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?
Answer:
"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!
The determinant of the transpose is the transpose of the determinant. So it's all about the conjugate coefficients.
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.