# Hermitian of A Determinant??

I'm a self-learner working through Gil Strangs Linear Algebra course. In the Chapter on Hermitian, Unitary Matrices, I came across this question:
" How are the eigenvalues of $$A^H$$ related to the eigenvalues of the square matrix A?
"The eigenvalues of $$A^H$$ are complex conjugates of the eigenvalues of A: det(A−λI)= 0 gives det($$A^H$$$$\overline{λ}I$$) = 0

I can't really see why this is the case. I'm assuming he is applying the conjugate transpose to det(A−λI) ?
If so how, I'm confused as to how the complex conjugate $$[det(A−λI)]^H$$ works its way inside the determinant.
I hope the question made sense and appreciate any insight someone might have! -Thanks!

• If $A$ is a square matrix, then $\det ⁡ ( A ^H ) = \overline{\det ⁡ ( A )}$ Jan 9 at 23:33
• Thanks for the Help Tanner! I appreciate your time! Jan 9 at 23:57

The determinant of a matrix can be written as $$\det A=\sum_\sigma \operatorname{sgn}(\sigma)\prod_{j=1}^n a_{j,\sigma(j)}.$$ Then $$\det \overline A=\sum_\sigma \operatorname{sgn}(\sigma)\prod_{j=1}^n \overline{a_{j,\sigma(j)}} =\overline{\sum_\sigma \operatorname{sgn}(\sigma)\prod_{j=1}^n a_{j,\sigma(j)}}=\overline{\det A}.$$ It's just the fact that the conjugate preserves sums, products, and real numbers.