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Math Help - direct sum question

  1. #1
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    direct sum question

    Given two sets
    U = \{p(x) \in R_4[x] { | } p(2) = p(-2) = 0\}
    W = \{p(x) \in R_4[x] { | } p(1) = p(-1) = p(0) \}

    Prove R_4[x] = U \oplus W

    Can someone please give some direction on how to solve this.
    Thanks!
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  2. #2
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    Quote Originally Posted by jayshizwiz View Post
    Given two sets
    U = \{p(x) \in R_4[x] { | } p(2) = p(-2) = 0\}
    W = \{p(x) \in R_4[x] { | } p(1) = p(-1) = p(0) \}

    Prove R_4[x] = U \oplus W

    Can someone please give some direction on how to solve this.
    Thanks!

    1) Prove first that \dim U=3\,,\,\,\dim W=2

    2) Now prove that U\cap W=\{0\}

    Tonio
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  3. #3
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    Thanks. I'll try to prove that. But I'm blanking out...

    Why is dimR_4[x]=5?

    I'm assuming you meant in your reply that dimU + dimW = 5 = dimR_4[x]
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  4. #4
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    Quote Originally Posted by jayshizwiz View Post
    Thanks. I'll try to prove that. But I'm blanking out...

    Why is dimR_4[x]=5?

    I'm assuming you meant in your reply that dimU + dimW = 5 = dimR_4[x]
    Because R_4[x] is the space of fourth degree polynomials with variable x: any member is of the form ax^4+ bx^3+ cx^2+ dx+ e which has the 5 polynomials, x^4, x^3, x^2, x, and 1, as basis.
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  5. #5
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    Quote Originally Posted by jayshizwiz View Post
    Thanks. I'll try to prove that. But I'm blanking out...

    Why is dimR_4[x]=5?

    I'm assuming you meant in your reply that dimU + dimW = 5 = dimR_4[x]

    If you didn't know that then perhaps it's be a good idea to first learn a little more about polynomials spaces:
    Linear Algebra: More Vector Spaces; Isomorphism - CliffsNotes

    Tonio
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  6. #6
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    Thanks, that's a great website.

    I'm just having one of those days where I seem to forget everything I've learned in Linear Algebra...no biggie
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  7. #7
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    Quote Originally Posted by tonio View Post
    1) Prove first that \dim U=3\,,\,\,\dim W=2

    2) Now prove that U\cap W=\{0\}

    Tonio
    But first don't I have to prove that R_4[x] = U + W??? That's the hard part and I have no idea how to do that.
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  8. #8
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    Quote Originally Posted by HallsofIvy View Post
    Because R_4[x] is the space of fourth degree polynomials with variable x: any member is of the form ax^4+ bx^3+ cx^2+ dx+ e which has the 5 polynomials, x^4, x^3, x^2, x, and 1, as basis.
    Actually, the way I am learning it is that R_4[x] = \alpha_0+\alpha_1x+\alpha_2x^2+\alpha_3x^3. It doesn't go to the 4th power. So dim R_4[x]=4

    Long story short, I proved R_4[x] = U+W.

    U=(4s, 4t, -s, -t)
    W=(i, j, 0, -j)

    How do I find dimU and dimV? Or how do I find a basis for U and for W?

    (and I'm pretty sure dimU=dimV=2 but I don't know how to prove it)

    **I think I got it...

    U=(4s,4t,-s,-t) = (4s, 0, -s, 0) + (0, 4t, 0, -t) \rightarrow dimU=2
    W=(i, j, 0, -j) = (i, 0, 0, 0) + (0, j, 0, -j) \rightarrow dimW = 2

    dimU + dimW = 4 = dim R_4[x]
    Last edited by jayshizwiz; October 5th 2010 at 05:42 AM.
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