Thread: Vector Subspaces

Hoping to get some help with questions that Im stuck on for exam revision
Show that:
a) U = (x,y,z) x+y=z is a subspace of R3
b) U= (x,y,z), y is rational and is not a subspace of R3
c) U= A|A2=A,AϵMmn is not a subspace of Mmn
d)U={F|f(x+pi)=f(x), FϵC(-infinity, + infinity) is a subspace of C(-inifity, +inifinity)
e) u= p(x)|p(1)=0, p(x)ϵPn is a subspace of Pn
Cheers

Originally Posted by rjd871

Hoping to get some help with questions that Im stuck on for exam revision
Show that:
a) U = (x,y,z) x+y=z is a subspace of R3
b) U= (x,y,z), y is rational and is not a subspace of R3
c) U= A|A2=A,AϵMmn is not a subspace of Mmn
d)U={F|f(x+pi)=f(x), FϵC(-infinity, + infinity) is a subspace of C(-inifity, +inifinity)
e) u= p(x)|p(1)=0, p(x)ϵPn is a subspace of Pn
Cheers

Have you tried to help yourself? If you have then post what you have done.
If you have not, then why do you think we can help if you do not tell us what you don't understand?
How does one prove a set is a subspace?

Some of those can't see because. . .well the type is all screwy. You might want to post a photo of the exam questions. And these all really follow from the definition, of which there are three conditions:

1. Subspace contains the 0-vector.
2. Subspace is closed under vector addition.
3. Subspace is closed under scalar multiplication.

With those three conditions to check, all of these are fairly boilerplate. Also, I'm fairly sure (b) isn't written correctly.

the reason the U of (b) is not a subspace is because we can consider (0,1,0) which is clearly in U (0 is an element of R, and 1 is rational) and then take e (the natural logarithm base) times it (since the scalar e is in the field R) to obtain e(0,1,0) = (0,e,0), which is not in U (since e is irrational), thus U is not closed under scalar multiplication.