What does 'M symmetric' mean in the context of matrices?

[AlphaNumeric2]

My dad came across this phrase in a book but neither of us are familiar with it. The statement is :

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

Does it just mean that M_{1}=M_{1}^{T} or something else? Obviously searching for "Matrix, symmetric" doesn't help in this question...

 

[Office_Shredder]

I think more context is needed. Consider you're never actually told what size the matrices are. It says diagonalizable in \mathbb{R}^{2}, so I would think they're 2x2, but usually you say diagonalizable over R or over C or over Q, etc. \mathbb{R}^{2} 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

 

[AlphaNumeric2]

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.

 

[Chris Hillman]

What book?

My dad came across this phrase in a book but neither of us are familiar with it...I'll ask him again and see if he's typed out something from memory or he copied it word for word.

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 N is indeed symmetric.

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