Determine subspace

This is a definition problem. Write down the properties of a subspace and pick them off, one at a time.

Go ahead - first list the properties.

Note: It strikes me as a little odd to show that something is a "valid subspace". Must we domonstrate that it is a Vector Space and then suggest a containing Vector Space of which ours is a subspace?
 
tkhunny said:
This is a definition problem. Write down the properties of a subspace and pick them off, one at a time.

Go ahead - first list the properties.

Note: It strikes me as a little odd to show that something is a "valid subspace". Must we domonstrate that it is a Vector Space and then suggest a containing Vector Space of which ours is a subspace?
Click to expand...

Well, those are my words. The problem says determine whether the following sets are subspaces. To show something is a subspace, you need to show that it's closed under vector addition and scalar multiplication. Do I need to worry about the additive inverse?
 
You do have to decide what containing Vector space, \(\displaystyle \mathbb{R}^{2}\) or something like that, will be sufficient.

A point of a subspace is its inheritance of properties from the contining Vector Space. You must demonstrate that it exists and that it is self-contained (closed). Other properties are inherited and require no demonstration.

Does it exist? Is it non-empty?
Is it closed under + (often addition)?
Is it closed under * (often multiplication)?
 
I'm not getting anywhere playing with (a[sub:2fs125hb]1[/sub:2fs125hb],a[sub:2fs125hb]2[/sub:2fs125hb],a[sub:2fs125hb]3[/sub:2fs125hb]) and trying to show it's closed under vector addition and scalar multiplication. This is a 4-week, MTWThF 10A - 12P Advanced Linear Algebra 1 class. I'm tired and cannot think very well. The class started today and this homework is due tomorrow. I'm going to hate this summer. I'm taking Abstract Algebra after this class.
 
a1 = 3a2
a3 = -a2

{a1,a2,a3} = {3a2,a2,-a2}

Pick 2

{3a2,a2,-a2}
{3b2,b2,-b2}

Add them

{3a2+3b2,a2+b2,-a2-b2}

Rewrite a little:

{3(a2+b2),(a2+b2),-(a2+b2)}

Since, from the containing vector space closure, a2+b2 = c2 is in the vector space, we have

{3c2,c2,-c2}

I think we have closure for vector addition.

Note: Sorry, I don't normally open attachments, so I didn't see the appropriate defintion and question. It no longer sounds odd. :D
 
Haha. I think we go through this every time I posted a link to my scanned work. :p Remember, I'm one of the good guys! 8-)

That was the general approach I was taking earlier. I just couldn't think. I'm still about to pass out. I'll upload my work, but for (b), I determined that it's not closed under vector addition. I get like c[sub:1bvla9vq]3[/sub:1bvla9vq] + 4 instead of 2, which breaks the closure.
 
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