Thread: subspace question

Determine if the following sets are subspaces of R3 or not and justify.

a) V={ [x1,x2,x3] | x3 = 1+x1 }

b) W={ [a-b, 2b, 3c-a] | a,b,c R(real) }

note: x1,x2,x3 and a-b,2b,3c-a are column vectors

I think for a) it is not because the zero vector is not present and for b) it is since W = span {[1,0,-1],[-1,2,0],[0,0,3]}

What do you think?

Originally Posted by dumbledore

Determine if the following sets are subspaces of R3 or not and justify.

a) V={ [x1,x2,x3] | x3 = 1+x1 }

b) W={ [a-b, 2b, 3c-a] | a,b,c R(real) }

note: x1,x2,x3 and a-b,2b,3c-a are column vectors

I think for a) it is not because the zero vector is not present

Good. That is correct. You could also note that and is NOT

and for b) it is since W = span {[1,0,-1],[-1,2,0],[0,0,3]}

What do you think?

Yes, that looks good and probably the simplest. You could also note that a= b= c= 0 gives the 0 vector, [0, 0, 0]. If k is any number, k[a- b, 3b, 3c- a]= [k(a- b), k(3b), k(3c- b)]= [ak- kb, 3(kb), 3(kc)- kb] which is the of the same for with ka in place of a, kb in place of b, and kc in place of c- this set is closed under scalar multiplication. Finally [a- b, 3b, 3c- a]+ [x- y, 3y, 3z- x]= [a- b+ x- y, 3b+ 3y, 3c-a+ 3z- x]= [(a+ x)- (b+ y), 3(b+ y), 3(c+ z)- (a- x)[/tex] which is of the same form with a+ x in place of a, b+y in place of b, and c+z in place of c so this set is closed under vector addition. But you are correct that the best way to do this is to find vectors that span the set.