Determinant of hermitian matrix

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Let $M=A+iB$ be a complex $n \times n$ Hermitian matrix. First of all we know that $$(\det M)^2=\det \begin{pmatrix} A & -B \\ B & A \end{pmatrix}.$$ Also $\det \begin{pmatrix} A & -B \\ B & A \end{pmatrix}$ is a polynomial in $n^2$ variables of degree $2n$. Is it true that $\det M$ is a polynomial, say D, of degree $n$ in this $n^2$ variables such that $D^2(M)=\det \begin{pmatrix} A & -B \\ B & A \end{pmatrix}$? The explicit calculations for $n=1,2,3$ suggest so, yet I can't find the information if this is true neither proof this.

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1 Answer

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The matrix is similar to

$$ S = T \hat{M} T^{-1}= \begin{pmatrix} I & iI \\ I & -iI \end{pmatrix} \begin{pmatrix} A & -B \\ B & A \end{pmatrix} \begin{pmatrix} \frac12I & \frac12 I\\-\frac i2I& \frac i2I\end{pmatrix} = \begin{pmatrix} A+Bi &\\ &A-Bi \end{pmatrix}= \begin{pmatrix} M &\\ &M^* \end{pmatrix}=\begin{pmatrix} M &\\ &M \end{pmatrix} $$ hence $S$ and $\hat M$ have the same characteristic polynomial.

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