Question

Guys GRE math prep help. I think this has something to do with linear algebra... Attaching..

Answer 1

that upside down u means intersection

Answer 2

oh okay..

Answer 3

there's a rule for intersecting vector spaces

Answer 4

okay..

Answer 5

so it's 2 only?

Answer 6

i think so..the intersection of 2 sets is the set of all the elements in both sets

Answer 7

oh wait

Answer 8

subspace can have smaller dimension

Answer 9

0, 1, 2

Answer 10

okay what if it were 2 and 3 dim subspaces.

Answer 11

the intersection?

Answer 12

yeah

Answer 13

it would still be 0,1,2 right?

Answer 14

i'm not sure if that's possible

Answer 15

0 is not possible anymore, because you work in a 4-dimensional vectorspace. So there's always overlap between a 2-dimensional subspace and a 3-dimensional subspace.

Answer 16

i thought zero was always a subspace?

Answer 17

Thomas is correct

Answer 18

guys i have trouble understanding the concept.. like what is a subspace?

Answer 19

if V was 2-dimentional and W was 3 dimensional and somehow \[V\cap W=\{0\}\]

then the basis vectors for V and W would have to be linearly independent and therefore span a space of dimension 5 (which we don't have in this problem)

Answer 20

oh because the intersection would have 2d and 3d elements?

Answer 21

no

Answer 22

Every element in both V and W is 4-dimensional.

Answer 23

ie. has four components.

Answer 24

anyway, bahrom needs to know what a subspace is

Answer 25

A is a subspace of B if $$A\subset B$$ and A is linear.

Answer 26

A is linear if it is closed under addition and scalar multiplication.

Answer 27

from wiki

Let K be a field (such as the field of real numbers), and let V be a vector space over K. As usual, we call elements of V vectors and call elements of K scalars. Suppose that W is a subset of V. If W is a vector space itself, with the same vector space operations as V has, then it is a subspace of V.

Answer 28

Since the space itself already satisfies the vector space axioms, a subspace which defines the same operations on subset of the same set need only satisfy the following axioms:

The subspace must have the additive identity element: 0 vector:
The subspace must be closed under addition and scalar multiplication.

Thomas is correct but missed the requirement of the 0 vector.

Answer 29

That follows from being closed under scalar multiplication: multiply by 0. Anyway, good to point that out.