$\det\left(I + A^TA^{-1}\right) = 2\left(1 + \operatorname{tr}\left(A^TA^{-1}\right)\right)$

Question

Let $A$ be an invertible $3\times3$ matrix with complex values. Prove that: $$\det\left(I + A^TA^{-1}\right) = 2\left(1 + \operatorname{tr}\left(A^TA^{-1}\right)\right)$$

I've tried to solve this problem with characteristic polynomial but in the end I got stuck in an expression involvind Newton's identities regarding that polynomial. I cannot relate $A^TA^{-1}$ with characteristic polynomial.

Answer 1

If $B = A^T A^{-1}$, we have $\det(B) = 1$. Moreover, $1$ is an eigenvalue of $B$ because $B - I = (A^T - A) A^{-1}$, $A^T - A$ is antisymmetric, and any antisymmetric matrix in odd dimensions must be singular. If the eigenvalues of $B$ are $1, \lambda, 1/\lambda$, then $\text{tr}(B) = 1 + \lambda + 1/\lambda$ and $\det(I+B) = 2 (1 + \lambda) (1 + 1/\lambda) = 4 + 2 \lambda + 2/\lambda$.

Answer 2

HINT: Every complex matrix can be put in upper-triangular form. Try it for any diagonal matrix $A$ first.

Answer 3

Let $S=A^TA^{-1}$. Then $$ \operatorname{tr}(S^{-1}) =\operatorname{tr}\left((S^{-1})^T\right) =\operatorname{tr}(A^{-1}A^T) =\operatorname{tr}(A^TA^{-1}) =\operatorname{tr}(S). $$ and $\det(S)=1$. Therefore the characteristic polynomial of $S$ is $$ \begin{aligned} p(x) &=x^3-\operatorname{tr}(S)x^2+\det(S)\operatorname{tr}(S^{-1})x-\det(S)\\ &=x^3-\operatorname{tr}(S)x^2+\operatorname{tr}(S)x-1. \end{aligned} $$ Consequently, $\det(I+S)=-p(-1)=2\left(1+\operatorname{tr}(S)\right)$.