The product of the eigenvalues of a square matrix $A$ equals

1. From the characteristic polynomial $p(\lambda) = \det(A - \lambda I)$, the constant term is $p(0) = \det(A)$. 2. Also, $p(\lambda) = (-1)^n (\lambda - \lambda_1)(\lambda - \lambda_2) \cdots (\lambda - \lambda_n)$. 3. At $\lambda = 0$: $p(0) = (-1)^n (-\lambda_1)(-\lambda_2)\cdots(-\lambda_n) = \lambda_1 \lambda_2 \cdots \lambda_n$. 4. So $\det A = \lambda_1 \lambda_2 \cdots \lambda_n$. 5. Consequence: $A$ is invertible iff all eigenvalues are non-zero. _Source: Sergei Treil, "Linear Algebra Done Wrong", §4.5 + §4.6 (Eigenvalue product = det)._