Question 7.E.3.6: Explain why det (e^A) = e^trace(A)....

Matrix Analysis and Applied Linear Algebra [1113429]

Explain why $det (e^A) = e^{trace(A)}.$

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If $\{λ_1, λ_2, . . . , λ_n\}$ are the eigenvalues of $A_{n×n},$ then $\{e^{λ_{1}}, e^{λ_{2}}, . . . , e^{λ_{n}}\}$ are the eigenvalues of $e^A$ by the spectral mapping property from Exercise 7.3.5. The trace is the sum of the eigenvalues, and the determinant is the product of the eigenvalues (p. 494), so $det (e^A) = e^{λ_{1}} e^{λ_{2}} · · · e^{λ_{n}} = e^{λ_{1} + λ_{2}+···+λ_{n}} = e^{trace(A)}.$

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