I am having difficulties in proving directly the following statement:
I know that similar matrices have the same eigenvalues and that implies that their traces are the equal (if eigenvalues exist).
I would like to know if to prove directly that $Tr(A)$= $Tr(U^{-1}AU)$ for a general × matrix, using the definitions of Trace and matrix multiplication.
I found this on the internet:
$Tr(AB)=Tr(BA)$ implies that if U is a square matrix n x n and it is invertible then
$Tr(A)$= $Tr(U^{-1}AU)$.
However I do not understand how can I prove directly that $Tr(A)$= $Tr(U^{-1}AU)$ using the fact that $Tr(AB)=Tr(BA)$
Can anyone help me on this?
Thanks
You only need to apply the properties of trace. If you set $A = U^{-1}$ and $B = AU$ then $$Tr(AB) = Tr(U^{-1}AU) = Tr(BA) = Tr(A U U^{-1}) = Tr(AI) = Tr(A)$$.