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Sorry about the vague subject but I really found some difficulties in calculating complex matrices.

Assume $Z$ is a square Hermitian non-singular complex matrix, then we denote $$F= \left[ \begin{array}{cc} Re(Z) & -Im(Z) \\ Im(Z) & Re(Z) \\ \end{array} \right], \tag{1}$$ where $Re(\cdot)$ and $Im(\cdot)$ represent the real part and imaginary part. In a paper, it shows that the inverse of $F$ is given by $$F^{-1}= \left[ \begin{array}{cc} Re(Z^{-1}) & -Im(Z^{-1}) \\ Im(Z^{-1}) & Re(Z^{-1}) \\ \end{array} \right]. \tag{2}$$ My question 1: How to prove (2)?

Another derivation in that paper, assume $Y$ also square complex matrix, then $$\left[ \begin{array}{cc} Re(Z^{-1}) & -Im(Z^{-1}) \\ Im(Z^{-1}) & Re(Z^{-1}) \\ \end{array} \right] \left[ \begin{array}{c} -Im(Y) \\ Re(Y) \\ \end{array} \right] = \left[ \begin{array}{c} -Im(Z^{-1}Y) \\ Re(Z^{-1}Y) \\ \end{array} \right] \tag{3} $$ My question 2: How to prove (3)?

Basically, as you can see, I have no idea how to calculate those $Re()$ and $Im()$. Do you have any idea how to search references about this subject? I searched "complex matrix computation", "real representation of complex matrices", no go. Any suggestion will be appreciated. Thanks.

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Theorem $$ \begin{align} Re(AB) &= Re(A)Re(B) -Im(A)Im(B) \\ Im(AB) &= Re(A)Im(B) + Im(A)Re(B) \end{align} $$

Proof $$ \begin{align} AB &= (Re(A)+iIm(A))(Re(B)+iIm(B)) \\ &= Re(A)Re(B) + i(Re(A)Im(B) + Im(A)Re(B)) - Im(A)Im(B). \end{align} $$ From which immediately follows that: $$ \begin{align} Re(AB) &= Re(A)Re(B) - Im(A)Im(B) \\ Im(AB) &= Re(A)Im(B) + Im(A)Re(B). \end{align} $$

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  • $\begingroup$ Thanks. It perfectly answers my question 2. Any idea of proving question 1? Many thanks. $\endgroup$ – MIMIGA Dec 12 '14 at 15:56
  • $\begingroup$ This answer all the questions. Think how. Hint: $Re(Id)= Id$ and $Im(Id)=0$. $\endgroup$ – hjhjhj57 Dec 12 '14 at 19:50
  • $\begingroup$ @MIMIGA If you think this answered your question, please click the checkmark on the left hand side of the answer to mark it as the right answer. $\endgroup$ – hjhjhj57 Dec 13 '14 at 18:55

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