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Linear Dependence and Subsets

  1. Jan 14, 2012 #1
    1. The problem statement, all variables and given/known data

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

    3. The attempt at a solutionRead post #2. This proof, I think, was incorrect.

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

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

    Sound about right?
    Last edited: Jan 14, 2012
  2. jcsd
  3. Jan 14, 2012 #2
    Actually, I have to revise my proof. I don't think I can write that [itex]\vec0 \in E[/itex]. In any case, we know that [itex]e_1 E_1 + ...+ e_n E_n = \vec0[/itex]. We can therefore say (I think) that [itex]\vec0 \in span{E}[/itex]. [itex]E \subseteq F[/itex] implies [itex]spanE \subseteq spanF[/itex]. Thus, [itex]\vec0 \in spanF[/itex]. Since E has a nontrivial solution to [itex]e_1 E_1 + ...+ e_n E_n = \vec0[/itex] and since [itex]\vec0 \in spanF[/itex], 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?
  4. Jan 14, 2012 #3


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    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 [itex]F=\{F_1,F_2,...,F_n\}[/itex] and let's say that the first m of those vectors are the ones that also belong to E, so [itex]E=\{F_1,F_2,...,F_m\}[/itex] where m<=n. Try expressing a proof using that notation.
  5. Jan 14, 2012 #4
    So if I write [itex]F=\{F_1,F_2,...,F_n\}[/itex] and [itex]E=\{F_1,F_2,...,F_m\}[/itex] then in E there exists [itex]F_1,...,F_m[/itex] distinct vectors such that [itex]f_b F_b + ... + f_m F_m = \vec0[/itex], where [itex]f_1,...,f_m[/itex] are not all zero vectors. Because [itex]E \subseteq F[/itex], F also contains these distinct vectors. Thus, in F we can say that there exist [itex]F_1,...,F_m[/itex] distinct vectors such that [itex]f_b F_b + ... + f_m F_m = \vec0[/itex]. Hence, we see that F is also linearly dependent.
  6. Jan 14, 2012 #5


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    The phrasing is still clumsy. Why are you starting with [itex]f_b[/itex]? Why not start with [itex]f_1[/itex]? And the f's aren't vectors, they are scalars. You've got the right idea though.
  7. Jan 14, 2012 #6
    I guess it doesn't matter to start with F1 or b. I thought Fb 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?
  8. Jan 14, 2012 #7


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    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.
  9. Jan 15, 2012 #8
    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 [itex]spanE + spanF[/itex], is this just the same as [itex]spanE \cap spanF[/itex]? I ask because I was doing some practice problems and one of them asked to show that [itex]L(E \cap F) \subseteq L(E) \cap L(F)[/itex]. The answer to this problem showed that [itex]L(E \cap F) \subseteq spanE + spanF[/itex], 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.
  10. Jan 15, 2012 #9


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    No. [itex]spanE + spanF[/itex], 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 [itex]spanE \cap spanF[/itex] and extend it to a basis for span(E) and span(F) to see the relation.
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