I would strongly assume that this must exist already somewhere but I couldn’t find the solution so I thought it would be interesting to post it here. The closest to this I could find is widely linear estimation (e.g., Picinbono, TSP, 1995), but it’s not quite the same.<\/p>\n
Consider the following widely linear system of equations:<\/p>\n
$$\\ma{A} \\cdot \\ma{x} + \\ma{B} \\cdot \\ma{x}^* = \\ma{c},$$<\/p>\n
where $\\ma{A},\\ma{B} \\in \\compl^{N \\times N}$ are square invertible matrices and $\\ma{x}, \\ma{c} \\in \\compl^{N \\times 1}$ vectors of corresponding dimension. We would like to find the vector $\\ma{x}$ which satisfies this equation given $\\ma{A}$, $\\ma{B}$, and $\\ma{c}$, if it exists.<\/p>\n
This is a system of equations but it is not linear in $\\ma{x}$. It is widely linear though and this implies that it is linear in\u00a0 ${\\rm Re}\\{\\ma{x}\\}$ and in ${\\rm Im}\\{\\ma{x}\\}$. Therefore, it can easily be rewritten as a set of linear equations by introducing the real and imaginary parts of all quantities, i.e., $\\ma{A} = \\ma{A}_{\\rm R} + \\jmath \\ma{A}_{\\rm I}$, $\\ma{B} = \\ma{B}_{\\rm R} + \\jmath \\ma{B}_{\\rm I}$, $\\ma{c} = \\ma{c}_{\\rm R} + \\jmath \\ma{c}_{\\rm I}$, and $\\ma{x} = \\ma{x}_{\\rm R} + \\jmath \\ma{x}_{\\rm I}$. Inserting this into the widely linear system of equations and separating real and imaginary parts we obtain<\/p>\n
$$\\left( \\left[ \\ma{A}_{\\rm R} + \\ma{B}_{\\rm R} \\right] \\cdot \\ma{x}_{\\rm R} + \\left[ -\\ma{A}_{\\rm I} + \\ma{B}_{\\rm I} \\right] \\cdot \\ma{x}_{\\rm I}\\right) + \\jmath \\cdot \\left( \\left[ \\ma{A}_{\\rm I} + \\ma{B}_{\\rm I} \\right] \\cdot \\ma{x}_{\\rm R} + \\left[ \\ma{A}_{\\rm R} – \\ma{B}_{\\rm R} \\right] \\cdot \\ma{x}_{\\rm I}\\right) = \\ma{c}_{\\rm R} + \\jmath \\ma{c}_{\\rm I}.$$<\/p>\n
As both sides of the equations are complex numbers, they are equal only if the real parts are equal and the imaginary parts are equal. Hence we have two real-valued systems of equation, which we can write in one larger system:<\/p>\n
$$ \\begin{bmatrix}
\n\\ma{A}_{\\rm R} + \\ma{B}_{\\rm R} & -\\ma{A}_{\\rm I} + \\ma{B}_{\\rm I}\u00a0 \\\\
\n\\ma{A}_{\\rm I} + \\ma{B}_{\\rm I} & \\ma{A}_{\\rm R} – \\ma{B}_{\\rm R}
\n\\end{bmatrix}
\n\\cdot
\n\\begin{bmatrix} \\ma{x}_{\\rm R} \\\\ \\ma{x}_{\\rm I} \\end{bmatrix}
\n= \\begin{bmatrix} \\ma{c}_{\\rm R} \\\\ \\ma{c}_{\\rm I} \\end{bmatrix}.$$
\n$$ \\ma{\\tilde{C}} \\cdot
\n\\begin{bmatrix} \\ma{x}_{\\rm R} \\\\ \\ma{x}_{\\rm I} \\end{bmatrix}
\n= \\begin{bmatrix} \\ma{c}_{\\rm R} \\\\ \\ma{c}_{\\rm I} \\end{bmatrix}.$$<\/p>\n
Consequently we have exactly one solution in $\\ma{x}$ if and only if the block matrix $\\ma{\\tilde{C}}$ is non-singular, which implies additional conditions on $\\ma{A}$ and $\\ma{B}$. For instance, a sufficient (but not necessary) condition is that $\\ma{A}_{\\rm R} + \\ma{B}_{\\rm R}$ and its Schur complement $\\ma{A}_{\\rm R} – \\ma{B}_{\\rm R} – (\\ma{A}_{\\rm I} + \\ma{B}_{\\rm I}) \\cdot (\\ma{A}_{\\rm R} + \\ma{B}_{\\rm R})^{-1} \\cdot (-\\ma{A}_{\\rm I} + \\ma{B}_{\\rm I})$ are both invertible.<\/p>\n
<\/p>\n