# subspace and basis

• Jan 2nd 2013, 03:08 AM
subspace and basis
can you help solve this? la | Flickr - Photo Sharing!
• Jan 2nd 2013, 05:26 AM
Deveno
Re: subspace and basis
sure we'll help...where are you stuck...show us what you've done so far.
• Jan 2nd 2013, 06:02 AM
HallsofIvy
Re: subspace and basis
The first problem is to show this is a subspace. Do you know what "subspace" means? What properties must a set have to be a subspace?

The second and third ask you to find a basis. Do you know what a "basis" is?
• Jan 2nd 2013, 06:33 AM
Re: subspace and basis
1)showing that W is a subspace was easy (closed under addition and scalar multiplication)
2)I took [1 2 0] and [0 0 1] as a basis( dimension 2) because [ a 2a b] is a linear combination of these two but is it correct or should I've done something else?
3)for the third question I setup the system
1 0 1 0|0
1 0 0 1|0
0 1 0 0|0
and deduced the linearly independent vectors which form the basis for R3
so was my work correct?
• Jan 2nd 2013, 07:06 AM
Deveno
Re: subspace and basis
for 2) you showed spanning, but not linear independence.

you only need three equations.

you want to find a 3rd vector (x,y,z) such that {(1,2,0),(0,0,1),(x,y,z)} is linearly independent.

one way to do this is to pick (x,y,z) so that it is orthogonal to (1,2,0) and (0,0,1).

if (1,2,0).(x,y,z) = 0, then x+2y = 0. that is: y = -x/2.

if (0,0,1).(x,y,z) = 0, then z = 0. can you combine these two conditions? prove the resulting set of 3 vectors is linearly independent.

the system you set up makes no sense at all, your vectors live in R3, why do you have a matrix with 4 columns?