# How to determine the eigenvalue of a $3\times 3$ matrix, given the other two complex eigenvalues

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?

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.

• 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? Jun 11 at 4:50
• Yes, it applies on every matrix no matter whether it can be diagonalized. You can verify that by calculating the constant term of $\det(A-xI)$. Jun 11 at 4:55
• 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. Jun 11 at 4:58

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.