MATH 3210 Class Notes 01-24-2025

notelinear-algebra

Defined Vector Spaces

Proposition 1

A vector space has a unique additive identity.

Proof 1

By the 3rd axiom of a vector space, we know there is at least one $0\in V$. Assume for that sake of contradiction there are two zeros: $0$ and $0^{\prime }$. Because $0^{\prime }$ is an additive identity, $0=0+0^{\prime }$, and since $0$ is an additive identity, $0=0^{\prime }$. Therefore, the additive identity is unique.

Proposition 2

Every element of $V$ has a unique additive inverse.

Proof 2

Assume $v\in V$ has 2 inverses: $w$ and $w^{\prime }$. Then $w=w+0=w+(v+w^{\prime })=(w+v)+w^{\prime }=0+w^{\prime }=w^{\prime }$, so $w=w^{\prime }$, and every element of $V$ has a unique additive inverse.

Notation

Now that we know additive inverses are unique, let

• $-v$ be the additive inverse of $v$.
• $w-v=w+(-v)$.

Defined Subspaces