Question

Vectorspace/Subspace: Verify that the set V(1) consisting of all scalar multiples of (1, -1, 2) is a subspace of R(3).

Help!! How do I show closed under addition and multiplication for this???

Answer 1

closure under scalar multiplication is pretty obvious since the definition of the subspace is all scalar multiples of the given vector

Answer 2

Try adding two of such multiples, and recalling what IS a MULTIPLICATION BY A SCALAR NUMBER

Answer 3

closure under addition can be shown by noting that\[a\vec v+b\vec v=(a+b)\vec v=c\vec v\]

Answer 4

if I let v(1) = (a(1), -a(1), 2a(1)) and let v(2) = (a(2), -a(2), 2a(2)), then the v(1) + v(2) = (a(1)+a(2), -a(1)-a(2), 2a(1)+2a(2)). how is that a scalar multiple of (1, -1, 2)?

Answer 5

what is a(1) ??

Answer 6

\[\vec v_1=\langle a,-a,2a\rangle\]\[\vec v_2=\langle b,-b,2b\rangle\]

Answer 7

the a(1), is a subscript 1, just meaning some number

Answer 8

then factor a(1)+a(2) out of your resultant vector and you will see it is a scalar multiple of the original.

Answer 9

(a(1)+a(2), -a(1)-a(2), 2a(1)+2a(2))=(a(1)+a(2))(1,-1,2)

Answer 10

is that a given that you can factor that out? is that by some scalar rule. I was thinking I might be able to do that but I couldn't find any rule for it.

Answer 11

well you can factor our any constant, and we can rewrite this\[\vec v=\langle(a_1+a_2)1,(a_1+a_2)(-1),(a_1+a_2)(2)\rangle\]

Answer 12

\[\vec v=\langle x,y,z\rangle\]\[c\vec v=\langle cx,cy,cz\rangle\]

Answer 13

so by let b = a(1) + a(2), then v(1) + v(2) = (b, -b, 2b) and therefore v(1) + v(2) is in W.

Answer 14

yep

Answer 15

thank you... got it... i hope maybe u can answer my next one.

Answer 16

I can try!