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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!

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

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  • $\begingroup$ Ok I think that makes sense! I really appreciate the help. I guess I could think about it as writing out all the terms in the big cofactor formula for determinant. Those are all sums and products and the complex conjugate only changes the imaginary parts. I was really stuck on this and am really grateful for the help. Thanks again! $\endgroup$
    – theMikus
    Jan 9 at 23:56
  • $\begingroup$ Glad I could help :) $\endgroup$ Jan 10 at 0:10

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