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Show that the set S is a subspace of R^3

  1. Apr 3, 2012 #1
    1. The problem statement, all variables and given/known data

    Determine if the set is a subspace:

    S = {(a, b, c) 2 R3 | a − 2b = 0 and 2a + b + c = 0};

    2. Relevant equations

    as above

    3. The attempt at a solution

    It is a subspace, I'm just not 100% sure how to write up the proof. So far I have this:

    The set is non-empty, as it contains the zero vector (0, 0, 0).

    0 - (2)0 = 0 and (2)0 + 0 + 0 = 0

    Let X = (2, 1, -5) and Y = (4, 2, -10) (element of) S and z be any real number.

    X + Y = (2 + 4, 1 + 2, -5 + -10) = (6, 3, -15) and 6 - (2)3 = 0 and (2)6 + 3 + -15 = 0

    (z)X = (z)(2, 1, -5) = ((z)2, (z)1, (z)(-5)) and (z)2 - (2)(z)1 = 0 and (2)(z)2 + (z)1 + (z)(-5) = 0

    Is it alright to use actual numbers in my proof? All the examples we did in class used letters only, and I wonder if I'm only proving it for two vectors, and not all the possible vectors in the subspace? Is it enough to show it like this? Also I'm a bit sketchy with my example of closure under scalar multiplication, I know it works for any z, but do I need a concrete example?

    Proofs are a bit confusing!
     
  2. jcsd
  3. Apr 3, 2012 #2

    tiny-tim

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    Hi Adyssa! :smile:
    Noooo.

    No no no no no!

    In other words: no.​

    (just got up! :zzz: :biggrin:)

    Use (a1,b1,c1) and (a2,b2,c2). :smile:

    (or (a,b,c) and (d,e,f) … but i think the former method is less confusing)
    Again, no (etc :wink:).

    Use an abstract "z".

    Remember, for all these proofs, you have to prove that it's true for all values, so technically your proof begins "For any vectors (a1,b1,c1) and (a2,b2,c2) …"
     
  4. Apr 3, 2012 #3
    OK thanks for the clarification. My new result is as follows:

    For any vectors X = (a1, b1, c1) and Y = (a2, b2, c2) [itex]\in[/itex] S and any real number z:

    X + Y = (a1 + a2, b1 + b2, c1 + c2) and (a1 + a2) - (2)(b1 + b2) = 0 and (2)(a1 + a2) + (b1 + b2) + (c1 + c2) = 0

    (z)X = (z)(a1, b1, c1) = ((z)a1, (z)b1, (z)c1) and (z)a1 - (2)(z)b1 = 0 and (2)(z)a1 + (z)b1+ (z)c1 = 0

    It seems to me that I haven't actually shown anything, particularly with the closure under addition, apart from the mechanics of vector addition. It's up to the reader to find vectors that fit.

    edit: subscripts
     
    Last edited: Apr 3, 2012
  5. Apr 3, 2012 #4

    tiny-tim

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    Hi Adyssa! :smile:

    (try using the X2 button just above the Reply box :wink:)
    No, you've defined those vectors (X + Y and zX) (by their coordinates) … the reader has no choice.

    Your proof is fine, I don't see what's worrying you about it. :confused:

    Rewrite it with the subscripts, so that it's clearer, and then keep reading it until you're convinced! :smile:
     
  6. Apr 3, 2012 #5
    I cleaned it up a bit. This part is the key huh:

    "For any vectors X = (a1, b1, c1) and Y = (a2, b2, c2) ∈ S"

    Not any vectors X, Y, just those X, Y ∈ S

    Thanks for your help!
     
  7. Apr 3, 2012 #6

    HallsofIvy

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    You haven't used the fact that these vectors must be in the subset- that [itex]a_1+ 2b_1= 0[/itex], [itex]a_2+ 2b_2= 0[/itex], [itex]2a_1+ b_1+ c_1= 0[/itex], and [itex]2a_2+ b_2+ c_2= 0[/itex] to show that the same is true of [itex](a_1+ a_2, b_1+ b_2, c_1+ c_2)[itex] and [itex](za_1, za_2, za_3)[/itex].
     
  8. Apr 3, 2012 #7

    tiny-tim

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    Yes, she has, it's a bit difficult to read, but …
     
  9. Apr 4, 2012 #8
    I see your point HallfofIvy and I've taken your advice, that is, showing first for each vector, and then for the sum of the two vectors. The scalar product I only show for a single vector but I think this is fine. I see in some examples that both axioms are proved together (scalar product of the sum of two vectors) but I will write it out the long way until I get a good grip on things.

    And next time I will use latex so it's more readable, I thought I might get away with but it in retrospect it looks awful. >.<

    Also, tiny-tim, I'm not a she, confusing handle I guess. For the record, it's a song that sort of defined an era for me, not a name, at least not a name that I know. :)
     
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