Determining if sets are subspaces of vector spaces

In summary: I see you have updated your post, so I'll stop here)In summary, the set of all vectors of the form (a,b,c) where a + b + c = 0 is a subspace of R3. The set of all vectors of the form (a,b,c) where ab = 0 is not a subspace of R3. The set of all vectors of the form (a,b,c) where ab = ac is also not a subspace of R3. This can be shown by finding counterexamples where addition and scalar multiplication are not closed.
  • #1
MoreDrinks
45
0

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?
 
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  • #2
MoreDrinks said:

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.
MoreDrinks said:
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.
MoreDrinks said:
Am I right or totally messing this up?
 
  • #3
Mark44 said:
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?
 
  • #4
MoreDrinks said:
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.
MoreDrinks said:
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.
MoreDrinks;4299304 Also said:
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.
 
  • #5
Mark44 said:
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.
 

1. What is a subspace?

A subspace is a subset of a vector space that contains all of the necessary properties to be considered a vector space itself. This means that it is closed under addition and scalar multiplication, and contains the zero vector.

2. How do you determine if a set is a subspace of a vector space?

To determine if a set is a subspace of a vector space, you must check if it satisfies the three properties of a vector space: closure under addition, closure under scalar multiplication, and contains the zero vector. If it meets all three criteria, then it is a subspace.

3. What are the necessary conditions for a set to be a subspace?

The necessary conditions for a set to be a subspace are: it must be closed under addition, closed under scalar multiplication, and contain the zero vector. If any of these conditions are not met, then the set is not a subspace.

4. Can a set be a subspace of more than one vector space?

Yes, a set can be a subspace of more than one vector space. As long as the set satisfies the necessary conditions for a subspace, it can be a subspace of any vector space that contains it.

5. What is the importance of determining if a set is a subspace of a vector space?

Determining if a set is a subspace of a vector space is important because it allows us to understand the properties and behavior of the set within the larger vector space. It also helps us to solve problems and make calculations involving the set more efficiently.

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