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

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

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