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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

1 Answers 1

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You only need to apply the properties of trace. If you set $A = U^{-1}$ and $B = AU$ then $$Tr(AB) = Tr(U^{-1}AU) = Tr(BA) = Tr(A U U^{-1}) = Tr(AI) = Tr(A)$$.

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    In this part 'Tr(AI)' is Trace of A*Identity matrix? – 2017-02-14
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    Yes, because $UU^{-1} = I$. – 2017-02-14
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    I understood. Thank you very much for your help – 2017-02-14