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

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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

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