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

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The discussion clarifies that GL(n, k) refers to the general linear group of n x n invertible matrices over a field k. It explains a key property of the trace function, stating that the trace of the product of two matrices is invariant under cyclic permutations. The proof shows that the trace of the matrix S is equal to the trace of the transformed matrix BSB^(-1) by applying this property. This confirms that Trace(S) equals Trace(BSB^(-1)) as requested. The explanation provides a concise proof for the original question posed by Music Freak.
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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:

Quick Proof about a Square Matrix? - Yahoo! Answers

I have posted a link there to this topic so the OP can find my response.
 
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Hello Music Freak,

$GL(n,K)$ 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, $$\mbox{tr }(BSB^{-1})=\mbox{tr }((BS)B^{-1})=\mbox{tr }(B^{-1}(BS))=\mbox{tr }((B^{-1}B)S)=\mbox{tr }(IS)=\mbox{tr }S$$
 
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