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?

(0,0) must be in X(why?)! Hence a*0+b*0+c!=0 - absurd!

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Originally Posted by Also sprach Zarathustra

(0,0) must be in X(why?)! Hence a*0+b*0=c!=0 - absurd!

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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...

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?

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.

Quote:

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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.

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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?

c is constant!

see again post #5 (1)

Quote:

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Originally Posted by Also sprach Zarathustra

c is constant!

see again post #5 (1)

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Thanks alot.

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