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Suppose that $A$ is a matrix $3\times 3$ with real entries and such that $\det(A)=1$. Also, suppose that $\lambda_1=(1+\sqrt{3}i)/2$ is a eigenvalue of $A$. The idea is to determine the remaining eigenvalue and so its characteristic polynomial.

I know, that since $A$ has real entries, the eigenvalues come as pairs of complex conjugates. Therefore $\lambda_2=(1-\sqrt{3}i)/2$ is another eigenvalue, and one is left to calculate the real eigenvalue $\lambda_3$.

Then once you have that, the characteristic polynomial is simply $p(\lambda)=(\lambda-\lambda_1)(\lambda-\lambda_2)(\lambda-\lambda_3)$. However, I don't know how to obtain that $\lambda_3$.

I have seen that since $\det(A)=1$ then $A$ is a unimodular matrix and so, a part of the group $GL_3(\mathbb{Z})$, and maybe one could say something with this... However, I think this question is supposed to be answered using linear algebra.

My main problem is not knowing how to relate the fact that $\det(A)=1$ with the eigenvalues and also, since $A$ is any matrix $3\times 3$ I don't think that to caculate the determinant by hand is the way to go.

Any ideas on how to proceed?

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For any $n \times n$ matrix $A$, the product of the eigenvalues is equal to the determinant. This is because the characteristic polynomial is $det(A-xI)$ and roots of the characteristic polynomial are the eigenvalues.

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  • $\begingroup$ I see it for this case since all the eigenvalues have multiplicity 1 and so one could diagonalize the matrix. However, what would happen if one can't diagonalize A? Does this still apply? $\endgroup$
    – leplata
    Jun 11 at 4:50
  • $\begingroup$ Yes, it applies on every matrix no matter whether it can be diagonalized. You can verify that by calculating the constant term of $\det(A-xI)$. $\endgroup$
    – Lapin
    Jun 11 at 4:55
  • $\begingroup$ Since the field of complex nos. is algebraically closed, the characteristic polynomial splits over it and you get $n$ roots. And the product of these roots (i.e. eigenvalues) is always equal to the determinant. This works even if the multiplicities aren't 1 and $A$ is not diagonalizable. $\endgroup$ Jun 11 at 4:58
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Specifically for a $3\times 3$ matrix you can relate the characteristic polynomial to matrix elements via

$$p(x)=x^3-\operatorname{tr}(A)x^2-\frac 12\left(\operatorname{tr}(A^2)-\operatorname{tr}(A)^2\right)x-\det(A)$$

And you get $(-1)^3\lambda_1\lambda_2\lambda_3=-\det(A)$

But as outofspace0105 indicated in his answer, this is known for any $n$ due to trigonalization being always possible in $\mathbb C$ (since characteristic polynomial of degree $n$ has $n$ roots). Therefore the matrix is similar to a triangular one with eigenvalues on the diagonal.

The determinant of a triangular matrix being equal to the product of its diagonal elements, you get the result as a consequence.

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