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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)

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  • $\begingroup$ 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? $\endgroup$ – Corp. and Ltd. Nov 16 '15 at 0:34
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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 $\mathbb{F}$.

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  • $\begingroup$ 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? $\endgroup$ – Corp. and Ltd. Nov 16 '15 at 0:42
  • $\begingroup$ 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. $\endgroup$ – Alekos Robotis Nov 16 '15 at 0:44
  • $\begingroup$ 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? $\endgroup$ – Corp. and Ltd. Nov 16 '15 at 0:45
  • $\begingroup$ 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. $\endgroup$ – Alekos Robotis Nov 16 '15 at 0:47
  • $\begingroup$ 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. $\endgroup$ – Alekos Robotis Nov 16 '15 at 0:48
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Adding the vectors gives

\begin{align}x + y &=\begin{pmatrix} x_1\\x_2\\x_1+x_2\end{pmatrix} + \begin{pmatrix} y_1\\y_2\\y_1+y_2\end{pmatrix}\\ &= \begin{pmatrix} x_1+y_1\\x_2+y_2\\x_1+x_2+y_1+y_2\end{pmatrix}\\ &= \begin{pmatrix} x_1+y_1\\x_2+y_2\\(x_1+y_1)+(x_2+y_2)\end{pmatrix} \end{align}

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.

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  • $\begingroup$ Thank you. I wish I could also choose this as best answer. $\endgroup$ – Corp. and Ltd. Nov 16 '15 at 0:53
  • $\begingroup$ No Problem. The key on these problems is that you are verifying the vector space axioms, so we don't know if V is a vector space yet. However, we do know how to add column vectors, and that addition is closed in the real numbers. The closure of addition for the set S "carries over" from the real numbers in this sense. $\endgroup$ – Kyle Funk Nov 16 '15 at 0:58
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After adding x and y, you can set (x+y) = z where z is a real number. This can be done since adding 2 real numbers will still result in a real. Then the set would be in V. You know for sure that it is an element of V because it is impossible for it not to. In simple terms you would get (z1,z2,z3+z4)^t. t=transpose. Which would look identical to S.

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