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Similar matrices help

  1. Jul 23, 2010 #1
    1. The problem statement, all variables and given/known data
    Let A [tex]\in M_{n x n }(F)[/tex] and let [tex]\gamma[/tex] be an ordered basis for [tex] F^{n} [/tex]. Then [tex] [L_{A}]_{\gamma} = Q^{-1} A Q [/tex], where Q is the n x n matrix whose j-th column is the j-th vector of gamma.

    3. The attempt at a solution

    I think I'm confused about some of the technical details.. :S And I don't really have much of a plan for proving this at all :S.. any help? thank you
  2. jcsd
  3. Jul 23, 2010 #2
    Well, you'll have to show us something.
    Do you understand all the notation of the statement?

    Maybe writing it out in words will help. Here's one way of writing the first sentence:

    Let A be an element of the set of nxn matrices with elements in the field F.

    Of particular interest: Do you understand what [LA]gamma means?
  4. Jul 23, 2010 #3

    I think the notation may be messing with me.. I'm also confused about the changing bases.. We need to change basis x to basis gamma. Are we just not given the basis x?

    and [LA]gamma is a linear transformation, so we need to decide the coordinate vectors A* x_i where x _ i are the vectors from gamma
  5. Jul 23, 2010 #4
    The original basis here is otherwise unspecified. Just assume that it's some other basis (besides [itex] \gamma [/itex]).

    As you say, [itex] \left[ L_A \right]_\gamma [/itex] is the linear transformation corresponding to the matrix A in the original basis, but with respect to the basis [itex] \gamma [/itex] instead.

    If you were in the original basis and somebody came up with a vector u and asked you to do the linear transformation, you could do that easily, right? Just give them back Au. Now someone is coming up to you with a vector in the new basis v, and wants the result of applying the linear transformation in the new basis ([itex] \gamma [/itex]). So if we can take v back to the old basis, apply A, then bring the result back to the new basis, we will have their answer.

    This is where Q comes in. So now, show that Q-1 and Q do these things.
  6. Jul 23, 2010 #5
    thanks, I have another question: it says that the j-th column of Q is the j-th vector of gamma.. but Q is a matrix , so it's the jth vector of gamma under what basis coordinates?
    the matrix Q would be in the coordinates of B' if Q is B -> B', so are we looking for gamma vectors with respect to gamma ? thanks :)
  7. Jul 23, 2010 #6
    The vectors in [itex] \gamma [/itex] would be in terms of the old basis.
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