System of equations in terms of an unknown and its conjugate

Question

I have a system of complex equations:

$$ A z + B \overline{z} = c$$

Where $A, B \in \mathbb{C}^{N \times N}$ and $z, c \in \mathbb{C}^{N \times 1}$.

I want to solve for $z$. If I could express it as just $Az = b$ there are a whole host of matrix solving techniques I could use. Or if it were just $A\overline{z} = b$, I could solve for $\overline{z}$ and take its conjugate to get $z$.

Answer 1

You can convert it into a problem of Linear Algebra over the reals. If $N=2$, $A=\left[\begin{smallmatrix}a_{11}&a_{12}\\a_{21}&a_{22}\end{smallmatrix}\right]$, $B=\left[\begin{smallmatrix}b_{11}&b_{12}\\b_{21}&b_{22}\end{smallmatrix}\right]$, $z=\left[\begin{smallmatrix}z_1\\z_2\end{smallmatrix}\right]$, and $c=\left[\begin{smallmatrix}c_1\\c_2\end{smallmatrix}\right]$, then your problem is equivalent to\begin{multline}\begin{bmatrix}\operatorname{Re}(a_{11})&-\operatorname{Im}(a_{11})&\operatorname{Re}(a_{12})&-\operatorname{Im}(a_{12})\\\operatorname{Im}(a_{11})&\operatorname{Re}(a_{11})&\operatorname{Im}(a_{12})&\operatorname{Re}(a_{12})\\\operatorname{Re}(a_{21})&-\operatorname{Im}(a_{21})&\operatorname{Re}(a_{22})&-\operatorname{Im}(a_{22})\\\operatorname{Im}(a_{21})&\operatorname{Re}(a_{21})&\operatorname{Im}(a_{22})&\operatorname{Re}(a_{22})\end{bmatrix}.\begin{bmatrix}\operatorname{Re}(z_1)\\\operatorname{Im}(z_1)\\\operatorname{Re}(z_2)\\\operatorname{Im}(z_2)\end{bmatrix}+\\+\begin{bmatrix}\operatorname{Re}(b_{11})&-\operatorname{Im}(b_{11})&\operatorname{Re}(b_{12})&-\operatorname{Im}(b_{12})\\\operatorname{Im}(b_{11})&\operatorname{Re}(b_{11})&\operatorname{Im}(b_{12})&\operatorname{Re}(b_{12})\\\operatorname{Re}(b_{21})&-\operatorname{Im}(b_{21})&\operatorname{Re}(b_{22})&-\operatorname{Im}(b_{22})\\\operatorname{Im}(b_{21})&\operatorname{Re}(b_{21})&\operatorname{Im}(b_{22})&\operatorname{Re}(b_{22})\end{bmatrix}.\begin{bmatrix}\operatorname{Re}(z_1)\\-\operatorname{Im}(z_1)\\\operatorname{Re}(z_2)\\-\operatorname{Im}(z_2)\end{bmatrix}=\begin{bmatrix}\operatorname{Re}(c_1)\\\operatorname{Im}(c_1)\\\operatorname{Re}(c_2)\\\operatorname{Im}(c_2)\end{bmatrix}.\end{multline}And, since\begin{multline}\begin{bmatrix}\operatorname{Re}(b_{11})&-\operatorname{Im}(b_{11})&\operatorname{Re}(b_{12})&-\operatorname{Im}(b_{12})\\\operatorname{Im}(b_{11})&\operatorname{Re}(b_{11})&\operatorname{Im}(b_{12})&\operatorname{Re}(b_{12})\\\operatorname{Re}(b_{21})&-\operatorname{Im}(b_{21})&\operatorname{Re}(b_{22})&-\operatorname{Im}(b_{22})\\\operatorname{Im}(b_{21})&\operatorname{Re}(b_{21})&\operatorname{Im}(b_{22})&\operatorname{Re}(b_{22})\end{bmatrix}.\begin{bmatrix}\operatorname{Re}(z_1)\\-\operatorname{Im}(z_1)\\\operatorname{Re}(z_2)\\-\operatorname{Im}(z_2)\end{bmatrix}=\\=\begin{bmatrix}\operatorname{Re}(b_{11})&\operatorname{Im}(b_{11})&\operatorname{Re}(b_{12})&\operatorname{Im}(b_{12})\\\operatorname{Im}(b_{11})&-\operatorname{Re}(b_{11})&\operatorname{Im}(b_{12})&-\operatorname{Re}(b_{12})\\\operatorname{Re}(b_{21})&\operatorname{Im}(b_{21})&\operatorname{Re}(b_{22})&\operatorname{Im}(b_{22})\\\operatorname{Im}(b_{21})&-\operatorname{Re}(b_{21})&\operatorname{Im}(b_{22})&-\operatorname{Re}(b_{22})\end{bmatrix}.\begin{bmatrix}\operatorname{Re}(z_1)\\\operatorname{Im}(z_1)\\\operatorname{Re}(z_2)\\\operatorname{Im}(z_2)\end{bmatrix},\end{multline}the previous system is equivalent to\begin{multline}\begin{bmatrix}\operatorname{Re}(a_{11})+\operatorname{Re}(b_{11})&-\operatorname{Im}(a_{11})+\operatorname{Im}(b_{11})&\operatorname{Re}(a_{12})+\operatorname{Re}(b_{12})&-\operatorname{Im}(a_{12})+\operatorname{Im}(b_{22})\\\operatorname{Im}(a_{11})+\operatorname{Im}(b_{11})&\operatorname{Re}(a_{11})-\operatorname{Re}(b_{11})&\operatorname{Im}(a_{12})+\operatorname{Im}(b_{12})&\operatorname{Re}(a_{12})-\operatorname{Re}(b_{12})\\\operatorname{Re}(a_{21})+\operatorname{Re}(b_{21})&-\operatorname{Im}(a_{21})+\operatorname{Im}(b_{21})&\operatorname{Re}(a_{22})+\operatorname{Re}(b_{22})&-\operatorname{Im}(a_{22})+\operatorname{Im}(b_{22})\\\operatorname{Im}(a_{21})+\operatorname{Im}(b_{12})&\operatorname{Re}(a_{21})-\operatorname{Re}(b_{21})&\operatorname{Im}(a_{22})+\operatorname{Im}(b_{22})&\operatorname{Re}(a_{22})-\operatorname{Re}(b_{22})\end{bmatrix}.\\.\begin{bmatrix}\operatorname{Re}(z_1)\\\operatorname{Im}(z_1)\\\operatorname{Re}(z_2)\\\operatorname{Im}(z_2)\end{bmatrix}=\begin{bmatrix}\operatorname{Re}(c_1)\\\operatorname{Im}(c_1)\\\operatorname{Re}(c_2)\\\operatorname{Im}(c_2)\end{bmatrix}.\end{multline}

