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Math Help - Find the formula for the projection on W along U?

  1. #1
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    Find the formula for the projection on W along U?

    Find the formula for the projection P:C^3--->C^3 on W along U, where:
    W=span{(1,i,0),(0,i,1)} and
    U=span{(0,1,0)}.

    Attempt at solution:
    According to my textbook: A function T: V -> V is called the projection on W1 along W2 if, for x = x1 + x2 with x1 E W1 and x2 E W2 we have T(x)=x1, but this requires that V be the direct sum of W1 and W2. For the above question W is taken over the real field F=R so for W to be a subspace of C^3, C^3 must be taken over the field R but then C^3 has dimension 6 (example of basis: {(1,0,0),(0,1,0),(0,0,1),(i,0,0),(0,i,0),(0,0,i)}) . Since the vectors in span W and span U are independent they form a basis for W and U respectively and although the intersection of the 2 basis is the empty set the sum of the dimensions is 3 and not 6 so C^3 is not the direct sum of W and U. Or do we just assume that it is the direct sum then x1= a(1,i,0)+b(0,i,1) and x2=c(0,1,0) and x=x1+x2=(a,ai+bi+c,b) and T(x)=T(a,ai+bi+c,b)=(a,ai+bi,b) where a,b,c are scalars and since T=T^2 this is a projection.

    This is the only definition the textbook gives for a projection so I'm really uncertain about how to approach the question and what to do. Any help would be greatly appreciated.Thanks in advance.
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  2. #2
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    Re: Find the formula for the projection on W along U?

    Quote Originally Posted by chocaholic View Post
    Find the formula for the projection P:C^3--->C^3 on W along U, where:
    W=span{(1,i,0),(0,i,1)} and
    U=span{(0,1,0)}.

    [...]

    For the above question W is taken over the real field F=R so for W to be a subspace of C^3, C^3 must be taken over the field R
    Why do you say that ?

    I think you must consider being in the \mathbb C-vector space \mathbb C^3.
    It's clear that W\cap U = \{0\}, you just have to prove that the sum is the space. The formula will follow.
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  3. #3
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    Re: Find the formula for the projection on W along U?

    I say that F=R because the complex number i forms part of the basis for W. If F=C wouldn't the basis only cantain real numbers because any complex number can then be multiplied in to get C^3?
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  4. #4
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    Re: Find the formula for the projection on W along U?

    In fact, if \mathbb F=\mathbb C, real numbers are sufficient to form a basis of the total space (e.g., \left((1,0,0),(0,1,0),(0,0,1)\right) for your exercise), but it's certainly not a necessity !

    Just think about the \mathbb C-vector space \mathbb C. Isn't (i) or (1+i\sqrt 5) a basis of this space ?

    Here is (given the context) the same. Even more, some subspaces of \mathbb C^3 (as a \mathbb C-vector space) have no basis formed by just real numbers. Like the \mathbb C-line \mathrm{Vec}((1,i,0)) (else, there's a z\in\mathbb C^\ast such that z \times 1 \in \mathbb R and z \times i \in \mathbb R, which isn't possible).
    And that the case (verify it) of your subspace W here. You have to understand that \mathbb C offers more directions than \mathbb R, hence the fact that W (exploiting a \mathbb C-direction that isn't a \mathbb R-direction) can not be expressed with a real basis.
    Last edited by pece; August 3rd 2011 at 06:08 AM. Reason: Completion
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  5. #5
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    Re: Find the formula for the projection on W along U?

    So even though there are complex numbers in the basis of W it can still be a subspace of C^3 taken over C? Then since (1,i,0),(0,i,1),(0,1,0) are linearly independent and since C^3 is 3 dimensional taken over C (1,i,0),(0,i,1),(0,1,0) is a basis for C^3 and C^3 is the direct sum of U and W and the formula for the projection is T(a,ai+bi+c,b)=(a,ai+bi,b) where a,b,c are elements of C? But shouldn't the formula have the form T(z1,z2,z3)=... where z1 z2 and z3 are elements of C? Or is it enough that (a,ai+bi+c,b) represent all elements of C^3 because it is made up of the basis of C^3?
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  6. #6
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    Re: Find the formula for the projection on W along U?

    Quote Originally Posted by chocaholic View Post
    So even though there are complex numbers in the basis of W it can still be a subspace of C^3 taken over C? Then since (1,i,0),(0,i,1),(0,1,0) are linearly independent and since C^3 is 3 dimensional taken over C (1,i,0),(0,i,1),(0,1,0) is a basis for C^3 and C^3 is the direct sum of U and W
    Yes.


    Quote Originally Posted by chocaholic View Post
    and the formula for the projection is T(a,ai+bi+c,b)=(a,ai+bi,b) where a,b,c are elements of C? But shouldn't the formula have the form T(z1,z2,z3)=... where z1 z2 and z3 are elements of C? Or is it enough that (a,ai+bi+c,b) represent all elements of C^3 because it is made up of the basis of C^3?
    That seems a little fuzzy.
    To find the formula of the projection, you have to express a vector of \mathbb C^3 with your known basis (the one of W added with the one of U).

    (a+ib,c+id,e+if) = A\times (1,i,0) + B\times (0,i,1) + C\times (0,1,0)

    Your formula is so T(a+ib,c+id,e+if) = A\times (1,i,0) + B\times (0,i,1). I let you find A, B and C.
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  7. #7
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    Lightbulb Re: Find the formula for the projection on W along U?

    Oh I get it we can solve for A,B,C are elements of C in terms of z1=a+ib z2=c+id z3=e+if because we have the basis!
    So we end up with A=a+ib B=e+if and C=c+id-ai+b-ei+f then from the initial definition that T(x)=x1 we have T(z1,z2,z3)=A(1,i,0)+B(0,i,1)=(a+ib,ia-b+ie-f,e+if)=(a+ib,i(a+e)-(b+f),e+if) I get it. Thank you so much!I really appreciate it.
    Last edited by chocaholic; August 3rd 2011 at 10:31 AM. Reason: incorrect calculation
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  8. #8
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    Re: Find the formula for the projection on W along U?

    Yes, that's it. But you made mistakes in the calculus.

    C = c+id-iA-iB = \dots
    and
    T(z_1,z_2,z_3) = A(1,i,0)+B(0,i,1) = (A,Ai+Bi,B) = \dots

    Verify it, what you give is wrong. (Well, for C, it doesn't really matter.)
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