# Subspace proof

• Nov 25th 2010, 10:10 AM
chr91
Subspace proof
Here attached is the question:

How do I prove it as an if and only statement.

We have learnt the definition of subspace being when the set is empty, closed under vector addition and closed under multiplication.

I can prove for when c=0 that it is a subspace. How do I prove for all other values, that it is not a subspace?
• Nov 25th 2010, 10:31 AM
Also sprach Zarathustra
(0,0) must be in X(why?)! Hence a*0+b*0+c!=0 - absurd!
• Nov 25th 2010, 10:40 AM
chr91
Quote:

Originally Posted by Also sprach Zarathustra
(0,0) must be in X(why?)! Hence a*0+b*0=c!=0 - absurd!

What?

I know that when c = 0, X is a subspace of R2 but I don't know how to prove that this is the case IF and ONLY IF c = 0

I can prove it for c=1, c=2 etc but not a general proof to show that for all non zero values for c, X is not a subspace...
• Nov 25th 2010, 10:44 AM
chr91
Oh I see what you mean, sorry. So do you have to start by proving that 0,0 is in X which shows that C has to equal 0?
• Nov 25th 2010, 10:50 AM
Also sprach Zarathustra
1. X subspace ==> (0,0) must be in X! Hence a*0+b*0+c=0 or c=0

2. c=0 ==> ax_1 + bx_2=0 ==> NOW prove that X is closed under vector addition and closed under multiplication.
• Nov 25th 2010, 11:02 AM
chr91
Quote:

Originally Posted by Also sprach Zarathustra
1. X subspace ==> (0,0) must be in X! Hence a*0+b*0+c=0 or c=0

2. c=0 ==> ax_1 + bx_2=0 ==> NOW prove that X is closed under vector addition and closed under multiplication.

Thanks, I can do that now.

The only question I have is, how does this prove that c can't be any other number in order for it to be a subspace of X?

Haven't we just taken an example to find a possible value for c, then prove c=0 is a subspace of X?
• Nov 25th 2010, 11:12 AM
Also sprach Zarathustra
c is constant!

see again post #5 (1)
• Nov 25th 2010, 12:02 PM
chr91
Quote:

Originally Posted by Also sprach Zarathustra
c is constant!

see again post #5 (1)

Thanks alot.

Sorry for being a bit 'slow.' Only introduced to subspaces yesterday so not fully understanding it all yet..