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Suppose $q$ is a quadratic form on $\mathbb{C}^n$: $q(x)=x^HAx$, with $H$ denoting the hermitian transpose. Since I am only interested in the real part of $q$, I am trying to determine a matrix $B$ so that

$$ \Re(x^HAx)=x^HBx $$

The real part of matrix $A$, defined as

$$ \Re\{A\} = \frac{1}{2}\left(A + A^H \right), $$

is a symmetric positive semidefinite matrix and $B$ is hermitian. I do know that $B$ exists - the question is, how do I get it from $A$? Thanks!

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Such a $B$ does not exist unless $A=0$.

Lemma. Suppose $U$ and $V$ are real symmetric matrices. Then for any two real vectors $x$ and $y$, $$(x+iy)^T(U+iV)(x+iy) = x^TUx - y^TUy - 2y^TVx + i( x^TVx - y^TVy + 2y^TUx ).\tag{1}$$

Since every quadratic form over $\mathbb{C}$ is induced by a symmetric bilinear form, we may assume without loss of generality that the $A$ and $B$ in your question are complex symmetric. If $\Re(z^TAz)\equiv z^TBz$, then $z^TBz\in\mathbb{R}$ for all $z\in\mathbb{C}^n$. By $(1)$, this implies that $B=0$. Hence $\Re(z^TAz)\equiv0$ and by $(1)$, we get $A=0$.

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  • $\begingroup$ This certainly cannot be true since, picking any hermitian $A$, we would just have $A = B \ne 0$ (or, similarly, picking any antihermitian $A$, we would just have $B = 0$). I believe that you may have forgotten to flip the imaginary part of the LHS in the lemma (e.g., the transpose, that is, since in the OP it is not just a transpose, but a hermitian conjugate). $\endgroup$ – Guillermo Angeris Feb 1 '19 at 19:47
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    $\begingroup$ @GuillermoAngeris This answer addressed his/her original question. The OP changed the question slightly before I posted my answer and I probably didn't notice that. As this is basically an abandoned question, I will leave my answer as it is. $\endgroup$ – user1551 Feb 2 '19 at 9:07
  • $\begingroup$ ah, I see. This makes more sense. $\endgroup$ – Guillermo Angeris Feb 2 '19 at 19:09
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Since $A$ can be written as the sum of its hermitian (or real, as you mentioned in the question) and anti-hermitian parts, then the inner product becomes $$ \Re\left\{x^H(A_h + A_a)x\right\} = \Re\left\{x^HA_hx + x^HA_ax\right\} = \Re\left\{x^HA_hx\right\} + \Re\left\{x^HA_ax\right\}. $$ An inner product with a hermitian matrix is always real, while the anti-hermitian part is purely imaginary (exercise!), so the anti-hermitian term cancels and we can drop the "real part" of the hermitian term, which gives $$ \Re\left\{x^H(A_h + A_a)x\right\} = x^HA_hx, $$ as you conjectured.

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