Is M a Vector Space Over Real Numbers?

[iloveannaw]

Homework Statement

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

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

 

[micromass]

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

Take $$(a_1,a_2,a_3)$$ in M and $$\alpha\in \mathbb{R}$$. Does $$(\alpha a_1,\alpha a_2, \alpha a_3)$$ 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?

 

[iloveannaw]

how could you tell if
$$(a_1+b_1,a_2+b_2,a_3+b_3)$$ was in M or not, in fact what is in M? I know at least vector $$(a_1,a_2,a_3)$$

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

thanks

 

[micromass]

I actually realized that you had 4 problems.

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

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

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

 

[iloveannaw]

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

$$a_{1} = 1 - a_{2}$$ and $$b_{1} = 1 - b_{2}$$

$$(a_{1} + b_{1}, a_{2} + b_{2}, a_{3} + b_{3}) = (1 - a_{2} + 1 - b_{2}, a_{2} + b_{2}, a_{3} + b_{3})$$

and

$$1 - a_{2} + 1 - b_{2} + a_{2} + b_{2} = 2$$ != 1

 

[micromass]

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 shouldn't 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...