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Linear transformation, subspace and kernel

  1. Nov 12, 2012 #1
    Hi

    We have a linear transformation g : ℝ^2x2 → ℝ g has U as kernel,

    U: the 2x2 symmetric matrices

    (ab)
    (bc)

    A basis for U is

    (10)(01)(00)
    (01)(10)(01)


    I thought this would be easy but I've been sitting with the problem for a while and I have no clue on how to solve it (maybe because I don't fully understand it).

    But this is how I understand it : we need to find a linear transformation that transforms symmetric 2x2 matrices (R^4) to 1x1 matrices (R), so we have

    g(a,b,b,c) = (a b b c) = 0

    ?
     
    Last edited: Nov 12, 2012
  2. jcsd
  3. Nov 12, 2012 #2

    Dick

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    How about this? For a general 2x2 matrix [[a,b],[c,d]] you can write a linear transformation to R as g(a,b,c,d)=w*a+x*b+y*c+z*d. You have to find w,x,y,z such that g=0 if and only if the matrix is symmetric.
     
  4. Nov 12, 2012 #3
    But how can we find the values for w,x,y and z so that g = 0 when we don't even know the values of a,b,b and c ?
     
  5. Nov 12, 2012 #4

    Dick

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    Try sample matrices. Like your basis for U.
     
  6. Nov 12, 2012 #5
    like this :

    a * [[1,0],[0,0]] + b * [[0,1],[1,0]] + c * [[0,0],[0,1]] = 0 ?
     
  7. Nov 12, 2012 #6

    Dick

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    No. Take a matrix in your basis. Like [[1,0],[0,0]]. Figure out what a,b,c and d are. Then put them into the form for g. You want g to give you a real number, not a matrix.
     
  8. Nov 12, 2012 #7
    a * [[1,0],[0,0]] + b * [[1,0],[0,0]] + c * [[1,0],[0,0]] +d* [[1,0],[0,0]] = 0

    (a,b,c,d) = 0

    But shouldn't it be a [[a,b],[b,c]] matrix ?
     
  9. Nov 12, 2012 #8

    Dick

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    No again. Let's back up. Let's take a basis of R2x2. e1=[[1,0],[0,0]], e2=[[0,1],[0,0]], e3=[[0,0],[1,0]], e4=[[0,0],[0,1]]. That's a basis for R2x2, right? What should g(e1) be?
     
  10. Nov 12, 2012 #9
    g(e1) = [[1,0,0,0]]

    g(e2) = [[0,1,0,0]]

    g(e3) = [[0,0,1,0]]

    g(e4) = [[0,0,0,1]]
     
    Last edited: Nov 12, 2012
  11. Nov 12, 2012 #10

    Dick

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    No no. g is a linear transformation from R2x2 to R! R is the real numbers, not matrices. g(e1) should be a number. Which number?
     
  12. Nov 12, 2012 #11
    1?

    I'm still confused because we don't have g and I'm not sure what the method is to find g.
     
  13. Nov 12, 2012 #12

    Dick

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    Nope, you don't have g yet. We are still working on that. But you do know e1 is symmetric, so e1 is in U the kernel of g. What does that tell you about g(e1)?
     
  14. Nov 12, 2012 #13
    But isn't e1 =[[1,0],[0,0]] ? How is this symmetric?
     
  15. Nov 12, 2012 #14

    Dick

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    I'm using a shorthand. I mean [[1,0],[0,0]] to be the matrix whose first row is 1,0 and the second row is 0,0. That's symmetric. Yes?
     
  16. Nov 12, 2012 #15
    Is g(e1)=a
     
  17. Nov 12, 2012 #16

    Dick

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    What is 'a'? What does 'kernel' mean? Explain 'kernel' in your own words.
     
  18. Nov 13, 2012 #17
    Could the linear transformation be [0,1,-1,0] ?
     
  19. Nov 13, 2012 #18

    Dick

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    It would if you can tell me what that means. I would like you to define g by telling me how g acts on a basis. What are g(e1), g(e2), g(e3) and g(e4)?
     
  20. Nov 13, 2012 #19
    We need to find a linear transformation g: R^2x2 -> R that has U as kernel.
    That means that all matrices of this type [a, b, b, c] will be transformed to zero since U is the kernel.
    We need to find a linear transformation that 'sends' U to zero. Since we have two b's one can have a negative value and the other a positive value, a and c will be zero because we're going to multiply it with zero.
    So no matter what the value of b is it will always be zero because +b-b = 0.

    So our linear transformation can be :

    [0,1,-1,0] or [0,-1,1,0]

    ?
     
  21. Nov 13, 2012 #20

    Dick

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    Absolutely right. But just writing [0,1,-1,0] doesn't say what you just told me. You mean that g([[a,b],[c,d]])=0*a+1*b+(-1)*c+0*d, right?
     
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