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Find all possible values of h

  1. Dec 29, 2015 #1
    1. The problem statement, all variables and given/known data
    Find all possible values of h.
    fh(a+bx+cx2+dx3) =
    [ a+b+c+hd b+c ]
    [ -b-c-hd hb ]


    2. Relevant equations


    3. The attempt at a solution
    1) Row reduce the matrix

    [ a+b+c+hd b+c ]
    [ -b-c-hd hb ]

    to (R1+R2)

    [ a hb+ b+c ]
    [ 0 hb ]

    to (R1-R2)

    [ a b+c ]
    [ 0 hb ]


    a+b+c = 0
    hb = 0

    From this row reduction I assume b=0, a=1 and c=-1
    h could still be any number though.
    So I got back to the original matrix.

    a+b+c+hd+b+c=0
    -b-c-hd+hb=0

    If I plug in the values I assumed (above) then I get d=-1 and h=-1

    Is there a better / quicker way to do this? I seem to come across a lot of similar problems.
     
  2. jcsd
  3. Dec 29, 2015 #2

    mfb

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    I don't understand the notation. Is fh a function? If yes, does the h in the name matter and what do you know about the function? Or is that a simple multiplication?
    The right side is a 2x2 matrix? If yes, is the equation true for a given x, or just for an arbitrary x?
     
  4. Dec 29, 2015 #3
    It's a linear transformation
     
  5. Dec 29, 2015 #4

    mfb

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    fh? Of a real number to 2x2 matrices of real numbers? An arbitrary one?
    My crystal ball isn't working properly today, so I cannot know that if you don't write it.
     
  6. Dec 30, 2015 #5

    haruspex

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    Just guessing here, is the matrix in the fh equation supposed to be surrounded by vectors involving x, maybe [1 x], [1 x]T?
     
  7. Dec 30, 2015 #6
    It's a P3 to M2x2 linear transformation. The first part of the question, which I'm having difficulty understanding, asks to find all possible values of h. I'm confused because h can have many many values, depending on what the values of x,y,z are and depending on what I set the polynomial of the matrix equations equal to.

    i.e.
    [ a+b+c+hd b+c ]
    [ -b-c-hd hb ]

    a+b+c+hd + b+c = 0
    -b-c-hd+hb = 0

    In this case I can row reduce the matrix to simplify both equations
    a + hb + b+c = 0
    0 + hb = 0

    My confusion here is that I can see hb = 0, but h could be 0 or b could be 0 and h could be any number. So I don't really know how I'm supposed to answer this question. 'Find all possible values of h' ... Well h = 0, h ≠ 0 (essentially saying h= 1,2,3,4,5,6,7,....etc)
     
  8. Dec 30, 2015 #7
    With my row reducing of the matrix (above) and setting both equations from the matrix = 0 I can show that h could = 0 or it could equal a positive number of a negative number. So is it enough for me to say all possible values of h are 0,1,-1 ?
     
  9. Dec 30, 2015 #8

    haruspex

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    I must be missing something. It seems to me that fh constitutes a linear transformation regardless of the value of h.
    You know the definition of a linear transformation, right? How would you check that fh is such?
     
  10. Dec 30, 2015 #9
    Yes. The second part of the question asks for the kernel and image. I know how to do that, but I'm confused with how i find all possible values of h.
     
  11. Dec 30, 2015 #10

    haruspex

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    Well, if you apply the definition of a linear transformation, can you either deduce something about h or show that it is a linear transformation regardless of h?
    For the kernel and image, they will depend on h, no?
     
  12. Dec 30, 2015 #11
    That's what confuses me. If I take the original matrix and split it into two equations and make them = 0 then my value of h depends on my value of a,b,c,d and vice versa.

    a+b+c+hd + b+c = 0
    -b-c-hd + hb = 0
     
  13. Dec 30, 2015 #12

    haruspex

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    I don't understand why two equations. If P is in the kernel, what matrix would fh(P) be?
     
  14. Dec 30, 2015 #13
    Isn't writing the matrix of the linear transformation:

    [ a+b+c+hd _ b+c | 0 ]
    [ -b-c-hd _ hb | 0 ]

    the same as writing them as two equations?

    a+b+c+hd + b+c = 0
    -b-c-hd + hb = 0
     
  15. Dec 30, 2015 #14

    haruspex

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    you've lost me. The linear transformation of interest is fh. That effectively maps a four vector to a four vector, so its matrix would be written as a 4x4.
    Try to answer my question. If P is in the kernel of fh what does the 2x2 matrix fh(P) look like?
     
  16. Dec 30, 2015 #15
    I don't understand your question. sorry.
     
  17. Dec 30, 2015 #16

    haruspex

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    For the purposes of the question, the set of polynomials of degree 3 is just a vector space here. Likewise the set of 2x2 matrices over R. fh is a linear transformation from the first to the second.
    When we say that an element X is in the kernel of a linear transformation T, what equation expresses that?
     
  18. Dec 30, 2015 #17
    fh: P3 → M2x2

    Ax = 0
     
  19. Dec 30, 2015 #18

    haruspex

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    Right, so write that last equation out for the case where x is the polynomial a+bx+cx2+dx3.
     
  20. Dec 30, 2015 #19
    I don't know how to format this, but it's the original 2x2 matrix * (1,x,x2,x3) = 0, only 1,x,x2,x3 would be a column.

    [ a+b+c+hd b+c ] * ((1,x,x2,x3) = 0
    [ -b-c-hd hb ]
     
  21. Dec 30, 2015 #20
    But we'd need a 4x4 matrix for A though for this to work...
     
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