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Here is the question I am trying to solve:

Let $A$ be an $n \times n$-matrix, with eigenvalues $\lambda_1, \dots , \lambda_n.$ Prove that $$\operatorname{det}A = \lambda_1 \dots \lambda_n.$$

My thoughts:

I know that the eigenvalues will give us an $n \times n$ diagonal matrix with the eigenvalues being the diagonal entries but I do not know how can I prove this. Also I know that the determinant of a diagonal matrix is the product of its diagonal entries but I do not know how to prove this either.

Could someone help me in solving this exercise please?

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    $\begingroup$ The eigenvalues are the roots of which polynomial? (Then their product has something to do with the free coefficient, Vieta.) $\endgroup$
    – dan_fulea
    Apr 6, 2022 at 14:50
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    $\begingroup$ Express $A$ in terms of a basis of eigenvectors $\endgroup$ Apr 6, 2022 at 14:53
  • $\begingroup$ they are the roots of the characteristic polynomial @dan_fulea but I do not know the relation between their products and the free coefficient ..... could you explain to me more details please? $\endgroup$
    – user965463
    Apr 6, 2022 at 15:14

1 Answer 1

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Let $A$ be an $n \times n$-matrix, with eigenvalues $\lambda_1, \dots , \lambda_n.$

$$P(\lambda)=\operatorname{det}(A - \lambda I_n)= (\lambda_1 - \lambda)(\lambda_2 - \lambda) \dots (\lambda_n - \lambda).$$ $$\operatorname{det}(A)=P(0)= \operatorname{det}(A - 0I_n)=(\lambda_1 - 0)(\lambda_2 - 0) \dots (\lambda_n - 0)= \lambda_1 \dots \lambda_n$$

If $\lambda$ = $0$, $\lambda I_n = O$, where $O$ is a null matrix. Then $A - 0I_n$=$A-O$=$A$

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  • $\begingroup$ I do not understand why $\operatorname{det} A = P(0),$ could you please explain this? $\endgroup$
    – user965463
    Apr 6, 2022 at 15:15
  • $\begingroup$ If $\lambda$ = $0$, $\lambda I_n = O$, where $O$ is a null matrix. Then $A - 0I_n$=$A-O$=$A$ $\endgroup$
    – Maria
    Apr 6, 2022 at 15:30
  • $\begingroup$ does not the determinant is expanded in terms of, say, the first row? should not there be some negative signs? $\endgroup$
    – user965463
    Apr 6, 2022 at 16:17
  • $\begingroup$ Do you know the signs of eigenvalues? $\endgroup$
    – Maria
    Apr 7, 2022 at 6:18
  • $\begingroup$ I do not understand you, could you clarify please? $\endgroup$
    – user965463
    Apr 7, 2022 at 8:07

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