Let $A \in \mathrm{Mat}_{n n}$ be a diagonalisable matrix with eigenvalues $\lambda_1, \lambda_2, \ldots, \lambda_n$, counted with multiplicity. Then $ \operatorname{det}(A)=\lambda_1 \lambda_2 \cdots \lambda_n $ Proof. We know by [[Real Square Matrix of Order n is Diagonalisable iff it has n Linearly Independent Eigenvectors (Diagonalisation Theorem)]] that there exists an invertible matrix $P \in \operatorname{Mat}_{n n}$ such that $ P^{-1} A P=\operatorname{diag}\left(\lambda_1, \lambda_2, \ldots, \lambda_n\right) $ Writing $B=\operatorname{diag}\left(\lambda_1, \lambda_2, \ldots, \lambda_n\right)$, we have that $A=P B P^{-1}$. Hence $ \begin{aligned} \operatorname{det}(A) & =\operatorname{det}\left(P B P^{-1}\right) \\ & =\operatorname{det}(P) \operatorname{det}(B) \operatorname{det}\left(P^{-1}\right) \\ & =\operatorname{det}(P) \operatorname{det}\left(P^{-1}\right) \operatorname{det}(B) \\ & =\operatorname{det}(B) \\ & =\lambda_1 \lambda_2 \cdots \lambda_n \end{aligned} $