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What is C^2?

  1. Nov 29, 2005 #1
    I was asked to find a basis beta for V such that [T]_beta is a diagonal matrix.
    V=C^2 and T is defined by T(z,w) = (z+iw, iz+w)
    for C = complex.
    I don't quite understand the expression C^2, I know C can be represented by a plane with the real and imaginery axis, but how abt C^2?
    and also, how do i figure out the basis for this? I used (1,0) and (0,1) as my basis and I got the right answer.. but the basis doesn't seem right to me intuitively.
     
  2. jcsd
  3. Nov 30, 2005 #2

    matt grime

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    C^2 is two dimensional complex vector space.

    Presumably you're happy with R^2


    T is not diagonal with respect to that standard basis.
     
    Last edited: Nov 30, 2005
  4. Nov 30, 2005 #3

    dextercioby

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    The cartesian product between [itex] \mathbb{C} [/itex] and [itex] \mathbb{C} [/itex] is written [itex] \mathbb{C}\times \mathbb{C} [/itex] and commonly written as [itex] \mathbb{C}^{2} [/itex].

    Daniel.
     
  5. Nov 30, 2005 #4
    "C^2 is two dimensional complex vector space."
    does this means that geometrically it has 4 axis in total?

    so, what is the basis for C^2?
    intuitively I thought the basis is {(1,0), (0,1), (i,0),(0,i)} , but i think there're way too many vectors..
    and (1+i , 0), (0, 1+i) seems like can't span the whole space..

    anyone can let me know what the basis for C^2 is?
     
  6. Dec 1, 2005 #5

    shmoe

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    You're considering C^2 as a vector space over what field?
     
  7. Dec 1, 2005 #6
    the question said for the linear operators T on a vector space V, if T is diagonalizable, find a basis beta for V suh that [T]_beta is a diagonal matrix.
    the answer for this quesiton is beta = {(1,1) , (1,-1)}
    but I still have no clue on the basis... please give me some hint!
     
  8. Dec 1, 2005 #7

    dextercioby

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    Well, we usually say that [itex] \mathbb{C}^2 [/itex] is a 2-dimensional vector space over the field of complex numbers, i.e. [itex] \mathbb{C} [/itex]. A basis in this vector space is made up of
    [tex] \left(\begin{array}{c} 1\\0 \end{array} \right) \ \mbox{and} \ \left(\begin{array}{c} 0\\1 \end{array} \right) [/tex]

    Daniel.
     
    Last edited: Dec 1, 2005
  9. Dec 1, 2005 #8

    matt grime

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    C^2 is two dimensional over C, you are already using a basis implicitly when you define T(z,w)=(z+iw,w+iz). It is the 'standard basis' and it is given by (0,1) and (1,0) which is a basis for C^2 because you're allowed to multiply these by any complex number ie (3+i,1+2i) = (3+i)*(1,0)+(1+2i)*(0,1)

    Suppose I asked you to diagonalize the REAL operator

    T(x,y)=(x+y,x-y)

    given with respect to the standard basis on R^2. Then you'd have no trouble doing it. This is exactly the same except that you're allowed to use complex numbers not just real ones.
     
  10. Dec 1, 2005 #9

    JasonRox

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    If you know how to find a basis for R^2, well then that's exactly how you do it. The same principle applies to C^2.

    Note: I try not to worry about the visual representation of it. That might be bad, but I'm doing fine without the visual representation. I have some for things like linear independence/depence, but it's still not necessary because it all comes from the definition.
     
  11. Dec 2, 2005 #10
    Thanks all for the reply! I got it now!
     
  12. Dec 2, 2005 #11

    matt grime

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    It is not bad to do without visualization, it is very necessary, and it is bad to rely on visualization; the amount of maths that it visualizable is tiny and you should not rely on visualizing it (which I disinguish from having geometric intuition)
     
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