3
$\begingroup$

If the set $V$ is defined by the points that go through the origin in $\mathbb{R}^2$ that satisfy the equation $ax+by=0$ then show $V$ is a vector space.

Resolution:Proving that $V$ is closed under addition and scalar multiplication, I know how to do this. How do I prove that addition is commutative $A+B=B+A$ , what resolution is correct?

Define $A=(x_1,y_1), B=(x_2,y_2)$

$1) \: A+B=(x_1,y_1)+(x_2,y_2)=(x_1+x_2,y_1+y_2)=(x_2+x_1,y_2+y_1)=(x_2,y_2)+(x_1,y_1)=B+A$

$2) \: A+B=(x_1+x_2,y_1+y_2)$, subs in to the line equation $a(x_1+x_2)+b(y_1+y_2)=a(x_2+x_1)+b(y_2+y_1)$

$ B+A=(x_2+x_1,y_2+y_1)$ sub that in to the line equation gives $a(x_2+x_1)+b(y_2+y_1)$ and that is what we have above so

A+B=B+A

Note: to prove that $v$ is closed under addition and scalar multiplication the second method is used.

What the right way to prove this?

(As an aside do I have to prove to each of the sets to prove that are vector spaces besides proving closed under addition and scalar multiplication do i have to prove 8 more ,like the one above and i.e. $c(u+v)=cu+cv$ being $u,v$ vectors and $c$ a scalar) (And hence memorinzing 8 more axioms)

$\endgroup$
4
  • $\begingroup$ I have edited your post to make it easier to read. Next time please be more careful with both grammar and math formatting. $\endgroup$
    – Arthur
    Dec 9, 2014 at 11:31
  • $\begingroup$ Where, what are those points? The real plane? The complex plane? Some space over a finite field, or over a field with positive characteristic...?? $\endgroup$
    – Timbuc
    Dec 9, 2014 at 11:37
  • $\begingroup$ you only need to show that V (which is a line of points through the origin) is closed under addition and scalar multiplication to show it is a vector subspace of ${{\mathbb{R}}^{2}}$. If $(x_1, y_1)$ is on this line then $ax_1+by_1 = 0$ is a hint on how to proceed. $\endgroup$
    – Paul
    Dec 9, 2014 at 11:44
  • $\begingroup$ Thank you all for your comments $\endgroup$
    – user173788
    Dec 9, 2014 at 14:57

2 Answers 2

Reset to default
3
$\begingroup$

To show that $V$ is closed under addition, we need to show that $A+B$ lies in $V$ (i.e. satisfies your linear equation) for any $A,B \in V$. Letting $A = (x_1,y_1)$ and $B = (x_2,y_2)$, we have to prove that $A+B = (x_1+x_2,y_1+y_2)$ satisfies the equation, i.e. $a(x_1+x_2) + b(y_1+y_2) = 0$. This is easy with our assumptions, namely we get $a(x_1+x_2) + b(y_1+y_2) = (ax_1 + by_1) + (ax_2 + by_2) = 0$. Proving that $V$ is closed under scalar multiplication is done in a similar way.

For your second question, you won't have to check all the axioms one by one. The reason is that $V$ is a subset of $\mathbb{R^2}$, which we already know is a vector space. You should know the following definition of a subspace.

Let $W$ be a vector space. Then a subset $V \subset W$ is a vector space (subspace of $W$) if and only if the following conditions hold:

  • $0 \in V$
  • $V$ is closed under addition
  • $V$ is closed under multiplication

Applying this with $W = \mathbb{R}^2$ should solve the problem much quicker than checking each of the axioms.

$\endgroup$
3
  • $\begingroup$ Thank you for your answer,i know that i do need to check every axiom to prove/disprove a subspace of vector space is a vector space. $\endgroup$
    – user173788
    Dec 9, 2014 at 15:01
  • $\begingroup$ However the text that i am reading says to verify the remaining axioms for the pratice, can you help me to know which of the answer above are correct or none, and add that to your answer.once again thank you $\endgroup$
    – user173788
    Dec 9, 2014 at 15:03
  • $\begingroup$ Correction:For the 1st comment type it wrong and is the oposite that i want to say ,what i meant was ...i don`t have to check every axiom... $\endgroup$
    – user173788
    Dec 9, 2014 at 15:12
0
$\begingroup$

Both 1) and 2) are correct, just differen ways of showing u+v=v+u being u,v vectors

$\endgroup$

You must log in to answer this question.