help with sub-spaces of functions

D

DarkSun

New member
Joined
Jan 3, 2009
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29
Hey,

a question about determining if a set of functions are a sub-space.
I know that basically given a vector space V and a set W, I have to find out if:
1) w is not empty (or contains zero vector, which is the same), 2) closed under addition, 3) closed under multiplication

Now let's say V is the space of real functions, W={F in V | F(1)+2F(2)+3F(3)=F(0)}
So...
1) is ok, if I take f0(1)+2f0(2)+3f0(3)=f0(0) I get 0=0, so the zero vector is there
But where do I go for 2) and 3)? adding 2 such functions doesn't seem to help much...
2) I believe is:
(f+g)(1)+2(f+g)(2)+3(f+g)(3) =
f(1)+g(1)+2f(2)+2g(2)+3f(3)+3g(3) =
(f(1)+2f(2)+3f(3))+(g(1)+2g(2)+3g(3))=f(0)+g(0)=(f+g)(0)
3) Here I am a bit stuck...
say x is a scalar and f function in W:
x(f(1)+2f(2)+3f(3)) =
(xf(1)+x2f(2)+x3f(3))
What now...?


Thanks
 
We have if F and G are in W then:
(1) - F(1)+2F(2)+3F(3)=F(0)
(2) - G(1)+2G(2)+3G(3)=G(0)

Adding (1) and (2) together:

(F+G)(1)+2(F+G)(2)+3(F+G)(3)=(F+G)(0)

Therefore F+G is in W/

..

For scalar multiplication:
Let F be in W, then:

F(1)+2F(2)+3F(3)=F(0)

Then:
a(F(1)+2F(2)+3F(3))=a(F(0))
aF(1)+a*2F(2)+a*3F(3)=aF(0)
(aF)(1) + 2(aF)(2) + 3(aF)(3) = (aF)(0)

So aF is also in W.
 
Hey,
I understand the stages of the solution, but something in the meaning escapes me....
When you say:
a(F(1) = aF(1) = (aF)(1) = (aF)(0)
and specifically: aF(1) = (aF)(1), what does it mean?
The left part multiply the result of the function by a which is cool, but the right one...? what does (aF) mean exactly?
 
I was showing the difference between a scalar times a function: a*F(x) and NEW functin (aF)(x). Yes, they are the "same thing" but we are concerned with whether the function aF belongs to W. That is why I wrote (aF)(x) as opposed to aF(x).

If you'd like, insert g(x) = a*F(x). Then show g is contained in W:

g(1)+2g(2)+3g(3) = a*F(0) = g(0)
 
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