# Proof for the necessity of conditions for a subspace

In [Axler 2015], Theorem 1.34 states that

A subset $U$ of $V$ is a subspace of $V$ if and only if $U$ satisfies the following three conditions:

• additive identity: $0\in U$;
• closed under addition: $u,w\in U$ implies $u+w\in U$;
• closed under multiplication: $a\in\mathbb{F}$ and $u\in U$ implies $au\in U$.

The proof says

If $U$ is a subspace of $V$, then $U$ satisfies the three conditions above by definition of vector space.

However, I cannot see why a vector space is closed under addition and multiplication by definition, which is given by (1.19)

A vector space is a set $V$ along with an addition on $V$ and a scalar multiplication on $V$ such that:

• commutativity: $u+v = v+u$, $\forall u,v \in V$;
• associativity: $(u+v)+w = u+(v+w)$ and $(ab)v=a(bv)$, $\forall u,v,w \in V$ and $a,b \in \mathbb{F}$;
• additive identity: $\exists 0 \in V$ such that $v+0=v$, $\forall v \in V$;
• additive inverse: for every $v \in V$, $\exists w \in V$ such that $v+w=0$;
• multiplicative identity: $1v=v$, $\forall v \in V$;
• distributivity: $a(u+v)=au+av$ and $(a+b)v=av+bv$, $\forall a,b \in \mathbb{F}$ and $u,v \in V$.

Any suggestions? Thanks!

UPDATE:

By Definition 1.18

• An addition on a set $V$ is a function that assigns an element $u+v\in V$ to each pair of elements $u,v\in V$.
• A scalar multiplication on a set $V$ is a function that assigns an element $\lambda v\in V$ to each $\lambda\in \mathbb{F}$ and each $v\in V$.

In other words, addition and scalar multiplication are defined to be closed, which answers the question.

Thank you very much for your help!

[Axler 2015] Axler, Sheldon. Linear algebra done right. Third Edition. Heidelberg: Springer, 2015.

• It would be helpful if you gave the definition of a vector space Axler uses for those that don't have access to the book – Hayden Apr 23 '15 at 10:04
• @Hayden Definition added. – B. Groeger Apr 23 '15 at 10:52
• Having an addition means being closed under addition. Similarly for scalar multiplication. – Gerry Myerson Apr 23 '15 at 10:56
• @Gerry Thank you for pointing that out. – B. Groeger Apr 23 '15 at 12:03

A vector space over a field $K$ is a non-empty set $V$ together with mappings $A: V \times V \to V$ and $M: K \times V \to V$ such that...
Here the two mappings $A$ and $M$ define what we want our addition and multiplication to look like, i.e. $x+y := A(x,y)$ and $\lambda x := M(\lambda, x)$.
Notice that by definition $A$ and $M$ take values (only) in $V$. This means that whenever we add to vectors $x,y$ or multiply a vector $x$ with a scalar $\lambda$, the result will still be in $V$. Hence, one says that $V$ is closed under addition and multiplication.
• Thank you for the answer. It turns out I've missed the point that addition and scalar multiplication on $V$ are defined to be closed. – B. Groeger Apr 23 '15 at 11:45