which of the following are linear subspaces ?

[Jamin2112]

Homework Statement

For each set below, which of the following sets are real linear subspaces (where addition and scalar multiplication are defined in the usual way for these sort of objects)? Justify your answers with an argument in which real linear space it is included, why it is closed under addition and scalar multiplication or a counterexample showing it is not.

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Homework Equations

Basically, something is a linear subspace if it contains the zero vector and has the closure property.

The Attempt at a Solution

I call on the literary experts at PF to help me decipher the above paragraph. Is it saying, for each of the 4 sets listed below, which of the other listed sets are linear subspaces? I'm a little confused. Once that is cleared up, I'll actually ask show which sets we're talking 'bout.

 

[Chantry]

They're just asking which are subspaces of R^n.

So you have to show:
a) cV, where V is a subset of the subspace, is an element of the subspace.
and b) v + w, where v,w are elements of the subspace, is again in the subspace.

It also wants to know what the domain of the subspace is from what I understand. This is fairly easy to figure out by inspection.

 

[Jamin2112]

There just asking which are subspaces of R^n.

It doesn't say that.



[PLAIN]http://turbo.inquisitr.com/wp-content/2009/11/skeptical-hippo-1.jpg

 

[Chantry]

Homework Statement

For each set below, which of the following sets are real linear subspaces (where addition and scalar multiplication are defined in the usual way for these sort of objects)? Justify your answers with an argument in which real linear space it is included, why it is closed under addition and scalar multiplication or a counterexample showing it is not.

Not a fan of sarcasm.

Particularly when I'm trying to help you out.

 

[micromass]

It doesn't say that.

Yes, it does...

 

[Jamin2112]

Not a fan of sarcasm.

Particularly when I'm trying to help you out.

So, for example, the vector [r - s; 2 + 2s; 0] (wrote it MATLAB-style), where r, s are real numbers, WOULD be a linear subspace because of those 3 properties. (?)

 

[HallsofIvy]

I don't know if the original post has been edited but right now it says nothing about "Rⁿ". It only talks about showing that a subset of some vector space is a subspace. So it is NOT correct to say that "They're just asking which are subspaces of R^n." For example, the set of all cubic polynomials, p(x), such that p(1)= 0, is a subspace of the vector space of all cubic polynomials which has nothing (directly) to do with Rⁿ. Of course, every finite dimensional vector space is isomorphic to some Rⁿ but there are other vector spaces than Rⁿ.

It would have been nice if Jamin2112 had said why he wrote "It doesn't say that", but he is corrrect.

To show that a subset of a vector space is a subspace you must show three things:
1) The set is non-empty. (Usually trivial.)
2) The set is closed under addition (if u and v are elements of the set so is u+ v).
3) The set is closed under multiplication (if u is an element of the set, and k is any scalar, so is ku).

Chantry mistyped (I am sure it was only that) when he wrote "where V is a subset of the subspace". He obviously mean "where V is an element of the subset".

 

[HallsofIvy]

So, for example, the vector [r - s; 2 + 2s; 0] (wrote it MATLAB-style), where r, s are real numbers, WOULD be a linear subspace because of those 3 properties. (?)

Well, you mean the set of such vectors, for any numbers r and s, as a subset of R³. Also, you should not just say "because of those 3 properties" but show that those 3 properties really do hold.

Yes, if you take r= s= -1, [r-s, 2+ 2s, 0]= [-1+ 1, 2- 2, 0]= [0, 0, 0]. (You originally wrote "contains the 0 vector" and I just said, above, "is non-empty". Obviously, if a set contains the 0 vector it is non-empty. If a set is non-empty, say, it contains v, and is closed under scalar multiplication, then it also contains (-1)v= -v and if it also is closed under addition, it contains v+ (-v)= 0. So, given closure, "is non-empty" and "contains the 0 vector" are equivalent.)

One possible vector in the set is [x- y, 2+ 2x, 0] and another is [a- b, 2+ 2b, 0]. If you add those, you get [x- y+ a- b, 2+ 2x+ 2+ 2y, 0+ 0]= [(x+a)- (y+ b), 4+ 2(y+ b), 0]. Is that in the same set? If so, what would the r and s be in [r- s,k 2+ 2s, 0] to be equal to [(x+a)- (y+b), 4+ 2(y+b), 0]?

If k is a scalar, then k[x- y, 2+ 2y, 0]= [k(x- y), k(2+ 2y), 0]= [kx- ky, 2k+ 2ky, 0]. Is that in the set? What would the r and s be in [r-s, 2+ 2s, 0] to e equal to [kx-ky, 2k+ 2ky, 0]?

 

[Jamin2112]

Well, you mean the set of such vectors, for any numbers r and s, as a subset of R³. Also, you should not just say "because of those 3 properties" but show that those 3 properties really do hold.

Yes, if you take r= s= -1, [r-s, 2+ 2s, 0]= [-1+ 1, 2- 2, 0]= [0, 0, 0]. (You originally wrote "contains the 0 vector" and I just said, above, "is non-empty". Obviously, if a set contains the 0 vector it is non-empty. If a set is non-empty, say, it contains v, and is closed under scalar multiplication, then it also contains (-1)v= -v and if it also is closed under addition, it contains v+ (-v)= 0. So, given closure, "is non-empty" and "contains the 0 vector" are equivalent.)

One possible vector in the set is [x- y, 2+ 2x, 0] and another is [a- b, 2+ 2b, 0]. If you add those, you get [x- y+ a- b, 2+ 2x+ 2+ 2y, 0+ 0]= [(x+a)- (y+ b), 4+ 2(y+ b), 0]. Is that in the same set? If so, what would the r and s be in [r- s,k 2+ 2s, 0] to be equal to [(x+a)- (y+b), 4+ 2(y+b), 0]?

If k is a scalar, then k[x- y, 2+ 2y, 0]= [k(x- y), k(2+ 2y), 0]= [kx- ky, 2k+ 2ky, 0]. Is that in the set? What would the r and s be in [r-s, 2+ 2s, 0] to e equal to [kx-ky, 2k+ 2ky, 0]?

Ah, I now see how to do these! And I did find and r and s in each of those 2 cases.

 

[HallsofIvy]

Ah, I now see how to do these! And I did find and r and s in each of those 2 cases.

Then you don't see at all- the set of vectors of the form [x-y, 2+ 2y, 0] is NOT a subspace- such r and s do not exist.

(IF the problem were {[x- y, 2x+ 2y, 0]} then it would be a subspace- but that missing "x" is important.)