# Linear Algebra Vector Space

From textbook definition of vector spaces, we know that it is a set V together with an operation of addition, x+y for any x,y $\in$ V, and an operation of scalar multiplication, sx for any x$\in$R.

This is straight-forward, but how would you know if x+y of x,y $\in$ V, would actually be in V?

To give some context, how would you know that S is closed under addition? Given $$S =\left\{\begin{pmatrix} x_1\\x_2\\x_1+x_2\\\end{pmatrix}\ :\ x_1x_2 \epsilon, R\right\}$$

I know that we should set $$\vec x = \begin{pmatrix} x_1\\x_2\\x_1+x_2\end{pmatrix},$$ $$\vec y = \begin{pmatrix} y_1\\y_2\\y_1+y_2\end{pmatrix},$$

and the sum of x and y would be closed under addition, but why is it so and how do we know for sure that the sum of $\vec x$ and $\vec y$ is going to be an element of V? (Reasoning behind this)

• I know that addition in this case is used to show preservation of addition, but how would you know that x+y is an element of V without first knowing that V is a vector space? Nov 16, 2015 at 0:34

Closure under addition of a vector space is an axiom of being a vector space, along with closure under scalar multiplication, and the presence of a zero vector. More formally, the canonical vector space axioms are as follows.

1) Additive Closure: $\forall x,y\in V, x+y\in V$.

2) Closure under Scalar Multiplication: $\forall x\in V$ and $\forall \lambda \in\mathbb{F}$, $\lambda x\in V$.

3) $\exists 0\in V: \forall x\in V, 0+x=x$.

Moreover, a vector space must satisfy commutativity under addition, associativity under addition, existence of additive inverses, existence of a multiplicative identity, and distributivity of scalar multiplication over vector addition.

So, to answer your question, it's by assumption that we know this property holds. That is, $x+y\not\in V$, for $x,y\in V$, then $V$ is not a vector space over $\mathbb{F}$.

• So is it that you ASSUME x+y = z, and see that z holds addition operation? Then later with the presence of closure under scalar multiplication and a zero vector, that you conclude that V is in fact a vector space? Nov 16, 2015 at 0:42
• See the edited post- it's a bit better written. The point being, that additive closure of vector addition is a necessary condition for being a vector space. If it isn't closed under addition, we wouldn't call it a vector space, so to speak. Nov 16, 2015 at 0:44
• So you would (1) assume x+y = z $\epsilon$ V, (2) see that addition property holds (3) see that multiplication and zero element are present (4) conclude that V is a vector space? Nov 16, 2015 at 0:45
• In order to verify that this is the case, you can not make the assumption- if that's what you were asking then I may have misunderstood. If you want to verify that this is indeed a vector space, suppose that $x,y\in V$. Then given your definition of the set $V$, show that $x+y\in V$. For the concrete method of doing this for your example, see Kyle's post above. Nov 16, 2015 at 0:47
• Likewise, do the same for the other axioms. All of this depends on the definition of the set you are considering, and the field over which it is defined. Nov 16, 2015 at 0:48

We want to compare this to the definition of $S$. If we let $a = x_1 + y_1$ and $b = x_2 + y_2$, we get
$$x+y = \begin{pmatrix} a\\b\\a+b\end{pmatrix}, \text{with } a, b\in \mathbb{R}$$.
This matches the definition of $S$, so we conclude that $x+y \in S$. We show closure under scalar multiplication the same way.