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

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!

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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 $$X=\text{diag}[(1-2au_{11})^{-1},\ldots,(1-2au_{nn})^{-1}].$$

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+1 nice. I think you could synopsize by writing that there is a solution iff $U$ has the form $U = \lambda ( 2 I - e e^T)$, with $\lambda \neq \frac{1}{2a}$ –  copper.hat Aug 30 '12 at 19:40
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 $U$.
$UX \neq [x_1 \ x_1; 0 \ x_2]$, it should be $UX \neq [x_1 \ x_2; 0 \ x_2]$ –  John Smith Aug 31 '12 at 1:24