ThmDex – An index of mathematical definitions, results, and conjectures.
Complex matrix determinant equals product of eigenvalues
Formulation 0
Let $A \in \mathbb{C}^{N \times N}$ be a D6159: Complex square matrix such that
(i) $\lambda_1, \ldots, \lambda_N$ is a D6192: Complex matrix eigenvalue sequence for $A$
Then \begin{equation} \text{Det} A = \prod_{n = 1}^N \lambda_n \end{equation}
Proofs
Proof 0
Let $A \in \mathbb{C}^{N \times N}$ be a D6159: Complex square matrix such that
(i) $\lambda_1, \ldots, \lambda_N$ is a D6192: Complex matrix eigenvalue sequence for $A$
Let $z \in \mathbb{C}$ be a complex number. Using the definition for the eigenvalue sequence, we have \begin{equation} f(z) : = \text{Det}(z I_N - A) = \prod_{n = 1}^N (z - \lambda_n) \end{equation} Therefore, using result R5569: Determinant of reflected complex matrix, we have \begin{equation} (-1)^N \text{Det} A = \text{Det} (- A) = f(0) = \prod_{n = 1}^N (0 - \lambda_n) = (-1)^N \prod_{n = 1}^N \lambda_n \end{equation} Diving both sides by $(-1)^N$, we conclude \begin{equation} \text{Det} A = \prod_{n = 1}^N \lambda_n \end{equation} $\square$