Proving something is a subset of another set in linear algebra

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Homework Statement



http://people.math.carleton.ca/~mezo/A8math1102-11.pdf

1b) please

1. Suppose F is a field, A ∈ Mmn(F), b ∈ Fm and v ∈ Fn is a particular solution to the equation Ax = b. Let S0  ⊆ Fn be the solution set to
the (homogeneous) equation Ax = 0, and S ∈ Fn be the solution set
to Ax = b.
(a) Prove that the set v + S0 = {v + w : w ∈ S0} is a subset of S.
(b) Prove that S is a subset of v + S0. (Hint: Suppose w ∈ S. Show that w - v ∈ S0. What does this say about w?)

Homework Equations





The Attempt at a Solution



How do I even know that w ε S? That's a huge supposition isn't it? Well to prove that one set is a subset of the other set, all you need to do is prove that each share a common element right?

And we know that the Set S has the following elements:

1. v+w [from the last proof]
2. v [as stated from our supposition that " v ε F^n is a particular solution to Ax=b"]
3. x [as stated from our supposition " S subset F^n be the solution set to Ax=b"

and S_0 contains the following elements in it's set:

1. 0 [trivial solution]
2. w [v+S_0) is defined as {v+w: w ε S_0}
3. x [from our supposition "Let S_0 subset F^n be the solution set to the (homogenous) equation Ax=0."

So since both have x in their elements, can't we say that S subset (v+S_0)?
 
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  • #2
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Homework Statement



http://people.math.carleton.ca/~mezo/A8math1102-11.pdf

1b) please

Homework Equations





The Attempt at a Solution



How do I even know that w ε S?
Unless S is empty, it has to have some elements. Any vector w in S satisfies Aw = b.
That's a huge supposition isn't it? Well to prove that one set is a subset of the other set, all you need to do is prove that each share a common element right?
No, not at all. Consider A = {1, 3, 5} and B = {1, 2, 3}. Both sets have 1 and 3 in common, but neither set is a subset of the other. For one set to be contained in another, each element of the first set must also be in the second set.
And we know that the Set S has the following elements:

1. v+w [from the last proof]
2. v [as stated from our supposition that " v ε F^n is a particular solution to Ax=b"]
3. x [as stated from our supposition " S subset F^n be the solution set to Ax=b"

and S_0 contains the following elements in it's set:

1. 0 [trivial solution]
2. w [v+S_0) is defined as {v+w: w ε S_0}
3. x [from our supposition "Let S_0 subset F^n be the solution set to the (homogenous) equation Ax=0."

So since both have x in their elements, can't we say that S subset (v+S_0)?
 
  • #3
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Unless S is empty, it has to have some elements. Any vector w in S satisfies Aw = b.


Thank you but w is also in S_0. So it has to be the same value as S_0 right? How do we know if w ε S?
 
  • #4
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Just so that other readers and I don't have to keep jumping from this page to the page you posted as a link, here is the problem statement. It would have been helpful for you to post it.
1. Suppose F is a field, A [itex]\in[/itex] Mmn(F), b [itex]\in[/itex] Fm and v [itex]\in[/itex] Fn is a particular solution to the equation Ax = b. Let S0  [itex]\subseteq[/itex] Fn be the solution set to
the (homogeneous) equation Ax = 0, and S [itex]\in[/itex] Fn be the solution set
to Ax = b.
(a) Prove that the set v + S0 = {v + w : w [itex]\in[/itex] S0} is a subset of S.
(b) Prove that S is a subset of v + S0. (Hint: Suppose w [itex]\in[/itex] S. Show that w - v [itex]\in[/itex] S0. What does this say about w?)
 
  • #5
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Just so that other readers and I don't have to keep jumping from this page to the page you posted as a link, here is the problem statement. It would have been helpful for you to post it.
1. Suppose F is a field, A [itex]\in[/itex] Mmn(F), b [itex]\in[/itex] Fm and v [itex]\in[/itex] Fn is a particular solution to the equation Ax = b. Let S0  [itex]\subseteq[/itex] Fn be the solution set to
the (homogeneous) equation Ax = 0, and S [itex]\in[/itex] Fn be the solution set
to Ax = b.
(a) Prove that the set v + S0 = {v + w : w [itex]\in[/itex] S0} is a subset of S.
(b) Prove that S is a subset of v + S0. (Hint: Suppose w [itex]\in[/itex] S. Show that w - v [itex]\in[/itex] S0. What does this say about w?)
Sorry mate.
 
