Cardinality of Generator of Vector Space is not Less than Dimension/Proof 2
Jump to navigation
Jump to search
Theorem
Let $V$ be a vector space over a field $F$.
Let $\BB$ be a generator for $V$ containing $m$ elements.
Then:
- $\map {\dim_F} V \le m$
where $\map {\dim_F} V$ is the dimension of $V$.
Proof
From Generator of Vector Space Contains Basis there exists a basis $B$ of $E$ such that $B \subseteq G$.
From Cardinality of Basis of Vector Space, $\card B = n$.
The result follows.
$\blacksquare$
Sources
- 1965: Seth Warner: Modern Algebra ... (previous) ... (next): Chapter $\text {V}$: Vector Spaces: $\S 27$. Subspaces and Bases: Theorem $27.14$
- 1969: C.R.J. Clapham: Introduction to Abstract Algebra ... (previous) ... (next): Chapter $7$: Vector Spaces: $\S 34$. Dimension: Theorem $67 \ \text{(iii)}$
- 1996: John F. Humphreys: A Course in Group Theory ... (previous) ... (next): $\text{A}.2$: Linear algebra and determinants: Theorem $\text{A}.7$