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- January 2nd 2013, 03:08 AMmohamadsubspace and basis
can you help solve this? la | Flickr - Photo Sharing!

- January 2nd 2013, 05:26 AMDevenoRe: subspace and basis
sure we'll help...where are you stuck...show us what you've done so far.

- January 2nd 2013, 06:02 AMHallsofIvyRe: 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? - January 2nd 2013, 06:33 AMmohamadRe: 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? - January 2nd 2013, 07:06 AMDevenoRe: 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 R^{3}, why do you have a matrix with 4 columns?