solving a matrix equation $X-I=a \cdot (X\cdot U^T + U \cdot X)$

Question

I am trying to solve the $n \times n$ diagonal matrix $X$ in the following equation:

$$X-I=a \cdot (X\cdot U^T + U \cdot X)$$

where $0<a<1$ is a given scalar, $U$ is a $n \times n$ given matrix, and $I$ is a $n \times n$ identity matrix.

Any suggestions are welcome!

Answer 1

Let $U=(u_{ij}), X=\text{diag}(x_1,\ldots,x_n) \in \mathbb{C}^{n\times n}$, and set $B=(b_{ij}):=a(XU^T+UX)$. We have to find $X$ such that $$\tag{1} X-I=a(XU^T+UX). $$ For every $i,j$ we have $$ b_{ij}=a\sum_{k=1}^n(x_i\delta_{ik}u_{jk}+u_{ik}x_k\delta_{kj})=ax_i(u_{ii}+u_{ji}). $$ Therefore $$ (1) \iff (x_i-1)\delta_{ij}=ax_i(u_{ii}+u_{ji}) \iff [\delta_{ij}-a(u_{ii}+u_{ji})]x_i=\delta_{ij}, $$ i.e. $$ (1-2au_{ii})x_i=1 \quad \forall 1 \le i \le n,\ \text{ and } (u_{ii}+u_{ji})x_i=0 \quad \forall j \ne i. $$

If $U$ is such that $1-2au_{ii} = 0$ for some $i$, there is no $X$ satisfying (1).

If $U$ is such that $1-2au_{ii}\ne 0$ for every $1 \le i \le n$, and $u_{ij}+u_{ii} \ne 0$ for some $j \ne i$, there is no $X$ satisfying (1).

If $U$ is such that $1-2au_{ii}\ne 0$ for every $1 \le i \le n$, and $u_{ij}+u_{ii}=0$ for every $j \ne i$, then $$ X=\text{diag}[(1-2au_{11})^{-1},\ldots,(1-2au_{nn})^{-1}]. $$

Answer 2

There is no solution in general.

For example, take $U = \begin{bmatrix} 1 & 1 \\ 0 & 1 \end{bmatrix}$, and $X = \begin{bmatrix} x_1 & 0 \\ 0 & x_2 \end{bmatrix}$. Then multiplying out gives: $$ X-I = \begin{bmatrix} x_1-1 & 0 \\ 0 & x_2-1 \end{bmatrix} = a\begin{bmatrix} x_1 & 0 \\ x_2 & x_2 \end{bmatrix} + a\begin{bmatrix} x_1 & x_2 \\ 0 & x_2 \end{bmatrix}= a (X U^T + U X).$$

This gives $ a x_2 = 0$, $x_1(1-2a) = x_2(1-2a) = 1$. The last two equations imply $a \neq \frac{1}{2}$, $x_1 \neq 0$, $x_2 \neq 0$, which contradicts the first equation. So there is no solution for this $U$.