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Show that if A is similar to B, then Tr(A) = Tr(B).

Show that if A is Diagonalizable, then Tr(A) is the sum of the eigenvalues.

Oct 21st, 2017

Thank you for the opportunity to help you with your question!

Tr(A) - Tr(B)=0

Tr(A - B)=0

but since A=B

therefore 0=0

showing that Tr(A) =Tr(B)

and for A Diagonalizable, Tr(A) is the sum of the eigenvalues as shown from the above solution

Please let me know if you need any clarification. I'm always happy to answer your questions.
Jul 24th, 2015

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Oct 21st, 2017
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Oct 21st, 2017
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