# Music Freak's question at Yahoo! Answers (Trace in the lnear group)

#### Fernando Revilla

##### Well-known member
MHB Math Helper
Here is the question:

So if S is a square matrix of order n, and B is in GL(n,k) where B is invertible, and the
Trace(S) is equal to Trace(BSB^(-1)).

I'm not sure what the GL(n,k) means...
Please help prove this. I know this is a short proof, but I can't seem to find it in my book.
Here is a link to the question:

$GL(n,K)$ is the general linear group of degree $n$ that is, the set of $n\times n$ invertible matrices over the field $K$, together with the operation of ordinary matrix multiplication. There is a well known property: for all $M,N$ matrices of $K^{n\times n}$ we have $\mbox{tr }(MN)=\mbox{tr }(NM)$ so, $$\mbox{tr }(BSB^{-1})=\mbox{tr }((BS)B^{-1})=\mbox{tr }(B^{-1}(BS))=\mbox{tr }((B^{-1}B)S)=\mbox{tr }(IS)=\mbox{tr }S$$