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

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if A is diagonalizable then let the diagonal entries be a(ii) for i=1, ..., n. then Tr(A) = Σa(ii) ie sum over i.

and the eigen values are roots of equation

(a(11)-λ) (a(22)-λ) ... (a(nn)-λ) = 0

which will give n roots same as a(11) , ... , a(nn)

so Tr(A) is sum of eigenvalues.

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