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Homework Help: U⊗V = U⊗W; find U,V,W

  1. Jun 26, 2012 #1


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    U⊕V = U⊕W; find U, V and W

    I need to give an example of different vectorspaces U, V, W so that [itex]U \oplus V = U \oplus W[/itex].

    Can anyone give a hint please? It's basically asking for V and W such that [itex]u_i + v_i = u_i + w_i[/itex] yet V and W have to be different. How?
    Last edited: Jun 27, 2012
  2. jcsd
  3. Jun 26, 2012 #2
    You are working with tensor products right?? You didn't mean to type [itex]\oplus[/itex] for direct sum??
  4. Jun 26, 2012 #3

    I like Serena

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    Hi srn! :smile:

    So... U, V and W have to be different, such as U=<(1,0)>, V = <(1,1)> and W=<(0,1)>?
  5. Jun 27, 2012 #4


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    Thanks for the replies. And sorry, clearly posted this too late because I messed up the symbol in the question. :( Meant to say direct sum indeed...

    Hey. :) Yes they do. But here the direct sums are not equal though, right? I guess you meant the tensor product? Sorry :(
  6. Jun 27, 2012 #5
    Are you sure they are not equal (when taking the direct sum)??
    What are the direct sums??
  7. Jun 27, 2012 #6


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    If [itex](R,S, +)[/itex] is a vectorspace with [itex]U, W[/itex] as subspaces, then [itex]U \oplus W = \{u + w | u \in U, w \in W\}[/itex] and every [itex]s \in S[/itex] can only be written in one possible way (as the sum of vectors of U and W). I.e. it's every possible combination of elements in [itex](R, U, +)[/itex] and [itex](R, W, +)[/itex].

    Suppose U=<(1,0)>, V = <(1,1)> and W=<(0,1)> are subspaces, then

    [itex]U \oplus W = R^2[/itex]. But how is [itex]U \oplus V = R^2[/itex]? I'm imagining [itex]R^2[/itex]. V is every possible vector through [itex]\stackrel{\rightarrow}{o}[/itex] with [itex]arg(v) = 1[/itex]. Then [itex]U \oplus V[/itex] would be the area under y = x for [itex]x > 0, y > 0[/itex]. How can you form (0,1) for example?

    edit: come to think of it, would [itex](R, V, +)[/itex] also contain (0,1) and (1,2) etc? I sort of assumed from "[itex]\forall v \in V[/itex] and [itex]\forall r \in R: rv \in V[/itex]" that [itex](R, V, +)[/itex] would only contain (1,1), (2,2) etc, is that incorrect?

    I'm sort of confused because my book says that if [itex]U \cap V \neq (0,0)[/itex] then [itex]U \oplus V[/itex] cannot exist. From the example, [itex]U \cap V[/itex] would be [itex]\{((x,0) | x \in R\}[/itex], but then (1,0) would be both in U and V?
    Last edited: Jun 27, 2012
  8. Jun 27, 2012 #7
    If V = <(1,1)> then how can (1,0) be in V? There is no scalar a such that a*(1,1) = (1,0). Similarly, there is no scalar b such that b*(1,0) = (1,1). So the intersection of U and V is indeed (0,0).
  9. Jun 27, 2012 #8


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    Uh, right. So the intersection is (0,0) but [itex]U + V \neq R^2[/itex]. There's no scalars so that [itex]a\cdot (1,0) + b\cdot (1,1) = (0,1)[/itex], for example. So U and V cannot form [itex]R^2[/itex] and the direct sums are hence not equal? edit: ooops, a = -1 and b = 1 :) so they do actually form R^2


    If U=<(1,0) then V=<(0,1)> and W=<(0,-1)> would form the same direct sum space, and that also answers the question I think? edit: eh, no, (R, V, +) is equal to (R, W, +) now.
    Last edited: Jun 27, 2012
  10. Jun 27, 2012 #9


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    Let U be the subspace of R2 spanned by <1, 0>. That is U is the set of all vectors of the form <x, 0> for any real number x. Let V be the vector space spanned by <0, 1> and let W be the subspace spanned by <1, 1>.

    You can then show that U⊕V= U⊕W.
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