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Math Help - Help in Linear Algebra(Similar Matrices)

  1. #1
    Newbie Zakaria007's Avatar
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    Lightbulb Help in Linear Algebra(Similar Matrices)

    Hi guys,
    I got in some trouble lately with a question in similar matrices. The question is:
    If we have two 2*2 matrices, A and B, then find an invertible matrix P that makes A similar to B. All that without diagonalization. I tried to solve this by assuming an arbitrary matrix P using the relation below:

    since, A=(P^-1)BP

    then, BP = AP

    The matrices A and B are as follows:

    A=[ 3 2 ]
    [-1 0 ]

    B=[-1 2]
    [ 0 1]

    P = [ a b]
    [ c d]


    Thanks guys I really would like a hand in this. I would appreciate it a lot.
    Last edited by Zakaria007; June 24th 2009 at 12:34 PM. Reason: Mistake with matrices
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  2. #2
    MHF Contributor Swlabr's Avatar
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    First things first - multiplication of matrices is not necessarily commutative, and so A=(P^{-1})BP does not in general imply that AP=BP. It does, however, imply that PA=BP.

    I have a slight problem with the matrices you give - was the question "find the matrix P" or "are A and B similar"? I ask this because I don't think they are similar. Trying to find a P such that AP=PB we discover that P has determinant zero and so is non-invertible.

    As a general rule for these questions, just hit them with a stick. What is AP? What is PB? just use a P with variables for entries. All the entries must tally, and so what must P look like? Even if you don't get the answer you will get some idea of what the matrix will look like, and so you can work from there.

    There will often be more subtle methods, and these are worth looking out for. But the stick will always work...especially on 2x2 matrices!
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  3. #3
    MHF Contributor

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    As Swlabr said, just go ahead and Do it!

    With A= \begin{bmatrix}3 & 2 \\ -1 & 0\end{bmatrix}, B= \begin{bmatrix}1 & 2 \\0 & 1\end{bmatrix}, and P = \begin{bmatrix}a & b \\  c & d\end{bmatrix}, AP= PB becomes

    \begin{bmatrix}3 & 2 \\ -1 & 0\end{bmatrix}\begin{bmatrix}a & b \\  c & d\end{bmatrix} = \begin{bmatrix}a & b \\  c & d\end{bmatrix}\begin{bmatrix}1 & 2 \\0 & 1\end{bmatrix}
    \begin{bmatrix}3a+ 2c & 3b+ 2d \\ -a & -b\end{bmatrix}= \begin{bmatrix}a & 2a+ b \\ c & 2c+ d\end{bmatrix}.

    So you must have 3a+ 2c= a, 3b+ 2d= 2a+ b, -a= c, and -b= 2c+ d. Solve those four equations for a, b, c, and d.
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  4. #4
    MHF Contributor Bruno J.'s Avatar
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    You can see right away that your matrices are not similar because they do not have the same determinant.
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