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Homework Help: Subset and subspace

  1. Mar 11, 2014 #1
    1. The problem statement, all variables and given/known data

    a)Which of these subsets contain the zero vector 0 = (0, 0, 0, 0) ?
    b)Which of these subsets are subspaces of R4 ?

    S = { (x1, x2, x3, x4) | x4 = -6 - 5 x1 }
    T = { (x1, x2, x3, x4) | x4 is an integer }
    U = { (x1, x2, x3, x4) | x1 + x4 ≤ -6 }

    3. The attempt at a solution

    If a set is closed under addition and closed under scalar multiplication, then it contains the zero vector. But isn't the definition of subset such that a vector space is a linear combination of all other vector. I'm terribly lost!

    I'm really struggling badly with this. Can someone give me a leg up?

    S = (x1,x2,x3,x4) and let W = (w1,w2,w3,w4)

    then S+W= (x1+w1,x2+w2,x3+w3,x4+w4)

    (x4+w4) = -6-5(x1+w1) (closed under addition?) (Where do I go from here?)

    Let k = scalar

    KW = (Kw1, kw2,kw3,kw4)

    kw4 = -6-5(kw1) (closed under scalar?)

    There are values for me to determine if the left hand side values equals the right hand side values.
    Last edited: Mar 11, 2014
  2. jcsd
  3. Mar 11, 2014 #2


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    a) fill in 0000:
    S yes/no ?
    T yes/no ?
    U yes/no ?
    b) Only have to look at the yesses ! Take a vector ##\vec x## and see if ##c\vec x## is always in the candidate.

    Why are you looking at S + T ?
    The question is: Is S a subspace of R4 or is it not
    Or am I the one misinterpreting the exercise ?
  4. Mar 11, 2014 #3


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    By editing my reply becomes somewhat floating in the air!

    Why don't you just check that 0000 is not in S and thereby discard S as a subspace?
  5. Mar 11, 2014 #4
    I'm equally confused too but I'm going with you.
    If I get you correctly:

    Let S = (x1,x2,x3,x4) = (0,0,0,0,)

    S = (x1,x2,x3,x4)| x4=-6-5x1

    x1 = 0, x2 = 0, x3=0,x4 = 0

    x4 = -6-5(x1) = -6-5(0)=-6
    -6=/= 0

    0→ is not an element of S so S does not contain the 0→

    Let T = (x1,x2,x3,x4) = (0,0,0,0)

    T = (x1,x2,x3,x4)| x4 is an integer
    x1=0, x2=0, x3=0, x4=0

    I presume that by 'integer', it implies relative integer. 0 is a relative integer.
    Hence (x4=0) = 0 = integer

    The 0→ is an member vector in T and so T contains the 0→

    Let U = (x1,x2,x3,x4) = (0,0,0,0)

    U = (x1,x2,x3,x4) | x1+x4 =< -6

    x1 = 0, x2=0,x3=0,x4=0
    x1 + x4 = 0, and so, 0 ~ -6 and ~< -6
    0→ is not a member vector of U.

    b) only T contains the zero vector.

    Span(T) = R4 (?)

    c1x1 + c2x2 + c3x3 +c4x4 = (x1,x2,x3,x4)
    c4x4 = 0 since in part (a) it has been established x4 =0 and any scalar, c, times zero = 0.
    Did I got it correctly?
    Last edited: Mar 11, 2014
  6. Mar 11, 2014 #5


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    No, S ≠ <0, 0, 0, 0>. S is a set of vectors in R4, subject to the condition you posted at the start of this thread. All you need to do is determine whether the zero vector satisfies the condition that x4 = -6 - 5x1
    Rather than use "→" to stand for "vector" it's better to just write vector.
    Span has nothing to do with showing that a subset of a vector space is a subspace of that vector space. Your textbook should show you the three things you need to check to demonstrate that a subset of a vector space is a subspace. Showing that the subset contains the zero vector is one of those things. What are the other two?
  7. Mar 11, 2014 #6
    One moment. Is part (a) correct?

    1) zero vector must be a member vector of the set.
    2) the set must be closed under addition operation
    3) the set must be closed under scalar multiplication

    How do I apply the above to the existing problem?
    Last edited: Mar 11, 2014
  8. Mar 11, 2014 #7


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    Yes part (a) is correct: Only T contains (0,0,0,0)
    Now for part (b). We don't have to look at S and U any more, because they don't contain (0,0,0,0).
    T only has as a criterion that x4 is an integer.
    It satisfies 1): (0,0,0,0) is a member.
    Now how about (2) ? If (a,d,c,d) and (e,f,g,h) both are members of T, what about the sum ?
  9. Mar 11, 2014 #8
    Is coming up with (a,b,c,d) and (e,f,g,h) a general strategy to this sort of problems?

    Let u = (u1,u2,u3,u4) and v = (v1,v2,v3,v4)

    By property (2): u+v = (u1+v1,u2+v2,u3+v3,u4+v4)

    x1 = u1+v1, x2 = u2+v2, x3 = u3+v3, x4 = u4+v4
    then, since x4 is an integer, u4+v4 is an integer because u4+v4 = 0?
  10. Mar 11, 2014 #9


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    Almost. A general strategy is a bit presumptuous: it's more like a beginning of the solution process to take an example and then generalize while going along.

    In this case if d and h are integer, then d+h is an integer too. So I am happy rule 2 holds.
    You are happy too, not because u4+v4=0 but because if u4 is an integer and v4 is an integer too, then u4+v4 is also an integer. Which means that the sum vector is indeed IN the subset.

    Now what about scalar multiplication ? Is criterion 3 satisfied ?
  11. Mar 11, 2014 #10

    Suppose U = (u1,u2,u3,u4) and k is a scalar an element of real number.

    kU = (ku1,ku2,ku3,ku4)

    x4 = ku4
    Since k is a scalar an element of real number and u4 an integer, then ku4 is an integer.
  12. Mar 11, 2014 #11


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    Are you actually thinking about this?
  13. Mar 11, 2014 #12


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    Back to the basic strategy: fill in something and see where it leads:

    (0,0,0,5) is a member. 1.5 is a scalar. 1.5 (0,0,0,5) = (0,0,0,7.5) So here we have found a counter example. One case where a scalar times a member does not give us a member. Conclusion ?
  14. Mar 11, 2014 #13

    Where did 0,0,0,5 came from?
  15. Mar 11, 2014 #14


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    I invented it. It is a member of T. Anything else would have been fine too. Point is that if the fourth coordinate is an integer and you multiply by a scalar that is not an integer, then you get something with a fourth coordinate that is not an integer and that means you are no longer in the subset. That means the subset is not a subspacce.
  16. Mar 11, 2014 #15
    Yes you're right!
    I forgotten that it was real number, k x integer.
    It's terrible having only 4 hours of sleep each day for the past 2 weeks..
  17. Mar 11, 2014 #16


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    Get some rest. We won't go away all at the same time, so someone's there when you get up and have more questions !
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