  • #6
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The hint is "suppose w is in S0". Just go from there. All this means is that w is a vector such that Aw = 0.
 
  • #7
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And we know that the Set S has the following elements:

1. v+w [from the last proof]
2. v [as stated from our supposition that " v ε F^n is a particular solution to Ax=b"]
3. x [as stated from our supposition " S subset F^n be the solution set to Ax=b"
I'm getting the idea that you think you are providing a complete list of the elements of S. IOW, that S = {v+w, v, x}. That is not the case.

In #1 above the hypothesis probably was something like, "suppose v and w are in S." Then you showed that v+w was also in S, thereby showing that S was closed under vector addition. v and w are symbols that can represent any of a possibly infinite number of vectors.
and S_0 contains the following elements in it's set:

1. 0 [trivial solution]
2. w [v+S_0) is defined as {v+w: w ε S_0}
3. x [from our supposition "Let S_0 subset F^n be the solution set to the (homogenous) equation Ax=0."

So since both have x in their elements, can't we say that S subset (v+S_0)?
 
  • #8
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In #1 above the hypothesis probably was something like, "suppose v and w are in S." Then you showed that v+w was also in S, thereby showing that S was closed under vector addition. v and w are symbols that can represent any of a possibly infinite number of vectors.
Here's what i know now:

A ε M_mn (F) where F is a field
b ε F^m
v ε F^n [is a solution to Ax=b]
x ε F^n [is a solution to both Ax=b and Ax=0]
w ε F^n [since the definition states v+S_0 = {v+w: w ε S_0}]

S_0 is the solution set to equation Ax=0
S is the solution to Ax=b

I also know that A(v+w)=b. Then from there it follows that (v+w) is an element of S

But i'm not sure if i can say w ε S since (v+w) is an element of S.

I'm not able to make that next step to come to that conclusion. This is where i'm stuck at.


Is it because (v+w) ε S, and S is a subset of F^n, then from the field axiom v ε F and w ε F?
 
  • #9
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Hint: Suppose w ∈ S. Show that w - v ∈ S0. What does this say about w?
kramer733 said:
But i'm not sure if i can say w ∈ S since (v+w) is an element of S.
Of course you can assume that w ∈ S. That's what "Suppose w ∈ S" means.
 
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  • #10
vela
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Here's what i know now:

A ε M_mn (F) where F is a field
b ε F^m
v ε F^n [is a solution to Ax=b]
Fine up to here.
x ε F^n [is a solution to both Ax=b and Ax=0]
No, what you wrote would imply that b=0.

The problem statement is saying that S is the solution set to the equation Ax=b. In other words, [itex]S = \{x \in F^n : Ax=b\}[/itex]. Similarly, you have [itex]S_0 = \{x \in F^n : Ax=0\}[/itex].
w ε F^n [since the definition states v+S_0 = {v+w: w ε S_0}]
This isn't really a definition of w, which is really just a dummy variable here. The problem statement is saying the notation "v+S0" means "the set of vectors of the form v+w where w is some vector in S0." You could have just as well written [itex]v+S_0 = \{v+Z : Z \in S_0\}[/itex] That no more defines Z as the original definition defines w. Both w and Z are dummy variables. They have no meaning outside of the context of the definition of v+S0.
I also know that A(v+w)=b. Then from there it follows that (v+w) is an element of S
You have to somehow define w first. Note that the w in part (a) and the w in part (b) have absolutely nothing to do with each other.
But i'm not sure if i can say w ε S since (v+w) is an element of S.

I'm not able to make that next step to come to that conclusion. This is where i'm stuck at.


Is it because (v+w) ε S, and S is a subset of F^n, then from the field axiom v ε F and w ε F?
You seem to be unclear on what you can assume.

To show S is a subset of v+S0, what you need to show is that for all vectors w in Fn, if w is in S, it is also in v+S0. So you assume w is in S and then show it follows that w is an element of v+S0. (If w is not in S, it doesn't matter because the implication is then automatically true.)
 

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