# Matrix N is 'M symmetric' ?

1. Dec 7, 2007

### 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...

2. Dec 7, 2007

### Office_Shredder

Staff Emeritus
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

3. Dec 7, 2007

### 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.

4. Dec 8, 2007

### Chris Hillman

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 $N$ 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
$$A = \left[ \begin{array}{cc} a & b \\ -b & d \end{array} \right]$$
while a Minkowski anti-symmetric matrix would take the form
$$A = \left[ \begin{array}{cc} 0 & b \\ b & 0 \end{array} \right]$$
which satisfies
$$\exp(A) = \left[ \begin{array}{cc} \cosh(b) & \sinh(b) \\ \sinh(b) & \cosh(b) \end{array} \right]$$
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