Prove that $\operatorname{det}A = \lambda_1 \dots \lambda_n.$

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?

• The eigenvalues are the roots of which polynomial? (Then their product has something to do with the free coefficient, Vieta.) Apr 6, 2022 at 14:50
• Express $A$ in terms of a basis of eigenvectors Apr 6, 2022 at 14:53
• 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?
– user965463
Apr 6, 2022 at 15:14

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 $$A - 0I_n$$=$$A-O$$=$$A$$

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