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Annihilator of a vector space

  1. Feb 2, 2009 #1
    1. The problem statement, all variables and given/known data
    If W1 and W2 are subspaces of V, which is finite-dimensional, describe A(W1+W2) in terms of A(W1) and A(W2). Describe A(W1 intersect W2) in terms of A(W1) and A(W2).

    A(W) is the annihilator of W (W a subspace of vector space V). A(W)={f in dual space of V such that f(w)=0 for all w in W}.

    3. The attempt at a solution

    All I have thought of so far is that
    A(W1+W2) is contained in A(W1) and
    A(W1+W2) is contained in A(W2)

    I'm very lost so any help is appreciated.
     
  2. jcsd
  3. Feb 2, 2009 #2
    I HATE THE ANNIHILATOR!!! I completely skipped that problem on my Final :) Still got a B+, thank goodness!
     
  4. Feb 3, 2009 #3

    quasar987

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    So you have convinced yourself that [itex]A(W_1+W_2)\subset A(W_1)\cap A(W_2)[/itex]. Could it be that the reverse inclusion is true as well, so that we have [itex]A(W_1+W_2)=A(W_1)\cap A(W_2)[/itex]?
     
  5. Feb 3, 2009 #4
    Well, wouldn't the equality be true if and only if W1=W2?
     
  6. Feb 3, 2009 #5

    quasar987

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    Suppose [itex]f\in A(W_1)\cap A(W_2)[/itex]. If [itex]w_1+w_2[/itex] is any element of [itex]W_1+W_2[/itex], what is [itex]f(w_1+w_2)[/itex]?
     
  7. Feb 3, 2009 #6
    Is this the right reasoning:

    If [itex]f\in A(W_1)\cap A(W_2)[/itex], then f is in [itex]A(W_1+W_2)[/itex]
    so if f brings some element in their intersection to zero, then f will bring any element in their sum to zero.

    ?
     
  8. Feb 3, 2009 #7

    quasar987

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    Mmh, no that doesn't make sense to me.

    But do you understand the reason why proving that if f is any element of [itex]A(W_1)\cap A(W_2)[/itex] such that for any element [itex]w_1+w_2[/itex] of [itex]W_1+W_2[/itex], then [itex]f(w_1+w_2)=0[/itex] is the same as proving that [itex]A(W_1)\cap A(W_2)\subset A(W_1+W_2)[/itex] ?
     
  9. Feb 3, 2009 #8
    Yes I think I understand the reason. Tell me if I'm wrong though. The reason is because if a map acts the same way on elements in the intersection of W1 and W2 as it does on elements in W1+W2 then it must be that the intersection is contained in the sum, since the dim of the intersection is less than or equal to the dim of the sum.
     
  10. Feb 3, 2009 #9

    quasar987

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    I do not know why you think the intersection of W1 and W2 has anything to do with the matter. Perhaps you think that [itex]A(W_1)\cap A(W_2) = A(W_1\cap W_2)[/itex]? Maybe this is so, but we will find out only in the second part of the question: "Describe A(W1 intersect W2) in terms of A(W1) and A(W2)."

    The reason is much simpler. It is simple set theory. If A and B are two sets, then by definition, we write [itex]A\subset B[/itex] is for any a in A, a is also in B.

    Here, [itex] A = A(W_1)\cap A(W_2)[/itex] and [itex]B= A(W_1+W_2)[/itex]. So in order to show that [itex]A(W_1)\cap A(W_2)\subset A(W_1+W_2)[/itex], we take f in [itex] A = A(W_1)\cap A(W_2)[/itex] and ask ourselves if it is also in [itex] A(W_1+W_2)[/itex]. But what does it mean for f to be in [itex] A(W_1+W_2)[/itex]? Well, it means that for any [itex]w_1+w_2\in W_1+W_2[/itex], we have [itex]f(w_1+w_2)=0[/itex].

    Do you have any doubts left?
     
  11. Feb 3, 2009 #10
    Then since f is a homomorphism f(w1+w2)=f(w1)+f(w2)=0. Does this imply that f is in A(W1) intersect A(W2)?
     
    Last edited: Feb 3, 2009
  12. Feb 3, 2009 #11

    quasar987

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    No, the function f is in A(W1) intersect A(W2) by hypothesis. Meaning we chose it to be in A(W1) intersect A(W2). And we want to show that it follows from this hypothesis that f is in A(W1+W2) also, since this is the same as saying that [itex]A(W_1)\cap A(W_2)\subset A(W_1+W_2)[/itex].
     
  13. Feb 4, 2009 #12
    Oh I get that inclusion now.

    I can't prove the reverse inclusion, that A(W1+W2) is contained in A(W1) intersect A(W2) though.

    What about the second part of the question that asks about A(W1 intersect W2) in terms of A(W1) and A(W2). Are they equal?
     
  14. Feb 4, 2009 #13

    quasar987

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    In your first post, you said

    "All I have thought of so far is that
    A(W1+W2) is contained in A(W1) and
    A(W1+W2) is contained in A(W2)"

    Well by definition of the intersection of two sets, if A is contained in B and A is contained in C, then A is contained in [itex]B\cap C[/itex].

    So what you have "thought of" (which I assume means you have proven) is actually that [itex]A(W_1+W_2)\subset A(W_1)\cap A(W_2)[/itex].
     
  15. Feb 4, 2009 #14
    Oh I see.

    And I think A(W1 intersect W2) = A(W1) + A(W2) but I don't know how to prove the inclusion from right to left (sum contained in intersection)
     
    Last edited: Feb 4, 2009
  16. Feb 4, 2009 #15

    quasar987

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    "sum contained in intersection" is easy as it follows from the definitions.

    How did you show that "intersection contained in sum" though?
     
  17. Feb 4, 2009 #16
    For that inclusion I did

    Assume f is in A(W1) intersect A(W2).
    This implies f(w1)=0 and f(w2)=0
    implies f(w1)+f(w2)=0
    implies f(w1+w2)=0 (since f is a homomorphism)
    implies f is in A(W1+W2)
     
  18. Feb 4, 2009 #17

    quasar987

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    But this shows [itex]A(W_1)\cap A(W_2)\subset A(W_1+W_2)[/itex].

    I though we were done with the first part of the question and we are talking about the second part of the question that asks to describe A(W1 intersect W2) in terms of A(W1) and A(W2).

    In post #14, you said you had [itex]A(W_1\cap W_2)\subset A(W_1)+A(W_2)[/itex]. How did you prove that?
     
  19. Feb 4, 2009 #18
    For that inclusion I did

    Assume f is in A(W1 intersect W2)
    This implies f is in A(W1) and f is in A(W2)
    implies f is in A(W1)+A(W2)
    implies A(W1 intersect W2) is contained in A(W1)+A(W2)
     
  20. Feb 6, 2009 #19

    quasar987

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    This does not follow. Consider for instance V=R³, W_1 = the x axis, W_2=the y axis. Then W_1 intersect W_2 is just the origin and so the map f(x,y,z)=x+y+z is in A(W1 intersect W2) but it certainly is not in A(W1) nor A(W2).
     
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