$\det (A^{-1})$ from eigenvalues of $A$ Suppose I have invertible square matrix $A$ in the complex field and I know all of its eigenvalues and they may be assumed to be non zero. Is there a way to write $\det(A)$ and $\det (A^{-1})$?
PS. This is related to my previous question Eigenvalues of $(A+B)^{-1}$ and is not homework. Thank you. 
 A: If $A$ is diagonalizable, it's easy. You can write $A=PDP^{-1}$ where $D$ is the diagonal matrix with the eigenvalues of $A$ in the diagonal. Therefore $\det(A)=\det(P)\det(D)\det(P)^{-1}=\det(D)=\lambda_1\cdots\lambda_n$, where $\lambda_i$ are the eigenvalues of $D$. Also $\det(A^{-1})=\det(A)^{-1}=1/(\lambda_1\cdots\lambda_n)$.
In general, we have that the characteristic polynomial of $A$ is of the form $p(t)=(t-\lambda_1)\cdots(t-\lambda_n)$. Evaluating in 0, we get that $p(0)=(-1)^n\lambda_1\cdots\lambda_n$. Now, $p(t)=\det(tI-A)$, and so by evaluating in 0, we get that $(-1)^n\lambda_1\cdots\lambda_n=\det(-A)=(-1)^n\det A$.  The determinant of $A^{-1}$ is just the inverse of the determinant of $A$.
A: The so-called Jordan form of a complex matrix $A$ has the eigenvalues of $A$ in the main diagonal and maybe some more stuff above the main diagonal. Now it is an elementary property of determinants that the determinant of an upper triangular matrix is the product of its diagonal elements. Furthermore the determinant of $A^{-1}$ is the reciprocal of the determinant of $A$. It follows that
$$\det(A)=\lambda_1\cdot\lambda_2\cdot\ldots\cdot\lambda_n\>,\qquad
\det(A^{-1})={1\over\lambda_1\cdot\lambda_2\cdot\ldots\cdot\lambda_n}\ .$$