Answer 2

You can split into the real and imaginary parts of $z$. Write $z = x + i y$, where $x, y$ are real $N$-vectors, and then you have

$$ Ax + Aiy + Bx - Biy = c $$ which you can rewrite as $$ (A+B)x + i(A-B)y = c $$ So forming an $N \times 2N$ matrix like this: $$ M = \pmatrix {(A+B) & i(A-B)} $$ by concatenating the two $N \times N$ matrices, and forming a real $2N$ vector $$ u = \pmatrix {x\\y} $$ by concatenating the vectors $x$ and $y$ vertically, you get a system you can solve by more ordinary techniques.

If you try this with, say $N = 2$, it'll become immediately obvious that you can instead solve two real problems by taking the real and complex parts of the matrix $M$ and the vector $c$. You can actually concatenate these vertically, to get $$ \pmatrix{\Re{(M)} \\ \Im{(M)}} u = \pmatrix{\Re{(c)} \\ \Im{(c)}}. $$ where $\Re(x + iy) = x$ and $\Im(x+iy) = y$, with the natural term-by-term extension to vectors and matrices.

Now you've got a $2N \times 2N$ real matrix problem to solve, and can apply all those techniques you've been yearning for.

Answer 3

Here is an approach that doesn't require building a larger real matrix.
Starting with the equation: \begin{equation} \begin{matrix} Az + B\bar{z} = c \end{matrix} \end{equation}

If $z=x +yi$ (where $x,y \in \mathbb{R}$), we have $ \bar{z} = z - 2yi$. Then we can write \begin{equation} \begin{matrix} \left(A+B \right)z ~~~&+ -2iBy &= c \\ Re \left (i I_{p}z \right) &+~~~~ I_{p}y &=0 \end{matrix} \end{equation} The second block equation just means $y = -Re(iz)$, where $Re()$ takes the real part of a complex expression. Note that you can scale the real and imaginary part of a single $z$ independently without changing the right hand side. With complex multiplication you can only scale both parts the same or rotate. Not complex linear.

We assume $A+B$ is invertible. Following the wiki page for Schur complement section "Application to solving linear equations" we can define the Schur complement \begin{equation} S = I_{p} - 2Re \left(I_{p} \left( A+B \right) ^{-1} B \right) \end{equation} Note that $S$ is real as it is the sum of two real matrices. We have made use of the fact that real variables can be factored out of the $Re()$ function. $Re(i \cdot w \cdot -2i ) = 2Re(w)$. where $w \in \mathbb{C} $.
But we can not do that for complex values, as the function is not linear over complex values. That is $ Re(i \cdot i) \neq i \cdot Re(i)$.

We arrive at the equation (If I did the math right): \begin{equation} z = \left( A+B \right) ^{-1} \left( c - 2iBS^{-1} Re \left( i I_{p} \left( A+B \right) ^{-1}c \right)\right) \end{equation}

In problem I originally worked on, wanted to add only a few real variables to linear system of complex variables. Then quite helpful. (In this case remove rows from lower left block and remove columns from upper right block. Left identity matrices in equations so easier to see importance.) I did answer a very similar question recently on solving equations with mixed real and complex variables. (Actually answered a couple of years ago, but just updated.) The link: https://math.stackexchange.com/a/4456883/1060562 If anyone has seen this before let me know.