Let \(V\) be a set with cardinality at least 2 and let \((V,+)\) be an Abelian group with respect to the operation \(+\). Let \(\mathbb{R}\) be the field of real numbers. For any \(c\in\mathbb{R}\) and \(v\in V\), define
\(c\cdot v = 0\), where \(0\) is the identity element in the Abelian group \((V,+)\).
Is \((V,+,\cdot)\) a vector space?
Solution
- Step1:For a vector space over \(\mathbb{R}\), scalar multiplication must satisfy the axiom \(1 \cdot v = v\) for all \(v \in V\).
- Step2:For every \(v \in V\) and every scalar \(c \in \mathbb{R}\), we have \(c \cdot v = 0\).
- Step3:For any \(v \in V\), \(1 \cdot v = 0\).
- Step4:The axiom requires \(1 \cdot v = v\), but we instead get \(1 \cdot v = 0\). This forces \(v = 0\) for all \(v\).
- Step5:That would mean \(V = \{0\}\), a singleton set.
- Step6:The problem states that \(|V| \geq 2\). Hence there exists some nonzero element \(v \neq 0\), but the scalar multiplication collapses it to zero, contradiction.
Conclusion: No — \((V,+,\cdot)\) is not a vector space (unless \(V=\{0\}\), which contradicts the given |V| ≥ 2).
