Finding X from the matrix equation which includes trace function

Question

Is there any way we can solve the following equation for $X$, where $X$, $A$, $B$ are square matrices of the same size and $\lambda$, $c$, and $a$ are scalar constants?

$\lambda I = B + \dfrac{cA}{1+a \text{ Tr} (AX)}$

Answer 1

Let us separate the matrix and the scalar part in your equation :

$$\lambda I - B = y A \ \ \ \ \ \text{with} \ \ \ \ \ y:=\dfrac{c}{1+a \text{ Tr} (AX)} \in \mathbb{R}\tag{1}$$

Let us take (see remark below) a matrix $X_0$ such that

$$\text{ Tr}(AX_0) \neq 0 \ \ \text{and} \ \ \neq -\frac{1}{a}\tag{2}$$

Conditions (2) imply that coefficient $y$ can take any real value by replacing $X_0$ by a convenient multiple $x X_0$ (with an exceptional case : it cannot take the value $c/a$).

Therefore, considering (1), the initial question is answered in this way : if $A$ is a multiple of $B$ minus a multiple of identity matrix, there is always a solution $X=xX_0$ obtained by inverting relationship $y=f(x)$ as defined by (1).

Remark : The fact that there exists an $X_0$ such that $\text{ Tr}(AX_0) \neq 0$ is central. In fact, it is a consequence of the following

Lemma : Let $E_{ij}$ be the classical matrix with all its entries equal to zero but one which is $e_{i,j}=1$ ; we have (please note the inversion of indices) :

$$trace(AE_{ij})=a_{ji}$$

(easy proof). As a consequence of the lemma, unless $A$ is the null matrix, it will always be possible to find such an $X_0$ [moreover, we will have a big amount of choice !].