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Subspace and subset

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

    S = { (x1, x2, x3, x4) | 4 x1 + x3 = 3 + 6 x2 + x4 }
    T = { (x1, x2, x3, x4) | x1 + x3 is an integer }
    U = { (x1, x2, x3, x4) | x1 x3 ≥ -5 }

    3. The attempt at a solution

    a) Which of these subsets contain the zero vector 0 = (0, 0, 0, 0) ?


    S = (x1,x2,x3,x4) = (0,0,0,0)
    4x1 =x3 = 3+6x2 + x4
    4(0) + 0 = 3+ 6(0) + 0
    0=3
    S is false

    T = (x1,x2,x3,x4) = (0,0,0,0)
    x1+x3 = integer and I assume that by 'integer' it implies relative integer such that 0 is an integer.
    If we assume so, then,
    x1 + x3 = 0 + 0 = 0

    T is true

    U = (x1,x2,x3,x4) = (0,0,0,0)
    x1.x3 =>-5
    0.0 = 0 but 0~= -5 but 0>-5

    U is true.

    b) Which of these subsets are subspaces of R4 ?
     
  2. jcsd
  3. Mar 11, 2014 #2

    Mark44

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    Your work for part a looks OK to me, but you conclude things like "T is true". S, T, and U are subsets of R4, so true/false isn't applicable. Instead, your answers should say whether the set contains the zero vector.

    For the b part, if you know that the containing space is a vector space (for your problems the space that contains each subset is R4, which is a vector space), what three things do you need to check to conclude that the subset is actually a subspace of the containing vector space? One of the things to check is whether the zero vector is an element of the subset. What are the other two?
     
  4. Mar 11, 2014 #3
    There is a persistent confusion regarding the definition of subset, subspace and vector space. In this case, is R4 the vector space or is, say, set A the vector space?
    It's all pretty relative isn't it? If R4 is the vector space and set A is the element of member vectors, then, set A is the subset of R4. The member vector in the set A is also a subset of A.
    Would be great if you could shed some light above.

    As to your question:

    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

    Another question, how does the concept of linear combination comes into the picture? I see the big picture but I'm having a hard time weaving the intermediate links.
     
  5. Mar 11, 2014 #4

    BvU

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    No, there is not. You can look at a few Khan videos to lift the fog! Subset, Subspace.

    R4 is generally a vector space. I don't know what A is. That is not the same as being confused about what A is.

    No, it is not.

    You can't say: "set A is the element of member vectors" without confusing yourself.
    You can say "A member vector in the set A is also a subset of A" because that is true. Something that is in A is in A.
     
  6. Mar 11, 2014 #5

    BvU

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    Re linear combination: If you can show a subset contains a zero element, is closed under addition and scalar multiplication, then that means that a linear combination of elements can not bring you out of the subset!

    If ##\vec x## and ##\vec y## are in the subset, then ##a\ \vec x + b\ \vec y## is also in the subset: ##a\ \vec x ## and ## b\ \vec y## by rule (3) and the sum by rule (2)
     
  7. Mar 11, 2014 #6

    Mark44

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    The confusion is on your part, right? All three terms are well defined.
    All of the vectors in your first post are 4-tuples, and the question in part b asks which of the three sets in part a are subspaces of R4. So clearly, R4 is the vector space being considered.
    As already said by BvU, no, it's not relative.
    A is not mentioned in your first post, so I don't know what you're asking.
    The rest of what you're saying doesn't make any sense.
    "set A is the element of member vectors" -- ??
    "set A is the subset of R4." -- It's given that S, T, and U are subsets of R4. Each vector in each of these sets has four coordinates. If these coordinates are real numbers, then these sets are subsets of R4, the vector space of 4-tuples of real numbers.

    What the problem is asking you is to identify which of these sets (S, T, and U) is also a subspace of R4.

     
  8. Mar 11, 2014 #7
    By part (a), T and U fulfills the condition of the zero vector.
    Hence, we test for the 2 other condition.

