# subspace question

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• Nov 12th 2012, 02:51 PM
dumbledore
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?
• Nov 12th 2012, 03:13 PM
HallsofIvy
Re: subspace question
Quote:

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 $2[x_1, x_2, x_1+ 1]= [2x_1, 2x_2, 2x_1+ 2]$ and $x_3= 2x_1+ 2$ is NOT $2x_1+ 1$

Quote:

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.