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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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1 Answer 1

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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