Vector Augend plus Addend equals Augend implies Addend is Zero
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Theorem
Let $V$ be a vector space over a field $F$.
Let $\mathbf a, \mathbf b \in V$.
Let $\mathbf a + \mathbf b = \mathbf a$.
Then:
- $\mathbf b = \bszero$
where $\bszero$ is the zero vector of $V$.
Proof
\(\ds \mathbf a + \mathbf b\) | \(=\) | \(\ds \mathbf a\) | ||||||||||||
\(\ds \leadsto \ \ \) | \(\ds \paren {-\mathbf a} + \paren {\mathbf a + \mathbf b}\) | \(=\) | \(\ds \paren {-\mathbf a} + \mathbf a\) | |||||||||||
\(\ds \leadsto \ \ \) | \(\ds \paren {\paren {-\mathbf a} + \mathbf a} + \mathbf b\) | \(=\) | \(\ds \paren {-\mathbf a} + \mathbf a\) | Vector Space Axiom $\text V 2$: Associativity | ||||||||||
\(\ds \leadsto \ \ \) | \(\ds \bszero + \mathbf b\) | \(=\) | \(\ds \bszero\) | Vector Space Axiom $\text V 4$: Inverses | ||||||||||
\(\ds \leadsto \ \ \) | \(\ds \mathbf b\) | \(=\) | \(\ds \bszero\) | Vector Space Axiom $\text V 3$: Identity |
$\blacksquare$
Sources
- 1969: C.R.J. Clapham: Introduction to Abstract Algebra ... (previous) ... (next): Chapter $7$: Vector Spaces: $\S 32$. Definition of a Vector Space: Theorem $64 \ \text{(i)}$