition of \(\mathcal{F}\), \(\mathcal{B}\) is linearly independent.
\(\mathcal{B} \in \mathcal{F}\).
3.2. Let \(\alpha \in S\) be any vector in the spanning set.If \(\alpha \in \mathcal{B}\), then \(\alpha \in \operatorname{span}(\mathcal{B})\).
elements of a set belong to its span.
3.3. If \(\alpha \notin \mathcal{B}\), then by maximality of \(\mathcal{B}\) in \(\mathcal{F}\), the set \(\mathcal{B} \cup \{\alpha\}\) is linearly dependent.
maximality of \(\mathcal{B}\).
3.4. Thus there exist scalars \(c_1, \dots, c_k, c \in F\), not all zero, such that:\[ c_1 \alpha_{i_1} + \cdots + c_k \alpha_{i_k} + c \alpha = 0. \]
definition of linear dependence.
3.5. If \(c = 0\), then \(\sum_{m=1}^k c_m \alpha_{i_m} = 0\), which forces \(c_1 = \cdots = c_k = 0\) since \(\mathcal{B}\) is linearly independent, contradicting that not all coefficients are zero.Thus \(c \neq 0\).
linear independence of \(\mathcal{B}\).
3.6. Dividing by \(c\) gives:\[ \alpha = -\sum_{m=1}^k \frac{c_m}{c} \alpha_{i_m} \in \operatorname{span}(\mathcal{B}). \]
field arithmetic.
3.7. Since every \(\alpha \in S\) lies in \(\operatorname{span}(\mathcal{B})\), we have \(S \subseteq \operatorname{span}(\mathcal{B})\).Since \(S\) spans \(V\), this implies \(V = \operatorname{span}(S) \subseteq \operatorname{span}(\mathcal{B}) \subseteq V\), so \(\operatorname{span}(\mathcal{B}) = V\).
span of a spanning set is \(V\).
4. Conclusion: \(\mathcal{B}\) is a finite basis for \(V\) containing \(k \le r\) elements, so \(V\) is finite-dimensional with \(\dim V = k \le r\).
1 and 3.
Q.E.D.