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

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

    Since u1+v1 and u3+v3 is an integer, then, (u1+v1) + (u3+v3) is an integer.


    Let u = (u1,u2,u3,u4)

    Scalar k, is a real number.

    k.u = (ku1,ku2,ku3,ku4)
    ku1 + ku3 = k(u1+u3)
    k(u1+u3) =/= integer

    so, T is not a subspace of R4.

    U = { (x1, x2, x3, x4) | x1 x3 ≥ -5 }
    Let w = (w1,w2,w3,w4)
    Let v = (v1,v2,v3,v4)

    w+v = (w1+v1,w2+v2,w3+v3,w4+v4)
    (w1+v1)(w3+v3) = 0
    0>-5

    Let w = (w1,w2,w3,w4)

    scalar k, is a real number.
    kw = (kw1,kw2,kw3,kw4)
    kw1.kw3= 0
    0>5

    U is a subspace of R4
     
  9. Mar 12, 2014 #8

    BvU

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    Hello, hope you had a good rest and are fresh and fruity again. Ready for some tough opposition again :smile:
    I agree that T is closed under addition. Not under scalar multiplication, so : no subset.

    Now over to U.
    Let w = (w1,w2,w3,w4) ##\in## U and v = (v1,v2,v3,v4) ##\in## U, can we prove that w+v ## \in## U ? ## (\in## U means "is in", ∈## )## ?

    From the U definition we know that w1 w3 ≥ -5 and that v1 v3 ≥ -5. That is all we know. Just for the fun of it, I drew a picture. Very helpful. Have some problems with 4 dimension drawings, so I restrict myself to the u1, u3 plane. Do you understand that u2 and u4 can be anything in ##R^4## and also in subset U ##\subset R^4## ?

    Blue hyperbolic lines are the limits of U. Anything on or in between ##\in ## U. Agree ?

    If we want to prove that U is closed under addition, we have to prove that
    $$ (w_1\ w_3 ≥ -5) \enspace \wedge \enspace (v_1\ v_3 ≥ -5) \enspace \Rightarrow \enspace (w_1+v_1)(w_3+v_3) ≥ -5 $$which is going to be very difficult:$$ (w_1+v_1)(w_3+v_3) = w_1 w_3+v_1 v_3 + w_1 v_3 + w_3 v_1 \geq -10 + w_1 v_3 + w_3 v_1 $$and then we know nothing.

    If we want to disprove that U is closed under addition, all we have to do is find one example that has ## (w_1+v_1)(w_3+v_3) < -5 ##. Much easier. Can you give an example ?
     

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  10. Mar 12, 2014 #9

    HallsofIvy

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    Wrong word. Subset, yes. Not a subspace.
     
  11. Mar 12, 2014 #10
    (u3 = 2, u1 = -6)?
    u1. u3 = -12
    -12 < -5.

    It's basically any u1 and u3 points that falls within the area enclosed by the hyperbolic, isn't it?
     
    Last edited: Mar 12, 2014
  12. Mar 12, 2014 #11

    BvU

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    Ivyleague is absolutely right.

    Now your candidate. You give me one vector that is outside U. What I was fishing for was two vectors inside U but with a sum outside U !

    Not true. For the whole first quadrant you can add as much as you like without leaving the quadrant ! Idem third quadrant! So you will have to search some more to come up with something that disproves U is closed under addition.

    Oh boy, I've managed to create confusion again, sorry. The hyperbola does not enclose anything. My sloppy language, sorry. But you did catch my meaning, so the damage is limited, I hope.​

    I seem to remember that for all vectors inside U one has x1 x3 ≥ -5. That definitely includes the whole first and third quadrants in the figure. The limits/boundaries of U, if you can call them that, are the blue lines with x1 x3 = -5.

    So the upper left and the lower right of the figure are outside U.

    Now I want you to find two vectors inside U but with a sum outside U.
    Big hint: they don't have to be two different vectors....
     
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