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

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- Thread starter alphaneutrino
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- #1

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

- #2

CompuChip

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The easiest way is to explicitly express the change of basis as

[tex]A \to A' = B A B^{-1}[/tex]

and use the cyclic property

[tex]\operatorname{Tr}(ABC) = \operatorname{Tr}(BCA) = \operatorname{Tr}(CAB)[/tex]

Or you can use that

[tex]\operatorname{Tr}(A) = \sum \lambda[/tex]

where [itex]\lambda[/itex] are the eigenvalues, and show that these are basis invariant

(for example, show that if v is a corresponding eigenvector, that [itex](B A B^{-1}) (B v B^{-1}) = \lambda v[/itex] as well).

- #3

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thanks CompuChip

- #4

CompuChip

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Actually, upon reading back my answer, I noted a little error. Instead of

[tex]

(B A B^{-1}) (B v B^{-1}) = \lambda v

[/tex]

I should of course have said

[tex]

(B A B^{-1}) (B v B^{-1}) = \lambda (B v B^{-1})

[/tex]

which simply says that if [itex]A v = \lambda v[/tex] then [itex]A' v' = \lambda v'[/itex] where the primes refer to matrices and vectors in the new basis (i.e. [itex]A' = B A B^{-1}[/itex] and [itex]v' = B v B^{-1}[/itex]).

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