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How are the eigenvalues of $A^H$ related to the eigenvalues of $A$?

Here $A^H$ is the conjugate transpose of $A$

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2 Answers 2

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Hint: For a matrix $X$, the determinant $det(X^H)=\overline{det(X)}$ where the overline indicates complex conjugation.

Examine $det((I\lambda-A)^H)$

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Lets say $\lambda_1$ is an eigenvalue of $A$ and $\lambda_2$ is an eigenvalue of $A^H$. We have: $$ Ax = \lambda_1 x $$ $$ A^Hy = \lambda_2 y $$ Doing conjugate transpose on both sides of above equation and since $(A^H)^H = A$, we get: $$ y^HA = \bar\lambda_2 y^H $$ Multiplying by $x$ to the right on both sides: $$ y^HAx = \bar\lambda_2 y^Hx $$ Now using $Ax = \lambda_1 x$ : $$ y^H \lambda_1 x = \bar\lambda_2 y^Hx $$ $$ \lambda_1 y^Hx = \bar\lambda_2 y^Hx $$ This gives us: $$ \lambda_1 = \bar \lambda_2 $$ or $\lambda_2 = \bar \lambda_1$ i.e the eigenvalues of conjugate transpose of a matrix are conjugates of the eigenvalues of the original matrix.