Determining if sets are subspaces of vector spaces

[MoreDrinks]

Homework Statement

Are the following sets subspaces of R³? The set of all vectors of the form (a,b,c), where

1. a + b + c = 0
2. ab = 0
3. ab = ac

Homework Equations

Each is its own condition. 1, 2 and 3 do not all apply simultaneously - they're each a separate question.

The Attempt at a Solution

I know from the back of the book that 1 is a subspace, while 2 and 3 are not. I know that the subset must be closed under addition and scalar multiplication for it to be a subspace, but I'm having trouble how seeing the conditions listed make (a,b,c) a subspace or not.

Here's me guessing

1. Who cares if they all equal zero when added together? (a,b,c)+(d,e,f) = (a+d,b+e,c+f), so we're fine. k(a,b,c) = (ka,kb,kc), so we're good.
2. a or b must equal zero. So we're left with (0,b,c) for example. Adding something of the form (a,b,c) could result in a non-zero in a, so we're not closed under addition, making us not a subspace. Correct?
3. b is equal to c. So we actually have something of the form (a,b,b) but we can add (d,e,f), and the sum does not necessarily leave us with b+e=b+f.

Am I right or totally messing this up?

 

[Mark44]

Homework Statement

Are the following sets subspaces of R³? The set of all vectors of the form (a,b,c), where

1. a + b + c = 0
2. ab = 0
3. ab = ac

Homework Equations

Each is its own condition. 1, 2 and 3 do not all apply simultaneously - they're each a separate question.

The Attempt at a Solution

I know from the back of the book that 1 is a subspace, while 2 and 3 are not. I know that the subset must be closed under addition and scalar multiplication for it to be a subspace, but I'm having trouble how seeing the conditions listed make (a,b,c) a subspace or not.

Here's me guessing

1. Who cares if they all equal zero when added together? (a,b,c)+(d,e,f) = (a+d,b+e,c+f), so we're fine. k(a,b,c) = (ka,kb,kc), so we're good.
2. a or b must equal zero. So we're left with (0,b,c) for example. Adding something of the form (a,b,c) could result in a non-zero in a, so we're not closed under addition, making us not a subspace. Correct?

You have the advantage of seeing the answer in advance, and this is shaping your response.

For this one a or b has to be zero, and possibly both. Two possible vectors in the set are (0, b, c) and (a, 0, d), where b is nonzero in the first vector and a is nonzero in the second.

3. b is equal to c. So we actually have something of the form (a,b,b) but we can add (d,e,f), and the sum does not necessarily leave us with b+e=b+f.

What you have is ab = ac, which doesn't necessarily imply that b = c. For example, if a = 0, then b and c can be any values.

To show that this set is not a subspace, take two vectors that are definitely in the set, and show that their sum is not in the set or that a scalar multiple of one of them is not in the set. What you have is more like arm-waving than an actual proof of something.

Am I right or totally messing this up?

 

[MoreDrinks]

You have the advantage of seeing the answer in advance, and this is shaping your response.

My answers preceded checking the answers - it should be useful to point out that the order of the post does not necessarily imply temporal precedence in the work and thinking on the part of the poster. But I felt so unsure about it and my capacities that I sought out help here.

For this one a or b has to be zero, and possibly both. Two possible vectors in the set are (0, b, c) and (a, 0, d), where b is nonzero in the first vector and a is nonzero in the second.

Was my reasoning, however, correct? (0,2,3)+(1,0,4) = (1,2,7) and ab≠0?

What you have is ab = ac, which doesn't necessarily imply that b = c. For example, if a = 0, then b and c can be any values.

To show that this set is not a subspace, take two vectors that are definitely in the set, and show that their sum is not in the set or that a scalar multiple of one of them is not in the set. What you have is more like arm-waving than an actual proof of something.

So, for example, one could say this:

ab=ac because a=0. Let's say a vector (0,2,5). This would also be true for (2,3,3).

(0,2,5)+(2,3,3) = (2,5,8)
10≠16

Does that work or have I delivered more hand waving.

Also, could you say something about number 1?

 

[Mark44]

My answers preceded checking the answers - it should be useful to point out that the order of the post does not necessarily imply temporal precedence in the work and thinking on the part of the poster. But I felt so unsure about it and my capacities that I sought out help here.



Was my reasoning, however, correct? (0,2,3)+(1,0,4) = (1,2,7) and ab≠0?

This is a perfectly good counterexample to show that this set (#2 in your list) is not a subspace of R³.

So, for example, one could say this:

ab=ac because a=0. Let's say a vector (0,2,5). This would also be true for (2,3,3).

(0,2,5)+(2,3,3) = (2,5,8)
10≠16

Does that work or have I delivered more hand waving.

This is also a perfectly good counterexample to show that this set (#3 in your list) is not a subspace of R³. This isn't handwaving at all - you have found a particular example of members of the set for which addition is not closed.

For #1, you have two vectors in your set - (a, b, c) and (d, e, f). Because they're in the set, you know that a + b + c = 0 and d + e + f = 0.

Show that their sum (a + d, b + e, c + f) is also in the set (i.e., that (a + d) + (b + e) + (c + f) = 0. This is not hard at all.

Also show that k(a, b, c) is also in the set; namely, that ka + kb + kc = 0. This is really easy, too.

 

[MoreDrinks]

For #1, you have two vectors in your set - (a, b, c) and (d, e, f). Because they're in the set, you know that a + b + c = 0 and d + e + f = 0.

Show that their sum (a + d, b + e, c + f) is also in the set (i.e., that (a + d) + (b + e) + (c + f) = 0. This is not hard at all.

Also show that k(a, b, c) is also in the set; namely, that ka + kb + kc = 0. This is really easy, too.

Oh, Jesus - this is one of those moments where you smack yourself for not realizing the ease of the problem earlier. Many thanks for the assistance.