Question

[X ∈ ℝ³ : AX=3X ] is a subspace?

Answer 1

It seems like a subspace to me.

Answer 2

yea but how do you know?

Answer 3

The zero vector is in there, if \(u, v\) are in there, then a linear combination is also, and finally, any scalar of a vector in this subspace is still in the subspace.

Showing these three things is enough to show it's a subspace.

Answer 4

thx

Answer 5

you're welcome

Answer 6

Is y a vector?

Answer 7

yes

Answer 8

It looks like the notation is saying that x is in the space spanned by y. If y is a single vector, this would mean that x is just y times some constant.

Answer 9

THX :)

Answer 10

You're welcome.

Answer 11

Not to rain on anyone's parade, but I'm not so sure. I don't see any specification on X except that it it a subset of R3. There are certainly lots of subsets that aren't subspaces. It's getting late, I could be wrong, but I don't think there is enough specified here to come to the conclusion that it's a subspace.

Answer 12

I suppose if we assume that X is a space to begin with, it's pretty clear.

Answer 13

Well, \(\mathbb{R^3}\) is a vector space, so we should be fine.

Answer 14

Since the 0-vector is in the subset, the subset is closed under addition, and it's closed under scalar multiplication, it must be a subspace.

Answer 15

How do we know that the zero vector is in the subset? I don't see any such specification at all.

Answer 16

Supppose \(X=\vec0\). Notice that for any transformation \(A, \;A\vec0=\vec0\) and \(3\cdot\vec0=\vec0\). Thus the zero vector is in the subset.