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Matrix N is 'M symmetric' ?

  1. Dec 7, 2007 #1
    My dad came across this phrase in a book but neither of us are familiar with it. The statement is :

    "Let [tex]M_{1}[/tex] and [tex]M_{2}[/tex] be matrices. [tex]N = M_{1}^{-1}M_{2}[/tex]. This matrix is [tex]M_{1}[/tex] symmetric and so it diagonalisable in [tex]\mathbb{R}^{2}[/tex]."

    Does it just mean that [tex]M_{1}=M_{1}^{T}[/tex] or something else? Obviously searching for "Matrix, symmetric" doesn't help in this question...
     
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  3. Dec 7, 2007 #2

    Office_Shredder

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    I think more context is needed. Consider you're never actually told what size the matrices are. It says diagonalizable in [tex]\mathbb{R}^{2}[/tex], so I would think they're 2x2, but usually you say diagonalizable over R or over C or over Q, etc. [tex]\mathbb{R}^{2}[/tex] isn't a standard field (and i'm not sure whether it's even possible to make it a field off the top of my head), so my first guess would be that you're missing something important
     
  4. Dec 7, 2007 #3
    I assumed that the matrices are 2x2, so I guess that refers to them being diagonalisable as a Real matrix. Unfortunately I don't have access to this book, he asked me over the phone and what he said differed a few times from what he then emailed me. I assume he's quoting directly, but that might be incorrect too...

    I'll ask him again and see if he's typed out something from memory or he copied it word for word. I agree, there does feel as if there's a vital bit of information missing.
     
  5. Dec 8, 2007 #4

    Chris Hillman

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    What book?

    Please make sure you make him tell you what book because I think this is vitally important information!

    My first guess was M-symmetric stands for "Minkowski-symmetric", but if your dad's book has nothing to do with relativistic physics that is fairly unlikely. Another guess is that the (extraneous?) symbol is a printer's error, since (particularly in the context of elementary linear algebra) the sentence with that symbol deleted appears to make sense if all matrices are nxn real matrices and if [itex]N[/itex] is indeed symmetric.

    If it helps, put [itex]L = \operatorname{diag} (-1,1,1, \dots 1)[/itex]; then the Minkowski adjoint can be taken to be [itex]A^{\ast} = L^{-1} \, A^T \, L[/itex] and then a Minkowski symmetric operator satisfies [itex]A^{\ast} = A[/itex]. For example in the 2x2 case a Minkowski-symmetric matrix would take the form
    [tex]
    A = \left[ \begin{array}{cc} a & b \\ -b & d \end{array} \right]
    [/tex]
    while a Minkowski anti-symmetric matrix would take the form
    [tex]
    A = \left[ \begin{array}{cc} 0 & b \\ b & 0 \end{array} \right]
    [/tex]
    which satisfies
    [tex]
    \exp(A) =
    \left[ \begin{array}{cc} \cosh(b) & \sinh(b) \\ \sinh(b) & \cosh(b) \end{array} \right]
    [/tex]
    which can be compared with the analogous facts for the usual transpose.

    I don't think we can offer any useful assistance until your dad supplies the missing context.
     
    Last edited: Dec 8, 2007
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