Suppose that A is diagonalizable and all eigenvalues of A are positive real numbers. Prove that det(A) > 0. Proof: Because A is diagonalizable, there is an invertible matrix P such that A1, A2,..., An of A. Since E. P-¹AP = D V = A. det (P) B. det (D) C. det (P-¹) det(A) det(P) D. det (P-¹AP) F. PAP = D where the diagonal entries of the diagonal matrix D are eigenvalues det(A), and X₁ > 0, ..., An > 0, we have det (A) = Q.E.D. v A₁... An > 0.

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