How can I show that trace is Invariant under the change of basis?

[alphaneutrino]

How can I show that trace is Invariant under the change of basis?

 

[CompuChip]

The easiest way is to explicitly express the change of basis as
$$A \to A' = B A B^{-1}$$
and use the cyclic property
$$\operatorname{Tr}(ABC) = \operatorname{Tr}(BCA) = \operatorname{Tr}(CAB)$$

Or you can use that
$$\operatorname{Tr}(A) = \sum \lambda$$
where $\lambda$ are the eigenvalues, and show that these are basis invariant
(for example, show that if v is a corresponding eigenvector, that $(B A B^{-1}) (B v B^{-1}) = \lambda v$ as well).

 

[alphaneutrino]

thanks CompuChip

 

[CompuChip]

Actually, upon reading back my answer, I noted a little error. Instead of
$$(B A B^{-1}) (B v B^{-1}) = \lambda v$$
I should of course have said
$$(B A B^{-1}) (B v B^{-1}) = \lambda (B v B^{-1})$$
which simply says that if [itex]A v = \lambda v[/tex] then $A' v' = \lambda v'$ where the primes refer to matrices and vectors in the new basis (i.e. $A' = B A B^{-1}$ and $v' = B v B^{-1}$).[/itex]