Determine if S is a subspace of V

  1. My answers aren't all correct and I am not sure why..

    Problem: Determine whether the given set S is a subspace of the vector space V.

    A. V is the vector space of all real-valued functions defined on the interval [a,b], and S is the subset of V consisting of those functions satisfying f(a)=f(b).
    B. V=R^n, and S is the set of solutions to the homogeneous linear system Ax=0 where A is a fixed m×n matrix.
    C. V=C^1(R), and S is the subset of V consisting of those functions satisfying f′(0)≥0.
    D. V=P_4, and S is the subset of P_4 consisting of all polynomials of the form p(x)=a(x^3)+(bx).
    E. V=R^n×n, and S is the subset of all symmetric matrices.
    F. V=R^4, and S is the set of vectors of the form (0,x2,3,x4).
    G. V=R^n×n, and S is the subset of all matrices with det(A)=0.

    I chose A,B,C, E, G.

    A: sine an cosine are real valued functions. in the case of sine,on the interval [0,2pi] f(0)=f(2pi).

    B:Just looking at the dimensions A has n columns which means( i think) n unknowns therefore falls in R^n

    C: if f(t) = t then f'(t)=1 which is greater than 0 at t=0.

    E:a symmetric matrix is a square matrix

    G:A must be square in order to take a determinate.

    I did not choose D and F:

    D: because P(x) is of degree 3 and V=P_4

    F: does not contain the 0 vector.

    Any insight would be greatly appreciated.
     
    Last edited: Feb 13, 2014
  2. jcsd
  3. In general, what sort of things do you need to check in order to determine whether a subset of a vector space is a subspace?
     
  4. HallsofIvy

    HallsofIvy 40,241
    Staff Emeritus
    Science Advisor

    This is not a multiple choice question. It is a "yes", "no" question. Do you mean "yes" for A, B, C, E, and G and no for the others.

    I have no idea what you mean here. Are you saying this is why the answer to A is "yes"? What you are saying here has nothing to do with being subspace.

    So you are saying is that the set of solutions is a subset of R^n. But the question was "is it a subspace?" You have not addressed the question.

    So there exist as function satisfying the conditions. That has NOTHING to do with the question: is this set a subspace?

    So a subset. But you have not proved it is a subspace.

    P_4 is the space of polynomials of degree less than or equal to 4.

    You did not, in most of these, address the question itself: is this subset a subspace. Do you know what a subspace is or what properties a subset must have to be a subspace?
     
  5. HallsofIvy

    HallsofIvy 40,241
    Staff Emeritus
    Science Advisor

    This is not a multiple choice question. It is a "yes", "no" question. Do you mean "yes" for A, B, C, E, and G and no for the others?

    I have no idea what you mean here. Are you saying this is why the answer to A is "yes"? What you are saying here has nothing to do with being subspace.

    So you are saying is that the set of solutions is a subset of R^n. But the question was "is it a subspace?" You have not addressed the question.

    So there exist as function satisfying the conditions. That has NOTHING to do with the question: is this set a subspace?

    So a subset. But you have not proved it is a subspace.

    P_4 is the space of polynomials of degree less than or equal to 4.

    You did not, in most of these, address the question itself: is this subset a subspace. Do you know what a subspace is or what properties a subset must have to be a subspace?
     
  6. subspace is a non empty vector space where the rules for vector addition and scalar multiplication are the same as the vector space it's a part of...so how would one prove that for these instances?
     
    Last edited: Feb 14, 2014




  7. Im supposed to pick the ones that are sub spaces. The ones under "I chose" are the ones a chose and the ones under" I did not choose", i did not choose. I hope that clears it up.

    I'm supposed to look to see if scalar multiplication and vector addition hold.
     
  8. Well, you pretty much answered your own question in another post; you need to check that vector addition and scalar multiplication "hold".

    In particular, you need to check that

    (1) ##u+v## is in ##S## whenever ##u## and ##v## are in ##S## and
    (2) ##\alpha u## is in ##S## for all real numbers ##\alpha## whenever ##u## is in ##S##.

    For instance, in (A), if ##f## and ##g## are in ##S##, then ##f(a)=f(b)## and ##g(a)=g(b)##. So

    (1) ##(f+g)(a)=f(a)+g(a)=f(b)+g(b)=(f+g)(b)##, so ##f+g## is in ##S##, and
    (2) for all real ##\alpha##, ##(\alpha f)(a)=\alpha f(a)=\alpha f(b)=(\alpha f)(b)##, so ##\alpha f## is in ##S##.

    So ##S## is a subspace of ##V## in (A).

    Based on what you wrote in the original post, I would strongly encourage you to, in each case, take a minute (or more) to figure out exactly what ##V## is, understand what it's members "look like", understand the vector addition and scalar multiplication, and verify that it is a vector space. Then look at the defining aspect of ##S## and truly understand what it means for a member of ##V## to also belong to ##S##. Then go ahead and check (1) and (2).

    Don't take this the wrong way, but almost nothing you said in the first post is really pertinent to this problem. I'm not trying to put you down. I just don't want you to waste time trying to salvage that work.

    Also, check to make sure that you copied (F) down correctly.
     
  9. No harm done. How could I possibly be offended? I have no experience with these problems. Thank you both for your time and patience. This has been enlightening.
     
    Last edited: Feb 14, 2014
  10. And in response to:

    check to make sure that you copied (F) down correctly.

    It's copy pasted. But checked anyway and it's correct.
     
    Last edited: Feb 14, 2014
  11. Math students (and I suppose teachers as well) are frequently completely unaware of their shortcomings, and many have fragile egos to boot. I usually (not always) err on the side of apologizing unnecessarily (often beforehand) rather than potentially offending. In other words, I try to nip it in the bud. It doesn't always work.

    OK. It just seems to me like the set of vectors of the form ##(0,x^2,x^3,x^4)## is a more interesting example. If what you have really is the problem, then you can actually keep what you have done there. Any subset that lacks the ##0## vector fails the (2) check.

    Also, there is a one-step check; for all ##u## and ##v## in ##S## and all real ##\alpha## and ##\beta##, ##\alpha u+\beta v## is also in ##S##. But I've never really found that to be very useful in practice. It's usually just as easy, if not easier, to check the conditions separately. The one-step check is more for when you completely understand the concept and just want to get through the problem as quickly as possible.
     
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