# On proving real vector spaces (subspaces)

1. Jan 31, 2004

### franz32

I hope someone can help me (guide) in this theorem.

How can I show that a "subset W of a vector space V is indeed
a subspace of V if and only if given u and v as vectors in W and
a and b are said to be scalars, then au + bv is in W."?

Can I assume a vector with my desired number of elements?
Also I am sure that W is a subset of V because it's given.

2. Jan 31, 2004

### HallsofIvy

Staff Emeritus
What do you mean by "number of elements"? The vector space is not stated to have a particular dimension (and in fact, the statement is true for infinite dimensional vectors spaces as well) so I wouldn't recommend trying to do this by looking at components.

What is the definition of "subspace". What you need to do is show that if "given u and v as vectors in W and
a and b are said to be scalars, then au + bv is in W", then all of the properties of the definition of subspace are satisfied.
(Hint: many of them, such as commutativity of addition, follow from the fact that u, v are in the V and those are true for V.)

You will also need to show(since this is "if and only if" that, IF W is a subspace of V, THEN "given u and v as vectors in W and a and b are said to be scalars, then au + bv is in W" but that's much easier.

3. Jan 31, 2004

### franz32

Hallsof Ivy

Hello there.

Yeah, what I mean about the "number of elements" is that the number of components in a vector. But anyway, ... it may not be the concern.

So au + bv is in W... does it mean that au + bv = bv + au or
au + (bv + cw) = (au + bv) + cw for w as a vector and c as a constant?

How about the converse of that statement? WHat is the approach?

4. Feb 2, 2004

### HallsofIvy

Staff Emeritus
Yes, of course. au+bv is in W so it is in V. au+ bv= bv+ au is true because V is a vector space and so addition is commutative.
au, bv, and cw are in W so they are in V. au + (bv + cw) = (au + bv) + cw because addition is associative in V.

(By the way, a, b, and c are "scalars" or "numbers". They are not necessarily "constant".)

5. Feb 2, 2004

### franz32

Thank you very much

Hello HallsofIvy!

Now, I get it... =) thank you. PArdon me for the thread on

the resistances. I didn't take a glance at it and I just

simply placed it there without knowing that it was easy.

6. Mar 12, 2009

### xiaobai5883

i want to similar question...
the question is shown as below...
Explain why the set w={(x,y)€R^2;|x|=|y|}, is not a real subspace.
anyone can help me??
thanks a lot...

7. Mar 12, 2009

### ThirstyDog

Consider vectors
$$(2,2), (-1,1) \in W$$
But
$$(2,2)+ (-1,1) = (1,3)$$
is not in W. Hence not a vector space.