Question

W={(a1, a2) in R^2| a1+2a2=0} is W a subspace?

Answer 1

It is the kernel of a linear map, hence it is a subspace.

Answer 2

how do you prove that it is a subspace?

Answer 3

\[ \mbox{Let}\;\;L:\mathbb{R}^2\rightarrow\mathbb{R}\;\mbox{with}\;L(x,y)=x+2y.\]

Answer 4

Then L is a linear map of R-vectorspaces.

Answer 5

Hence, \[ ker L:=\{ (x,y)\in\mathbb{R}^2\;|\;L(x,y)=0 \}=W \] is a subspace of |R^2.

Answer 6

how do you prove that it is a subspace?

Answer 7

I just proved it to you! The preimage of 0 under a linear map is a subspace. That is a theorem of linear algebra which you can use.

Answer 8

In case you don't know that theorem, you may also prove the assertion directly.

Answer 9

Would you like me to show you how to do it without use of linear maps?

Answer 10

yes please, we haven't learned linear maps yet

Answer 11

Ok, then take (x,y) in W and lambda in |R. Then (lambda*x, lambda*y) is also in W because if x+2y=0, then also 0=lambda*(x+2y)=lambda*x+2*(lambda*y). Now take (x,y) in W and (v,w) in W. Then (x+v, y+w) is also in W, because if x+2y=0 and v+2w=0, then also 0=(x+2y)+(v+2w)=(x+v)+2*(y+w). So, W is closed under scalar multiplication and vector addition and is thus a linear subspace of |R^2.