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?
Best Answer
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.