Vector spaces

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Homework Statement



show whether the following set of vectors [tex]M = \left\{\left(a_{1},a_{2},a_{3}\right) with a_{1},a_{2},a_{3} \in \Re\right\}[/tex]

with the following limitations:
1) a1 is rational
2) a1 = 0
3) a1 + a2 = 0
4) a1 + a2 = 1

is a vector space over the field of real numbers.


Homework Equations


various axioms
* x+y = y+x.
* (x+y)+z = x+(y+z).
* 0+x = x+0 = x.
* (-x) + x = x + (-x) = 0.

For every x in X and real numbers c,d, we have

* 0x = 0
* 1x = x
* (cd)x = c(dx)

* c(x+y) = cx + cy.
* (c+d)x = cx +dx.


The Attempt at a Solution



I just don't get it, I really wish I could. I understand the axioms but when I apply them i find that M is a vector space regardless of the limitations.

somebody please help
 
Last edited:

Answers and Replies

  • #2
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You'll need to check a few things, first of all you'll need closure of the operations.
What do I mean with that?

Take [tex](a_1,a_2,a_3)[/tex] and [tex](b_1,b_2,b_3)[/tex] in M. Does [tex](a_1+b_1,a_2+b_2,a_3+b_3)[/tex] lie in M. In other words, does this triple satisfy the three limitations. This means that the addition is closed in M.

Take [tex](a_1,a_2,a_3)[/tex] in M and [tex]\alpha\in \mathbb{R}[/tex]. Does [tex](\alpha a_1,\alpha a_2, \alpha a_3)[/tex] lie in M. In other words, does this triple satisfy the three limitations. This means that scalar products are closed in M.

I conjecture that neither addition nor multiplication is closed in M (thus we are not dealing with a vector space). Can you give me examples why not?
 
  • #3
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how could you tell if
[tex](a_1+b_1,a_2+b_2,a_3+b_3)[/tex] was in M or not, in fact what is in M? I know at least vector [tex](a_1,a_2,a_3)[/tex]

and there are four limitations now - i just edited my op

thanks
 
  • #4
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I actually realized that you had 4 problems.

Let me solve problem 1 fo you:
So let M be the collection of triples [tex](a_1,a_2,a_3)[/tex] such that [tex]a_1[/tex] is rational.

First we check closure under addition. Take two triples [tex](a_1,a_2,a_3)[/tex] and [tex](b_1,b_2,b_3)[/tex] in M. This means that [tex]a_1[/tex] and [tex]a_2[/tex] are rational. We need to check that [tex](a_1,a_2,a_3)+(b_1,b_2,b_3)=(a_1+b_1,a_2+b_2,a_3+b_3)[/tex] is in M. For this, we just need to check that [tex]a_1+b_1[/tex] is rational. But since [tex]a_1[/tex] and [tex]a_2[/tex] are rational, so is their sum.
So closure under addition is checked.

Now, closure under multplication. Take a triple [tex](a_1,a_2,a_3)[/tex] in M (this means that [tex]a_1[/tex] is rational) and [tex]\alpha\in \mathbb{R}[/tex]. We need to check that [tex](\alpha a_1,\alpha a_2,\alpha a_3)[/tex] is in M. For this, we just need to check that [tex]\alpha a_1[/tex] is rational. But this is not always the case. For example, [tex](a_1,a_2,a_3)[/tex] could be [tex](1,0,0)[/tex] (this is indeed a vector in M) and [tex]\alpha[/tex] could be [tex]\pi[/tex], then [tex](\alpha a_1,\alpha a_2,\alpha a_3)=(\pi, 0,0)[/tex]. This is not in M, since [tex]\pi[/tex] is not rational.
So we do not have closure under multiplication, so we do not have a vector space.

Can you check the other examples? this is quite analogous.
 
  • #5
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thanks, that seems so much clearer. I knew that the result of the operation had to remain in M but I had no idea what that meant.
hmm, if I've understood it 2) a1 = 0 and 3) a1 + a2 = 0 are vector spaces over R and
4) a1 + a2 = 1 isn't because

[tex]a_{1} = 1 - a_{2}[/tex] and [tex]b_{1} = 1 - b_{2}[/tex]

[tex](a_{1} + b_{1}, a_{2} + b_{2}, a_{3} + b_{3}) = (1 - a_{2} + 1 - b_{2}, a_{2} + b_{2}, a_{3} + b_{3}) [/tex]

and

[tex]1 - a_{2} + 1 - b_{2} + a_{2} + b_{2} = 2 [/tex] != 1
 
  • #6
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3,297
Yes! I've think you've got it.

But note, for 2 and 3, you've just checked that the space is closed under the operations. You'll need to check the other axioms of a vector space to (associativity, neutral element,...). But this shouldnt be to much of a problem...

For 1 and 4, you've seen that they are not closed under the operations, so they cannot be vector spaces...
 

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