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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.

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  • $\begingroup$ What have you attempted for the problem? Where are you getting stuck? $\endgroup$ Jun 10, 2020 at 2:40
  • $\begingroup$ expanding the det(A-𝜆I) I got (a11-𝜆)(a22-𝜆)...(ann-𝜆)+p(𝜆) but don't really know where to go from there. @Carlo $\endgroup$ Jun 10, 2020 at 2:50
  • $\begingroup$ That is very bad. Recall that if $A$ is an $n \times n$ matrix, then $\det(A - xI)$ is a polynomial in $x$ of degree $n.$ Can you use this to finish the problem? (I imagine from this that you can assume that $A$ has entries in some algebraically closed field -- probably the complex numbers $\mathbb C.$) $\endgroup$ Jun 10, 2020 at 2:53

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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} $ $\square$

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