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Suppose $A=B+jC$ is a positive semi-definite (psd) matrix in $\mathbb C^{n\times n}$, where $B$ and $C$ are its real and imaginary parts. Is the real matrix $\left[\begin{array}{cc}B & C^\top \\ C & B\end{array}\right]$ positive semidefinite?

syeh_106
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    Without thinking too much, the logic should presumably be: Let $v=\binom{x}{y}$ with $x,y$ as real vectors. If $v$ violates the positive-definiteness of the block matrix, then the complex vector $z=x+jy$ should do the same for $A$. – Semiclassical Feb 15 '23 at 07:34
  • @Semiclassical Indeed. Thanks a lot for the pointer. – syeh_106 Feb 15 '23 at 07:46
  • If $A$ is positive semi-definite, then $C=0$ (see here: https://math.stackexchange.com/questions/267300/positive-definite-matrix-must-be-hermitian). This makes the problem considerably easier. – MaoWao Feb 15 '23 at 08:17
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    @MaoWao Wait, what? That is not true! If $A$ is PSD, then $A$ is Hermitian which only implies that $C^T=-C$, but not $C=0$. – Levent Feb 15 '23 at 11:39
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    @Levent I see. You used a different (in my experience non-standard) notion of real and imaginary part. At least in operator algebras, it is common to call $(A+A^\ast)/2$ the real part and $(A-A^\ast)/2i$ the imaginary part of $A$. But I guess you used the entrywise real and imaginary part, right? – MaoWao Feb 15 '23 at 13:08
  • @MaoWao Yes, I used the entrywise real and imaginary part. I would call $(A+A^)/2$ the Hermitian part and $(A-A^)/2$ the skew Hermitian part of $A$, respectively. I didn't know that they were called the real and the imaginary part in different contexts. It does make sense though! – Levent Feb 15 '23 at 13:12

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First, I will assume that $B^\top = B$ and $ C^\top =-C$ so that $A$ is hermitian. (Otherwise, I don't know what PSD means for a complex matrix.)

For the complex matrix $B+jC$, the quadratic form is

\begin{align} Q(x,y) &=(x+j y)^* (B+jC)(x+jy)\\ &=(x^\top -j y^\top)[(Bx-Cy)+j(By+Cx)]\\ &=x\cdot (Bx-Cy)+j x\cdot (By+Cx)-jy\cdot (Bx-Cy)+y\cdot(By+Cx)\\ &=(x\cdot Bx-x \cdot Cy+y\cdot By+y\cdot Cx)+j(x\cdot By+x\cdot Cx-y\cdot Bx+y\cdot Cy) \end{align} From here we may use the assumptions on $B,C$ to write \begin{align} x\cdot By&=By\cdot x=(By)^\top x=y^\top Bx=y\cdot Bx,\\ x\cdot Cx&=Cx\cdot x=(Cx)^\top x=-x^\top (Cx)=-x\cdot Cx=0,\\ y\cdot Cy&=Cy\cdot y=(Cy)^\top y=-y^\top (Cy)=-y\cdot Cy=0,\\ x\cdot Cy&=(Cy)\cdot x=(Cy)^\top x=- y^\top Cx=-y\cdot Cx \end{align} Hence the imaginary part of the quaratic form vanishes and the real part simplifies to $$Q(x,y)=x\cdot Bx+2(Cx)\cdot y+y\cdot By.$$ But this is equivalent to $$ \begin{bmatrix} x^\top & y^\top \end{bmatrix} \begin{bmatrix} B & C^\top \\ C & B\end{bmatrix} \begin{bmatrix} x \\ y \end{bmatrix}$$ That is, the quadratic form on $B+jC$ is equivalent to the quadratic form on $\begin{bmatrix} B & C^\top \\ C & B\end{bmatrix}$. Thus each is PSD if and only if the other is as well.

Semiclassical
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