The Determinant of a Matrix Is the Product of Its Eigenvalues

Question

Prove that determinant of a matrix $A$ is the product of its eigenvalues (counting multiplicities).

We are given the following hint: first, show that $\det(A - x I) = (\lambda_1 - x)(\lambda_2 - x) \cdots (\lambda_n - x),$ where the $\lambda_i$ are (not necessarily distinct) eigenvalues of $A;$ then, compare the free terms (i.e., the terms without $x$), or plug in $x = 0$ to get the conclusion.

Answer 1

Let $A \in M_{n \times n} (\mathbb{K}) $ such that $\mathbb{K} =\overline{\mathbb{K} } $ then let $|A-\lambda id| =\Pi_{i=1}^n{(\lambda_i-\lambda) }$ counting multiplicity of $\lambda_i$, so now Since $|A|$=$|A-\lambda id|$ with $\lambda=0$ and remplazing $\lambda=0$ $|A|=\Pi_{i=1}^n{\lambda_i} $ $\square$