Linear Dependence and Subsets: Proving Linear Dependence in Sets of Vectors

[TranscendArcu]

Homework Statement

Suppose that E,F are sets of vectors in V with E \subseteq F. Prove that if E is linearly dependent, then so is F.

The Attempt at a Solution

Read post #2. This proof, I think, was incorrect.

If we suppose that E is linearly dependent, then we know that there exists E_1,...,E_n distinct vectors such that,
e_1 E_1 + ... + e_n E = \vec0, where e₁,...,e_n are numbers. Thus, we know \vec0 \in E. Since E \subseteq F, \vec0 \in F. Any set containing the zero-vector must certainly be linearly dependent since n \vec0 = \vec0, where n is any number.

Thus, if E is linearly dependent, then F is linearly dependent.

Sound about right?

 

[TranscendArcu]

Actually, I have to revise my proof. I don't think I can write that \vec0 \in E. In any case, we know that e_1 E_1 + ...+ e_n E_n = \vec0. We can therefore say (I think) that \vec0 \in span{E}. E \subseteq F implies spanE \subseteq spanF. Thus, \vec0 \in spanF. Since E has a nontrivial solution to e_1 E_1 + ...+ e_n E_n = \vec0 and since \vec0 \in spanF, F must also have a nontrivial solution to get the zero vector. In one such case, F's nontrivial solution is identical to E's.

Is that better?

 

[Dick]

Actually, I have to revise my proof. I don't think I can write that \vec0 \in E. In any case, we know that e_1 E_1 + ...+ e_n E_n = \vec0. We can therefore say (I think) that \vec0 \in span{E}. E \subseteq F implies spanE \subseteq spanF. Thus, \vec0 \in spanF. Since E has a nontrivial solution to e_1 E_1 + ...+ e_n E_n = \vec0 and since \vec0 \in spanF, F must also have a nontrivial solution to get the zero vector. In one such case, F's nontrivial solution is identical to E's.

Is that better?

Nope. The zero vector is in ANY span. F is a set of vectors. E is a subset of those vectors. Let's list the elements of F as F=\{F_1,F_2,...,F_n\} and let's say that the first m of those vectors are the ones that also belong to E, so E=\{F_1,F_2,...,F_m\} where m<=n. Try expressing a proof using that notation.

 

[TranscendArcu]

Nope. The zero vector is in ANY span. F is a set of vectors. E is a subset of those vectors. Let's list the elements of F as F=\{F_1,F_2,...,F_n\} and let's say that the first m of those vectors are the ones that also belong to E, so E=\{F_1,F_2,...,F_m\} where m<=n. Try expressing a proof using that notation.

So if I write F=\{F_1,F_2,...,F_n\} and E=\{F_1,F_2,...,F_m\} then in E there exists F_1,...,F_m distinct vectors such that f_b F_b + ... + f_m F_m = \vec0, where f_1,...,f_m are not all zero vectors. Because E \subseteq F, F also contains these distinct vectors. Thus, in F we can say that there exist F_1,...,F_m distinct vectors such that f_b F_b + ... + f_m F_m = \vec0. Hence, we see that F is also linearly dependent.

 

[Dick]

So if I write F=\{F_1,F_2,...,F_n\} and E=\{F_1,F_2,...,F_m\} then in E there exists F_1,...,F_m distinct vectors such that f_b F_b + ... + f_m F_m = \vec0, where f_1,...,f_m are not all zero vectors. Because E \subseteq F, F also contains these distinct vectors. Thus, in F we can say that there exist F_1,...,F_m distinct vectors such that f_b F_b + ... + f_m F_m = \vec0. Hence, we see that F is also linearly dependent.

The phrasing is still clumsy. Why are you starting with f_b? Why not start with f_1? And the f's aren't vectors, they are scalars. You've got the right idea though.

 

[TranscendArcu]

The phrasing is still clumsy. Why are you starting with f_b? Why not start with f_1? And the f's aren't vectors, they are scalars. You've got the right idea though.

I guess it doesn't matter to start with F₁ or _b. I thought F_b gave the impression that E didn't necessarily have to start with the first element of F. But I seem to recall that ordering doesn't matter in sets anyway (isn't that right?)

Besides these problems (which I think will be pretty trivial to fix) is there anything else that is making the proof clumsy?

 

[Dick]

I guess it doesn't matter to start with F₁ or _b. I thought F_b gave the impression that E didn't necessarily have to start with the first element of F. But I seem to recall that ordering doesn't matter in sets anyway (isn't that right?)

Besides these problems (which I think will be pretty trivial to fix) is there anything else that is making the proof clumsy?

In setting it up we designated that the first m elements of F were to be the elements of F n E. Sure, sets are unordered, but our labeling of the elements of the sets has some useful order. You kept the last element of E as F_m after all. Don't ponder this deeply, just change the 'b' to '1' and see if it reads more simply.

 

[TranscendArcu]

Okay. Thanks!

I won't rewrite the work here since there isn't much need if the changes you recommend are so easy to make.

On the other hand, I also want to ask another question regarding sets. If I have a statement like spanE + spanF, is this just the same as spanE \cap spanF? I ask because I was doing some practice problems and one of them asked to show that L(E \cap F) \subseteq L(E) \cap L(F). The answer to this problem showed that L(E \cap F) \subseteq spanE + spanF, but stopped there. It isn't immediately obvious to me how the stuff on the right hand side of the subset sign relates to what I was asked to show.

 

[Dick]

Okay. Thanks!

I won't rewrite the work here since there isn't much need if the changes you recommend are so easy to make.

On the other hand, I also want to ask another question regarding sets. If I have a statement like spanE + spanF, is this just the same as spanE \cap spanF? I ask because I was doing some practice problems and one of them asked to show that L(E \cap F) \subseteq L(E) \cap L(F). The answer to this problem showed that L(E \cap F) \subseteq spanE + spanF, but stopped there. It isn't immediately obvious to me how the stuff on the right hand side of the subset sign relates to what I was asked to show.

No. spanE + spanF, is defined as the set of all vectors e+f where e is in span(E) and f is in span(F). Pick a basis for spanE \cap spanF and extend it to a basis for span(E) and span(F) to see the relation.