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Similar matrices

  1. Nov 14, 2005 #1
    [SOLVED] Similar matrices

    I was given 2 matrices and need to prove that they are similar,
    after i performed row operations on it, i got
    A =
    [100]
    [040]
    [006]
    and B =
    [600]
    [040]
    [001]
    I was stupid enough for not using the fact that their trace are equal to prove it. and instead, keep figuring out the invertible matrix Q that satisfies A=(Q^-1)BQ. But still, i can't figure out the matrix Q that does the work..
    Can anyone tell me if there's any other way to prove the 2 matrices ar similar but do not deal with the trace? or any fast way to figure out Q?
     
  2. jcsd
  3. Nov 14, 2005 #2

    Hurkyl

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    Having equal trace isn't sufficient for two matrices to be similar! Consider the diagonal matrices A and B, with A's diagonal entries being (2, 0) and B's diagonal entries being (1, 1).

    I think this is a case of not understanding the geometric meaning of similarity! (Which, in this case at least, makes the similarity transformation obvious)

    Suppose A and B are similar, so that [itex]A = Q B Q^{-1}[/itex]. This is equivalent to saying that [itex]AQ = QB[/itex].

    Now, for any vector v, we have [itex]A(Qv) = Q (Bv)[/itex].

    Can you interpret that equation geometrically? (This question is an important one! If you cannot answer it, keep in mind that one of the things you're supposed to get out of this class is to develop your geometric intuition, so think about it)

    Even if your geometric intuition is failing, you could appeal to your algebraic intuition! (Same comment applies) When you want to use algebra on matrices, it is often a very good idea to study what they do to certain vectors...


    This approach won't work for all matrices, incidentally, but it will for all "good" ones.
     
  4. Nov 15, 2005 #3
    Well, i think A(Qv) means you first express v with respect to a basis and then A transform Qv to another space? and Q(Bv) is the reverse of it?........
     
  5. Nov 15, 2005 #4

    Hurkyl

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    Right -- so, it sounds like the fact that A and B are similar means that we can find a change of basis Q such that the following two procedures are the same:

    Change the basis, then apply A.
    Apply B, then change the basis.

    So that A and B are really doing the "same thing" (thus the term similar), just with regards to different bases.


    Your two given matrices are really easy to understand geometrically -- what change of basis is suggested here?
     
  6. Nov 16, 2005 #5
    Thanks for prompt reply!
    so, i guess i can let Q to be a permutation matrix that reorders the row/col
    [001]
    [010]
    [100]
    just want to make sure, so as long as I can find a Q that fits AQ=QB, that automatically shows A ~ B and Q doesn't have to be unique..
     
  7. Nov 16, 2005 #6

    Hurkyl

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    Almost: you have to find an invertible Q. Q can't be unique, because if

    Q couldn't possibly be unique, because 2Q wouls also suffice. However, there do exist matrices that are similar in more substantially different ways. How many ways is I similar to I?
     
  8. Dec 2, 2005 #7

    JasonRox

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    I have to use my geometric intuition. :frown:

    I must be stupid because I have no clue how that is presented. I know about similiar matrices, and how to determine whether or not they are similiar, and how to find a matrix P, such that A = P^-1 B P.

    I just read the theorems, and they all make sense. The thought of how the proof is going to go is something I can usually think of because they even go on with it. Sometimes the proof is tricky, which I can't get on the spot, but I still know intuitively it is true and what not.
     
  9. Dec 2, 2005 #8
    Geometrically, Hurkyl implies that two linear transformations are similar if they perform the same action within linear transformations of the space. Ie., if A only dilates one subspace by t and B only dilates one subspace by t, they are similar, because A can be carried out by simply rotating the relevant subspace into the subspace dilated by B, then rotating it back. Also, a scaling of a single axis by a negative value is similar to a rotation of the axis by [itex]\pi[/itex] (where the intervening linear transformation is the identity).
    Similarly, all matrices of the same linear transformation with respect to different bases are similar. Looking at similarity as an equivalence relation will pick out the set of unique linear transformations on a space.
    In this respect, we can suppose A in the original problem acts on the standard basis, which then is transformed to the basis listed in your matrix (column vectors). B is then obviously similar; it can carry out A by transforming the standard basis, dilating the relevant subspaces, then transforming it back. The y-axis remains the same, the x-axis is transformed to the positive z-axis and the z-axis is transformed to the positive x-axis. This can only be done by an inversion of the xz-plane, whose matrix you can then derive.
     
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