Linear Algebra - Subspaces proof

[mattmns]

Hello, just wondering if my proof is sufficient.

Here is the question from my book:

Show that the following sets of elements in R2 form subspaces:
(a) The set of all (x,y) such that x = y.
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So if we call this set W, then we must show the following:
(i) $$0 \in W$$
(ii) if $$v,w \in W$$, then $$v+w \in W$$
(iii) if $$c \in R$$ and $$v \in W$$ then $$cv \in W$$

Pf:
(i) $$0 \in W$$ because we can take x = 0 = y
(ii) if $$v,w \in W$$ then $$(v,v) \in W$$ and $$(w,w) \in W$$ and $$(v,v) + (w,w) = (v + w, v + w) \in W$$ so $$v+w \in W$$ because v + w = x = y = v + w
(iii) if $$c \in R$$ and $$v \in W$$, then $$(v,v) \in W$$ and $$c(v,v) = (cv,cv) \in W$$ so $$cv \in W$$ because cv = x = y = cv
Therefore W is a subspace.

Does that look just fine? Thanks.

 

[Yoss]

Yah, looks good to me.

 

[mattmns]

Cool, thanks!

 

[matt grime]

No, that's not right, but the idea is correct. You state v is in W but then write the vector (v,v) is in W. The v in the brackets cannot be the v outside the brackets. You need to change that, v canont simultaneously be a vector in R^2 and elements of R.

 

[mattmns]

Thanks matt, I think I am seeing what you are saying. So if I said V is in W, then (v,v) is in W, that would be correct? Because big V is not the same as little v. Thanks.

 

[HallsofIvy]

Yes, that would work. Although it might be even better to say:
V= (x,x), U= (y,y).

 

[matt grime]

Thanks matt, I think I am seeing what you are saying. So if I said V is in W, then (v,v) is in W, that would be correct? Because big V is not the same as little v. Thanks.

well, (v,v) is in W because of the definition of W, and is not deduced from V is in W. what you ought to say is if V in W then V=(v,v) for some v. there is no need to actually use V. W is closed under addition because (v,v)+(x,x)=(v+x,v+x).

 

[JasonRox]

Just for future reference, you can cut down the work for the proof for a subspace, simply by showing...

0 e W
cx+y e W, for any scalar c (depending on the field).