<s>[INST] How to prove that trace$(ABA^{-1}B^{-1})$=$3$ If $A,B$ are two $3 \times 3$ square matrices and trace(A) is defined as the sum of all diagonal elements. trace$(ABA^{-1}B^{-1})$=$3$
I could easily verify the above for the identity matrix.But I couldn't generalise it.
Please help me in this regard.thanks.
 [/INST] Counterexample:
$$\begin{align}
A&=\begin{pmatrix}1&1&0\\0&1&0\\0&0&1\end{pmatrix}
&B&=\begin{pmatrix}1&0&0\\1&1&0\\0&0&1\end{pmatrix}
\\A^{-1}&=\begin{pmatrix}1&-1&0\\0&1&0\\0&0&1\end{pmatrix}
&B^{-1}&=\begin{pmatrix}1&0&0\\-1&1&0\\0&0&1\end{pmatrix}
\end{align}$$
results in
$$ABA^{-1}B^{-1}=\begin{pmatrix}3&-1&0\\1&0&0\\0&0&1\end{pmatrix}$$
with trace $4$.
</s>