Prove that matrix is symmetric and positive definite given the fact that $A+iB$ is.

Question

I have some questions regarding the following problem

Let $ A + iB $ - hermitian and positive definite, where $A, B \in \mathbb R^{n\ \times\ n} $ show that the real matrix $$C =\begin{pmatrix} A & -B \\ B & A \end{pmatrix} $$ is symmetric and positive definite.

How can the following system of linear equations be solved using Cholesky decomposition of C: $ (A+iB)(x+iy)=B+iC$

I've tried to transform C and get to $A + iB$, I've tried to compute the determinants to get to $A + iB$ but I had no result.

Moreover, I have no idea how to approach the second part of the problem.

Thank you

Answer 1

Since $A+iB$ is Hermitian, $A$ is symmetric and $B$ is antisymmetric (the latter implying that $x^TBx=0$ for all real $x$) so the block matrix $C$ is symmetric. If $x=y+iz$ with real $y$ and $z$, $$ \begin{split} x^*(A+iB)x&=(y-iz)^T(A+iB)(y+iz) \\&= y^TAy+z^TBy-y^TBz+z^TAz +iy^TAz+iz^TBz+iy^TBy-iz^TAy \\&= y^TAy+z^TBy-y^TBz+z^TAz \\&= \pmatrix{y\\z}^T\pmatrix{A&-B\\B&A}\pmatrix{y\\z}. \end{split} $$ This implies that

$$ \text{$A+iB$ is HPD}\quad \Leftrightarrow \quad \pmatrix{A&-B\\B&A}\;\text{is SPD}. $$

We seek the solution of $(A+iB)(y+iz)=b+ic$. Since $$ (A+iB)(y+iz)=Ay-Bz+i(By+Az), $$ we have two equivalent (real) equations $$ Ay-Bz=b,\quad By+Az=c. $$ Therefore,

$$ (A+iB)(y+iz)=b+ic\quad\Leftrightarrow\quad \pmatrix{A&-B\\B&A}\pmatrix{y\\z}=\pmatrix{b\\c}. $$

Since we already know, that the real block matrix is SPD, it has a Cholesky factorization.

Answer 2

$A+iB= (A+iB)^\ast=A^T-iB^T$ so that $$ A=A^T,\ B=-B^T $$ Hence given block matrix in OP is symmetric.

Note that if $H$ is hermitian and $V$ is complex vector then $(V,HV)$ is real : $$ (V,HV)- \overline{(V,HV)}=(V,HV)-(HV,V)=0 $$

So $$ 0< (u+iv,(A+iB)(u+iv))=(u+iv, Au-Bv + i(Av+Bu )) $$ $$ =(u,Au - Bv) + (v,Av+ Bu) =(u,Au) + (v,Av) +2(v,Bu)$$ $$ = (u\ v)\bigg(\begin{array}{cc} A&-B\\B& A \end{array}\bigg)\bigg(\begin{array}{c} u\\ v\end{array} \bigg) $$