Express $\mathrm{Tr}(X)$ in terms of $A$, given that $X=A^TX(I+X)^{-1}A$ Given real non-singular $n\times n$ matrix $A$, with all eigenvalues larger that $1$.

Express $\mathrm{Tr}(X)$ in terms of $A$, given that
$X=A^TX(I+X)^{-1}A$.    $\quad$($X$ is sym. pos. def.)

It is allowed to assume that $A$ is in any special form that can be obtained using similarity transformation, i.e. $A$ can be changed by $\hat{A}$, if $A=P^{-1}\hat{A}P$, for some nonsingular $P$.

My attempt: For symmetric $A$ case, WLOG we can assume $A$ is diagonal, then
\begin{align}
X&=AX(I+X)^{-1}A\\
AX^{-1}AX&=I+X\\
AYA&=Y+I\\
(A\otimes A-I)\mathrm{vec}(Y)&=\mathrm{vec}(I)\\
\mathrm{vec}(Y)&=(A\otimes A-I)^{-1}\mathrm{vec}(I),
\end{align}
where $Y=X^{-1}$ and $\otimes$ denotes Kronecker product.
If $\mathrm{vec}(Y)=(A\otimes A-I)^{-1}\mathrm{vec}(I)$, then $\mathrm{vec}(X)=(A\otimes A-I)\mathrm{vec}(I)$, using this result.
From last eq. it is easy to see that $\mathrm{Tr}(X)=\sum_{i=1}^na^2_i-n$, where $a_i$ are diagonal elements of $A$.

Assume that all eigenvalues of $(A)$ are equal to $\lambda$ and that Jordan canonical form of $A$ consists of single Jordan block $J$. Then
\begin{align}
\mathrm{vec}(Y)&=(A\otimes A-I)^{-1}\mathrm{vec}(I)\\
&=\begin{bmatrix} 
\lambda J-I & J &  &&  \\ 
&\lambda J-I&J& &\\
 &  & \ddots & \ddots& \\
&  &  & \lambda J-I & J \\
& &  &  & \lambda J-I
\end{bmatrix}^{-1}\mathrm{vec}(I)\\
&=\begin{bmatrix} 
M & -MJM & MJMJM & \dots & (-1)^{n-1}MJM\cdots M \\ 
 & \ddots & \ddots & \ddots & \ddots\\
  &  & M & -MJM & MJMJM\\
&  &  & M& -MJM\\
&&&&M
\end{bmatrix}\mathrm{vec}(I)\\
&=\begin{bmatrix} 
TJ^{-1} & -T^2J^{-1} & T^3J^{-1} & \dots & (-1)^{n-1}T^nJ^{-1} \\ 
 & \ddots & \ddots & \ddots & \ddots\\
  &  & TJ^{-1} & -T^2J^{-1} & T^3J^{-1}\\
&  &  & TJ^{-1}& -T^2J^{-1}\\
&&&&TJ^{-1}
\end{bmatrix}\mathrm{vec}(I)
\end{align}
where $M=(\lambda J-I)^{-1}$ and $T=MJ=(\lambda I-J^{-1})^{-1}$. Second last equation is obtained using Block matrix inversion formula. I am not sure what to do next, the expression looks too complicated. Maybe there is an easier way to do it?

EDIT:    $\qquad$ For $A=\begin{bmatrix} 
a_1 & 1\\ 
 0&a_2
\end{bmatrix}$, we will get $\mathrm{Tr}(X)=(a_1^2+a_2^2-2)+\frac{(a_1^2-1)(a_2^2-1)}{(a_1a_2-1)^2+1}$.
$\qquad$ While For $A=\begin{bmatrix} 
a_1 & 0\\ 
 0&a_2
\end{bmatrix}$, we will get $\mathrm{Tr}(X)=(a_1^2+a_2^2-2)$.
The original problem is equivalent to the following problem:

Find $\mathrm{Tr}(Z)$, where $A'Z^{-1}A+Z=A'A+I$ and $Z-I>0.$

 A: Disclaimer : This solution is wrong, however I will leave it in case it can be useful to find a definitive solution.
Let $AA^T = U\Sigma U^T$ with $\Sigma$ diagonal. Let $Y=U (\Sigma-I) U^T$ (with eigenvalues larger than 0), then
\begin{align*}
A^T Y(I+Y)^{-1}A&=A^TU(\Sigma-I)\Sigma^{-1}U^TA\\
&=A^TUU^TA-A^TU\Sigma^{-1} U^TA\\
&=A^TIA-A^T(AA^T)^{-1}A\\
&=Y^T\\
&=Y
\end{align*}
So $Y$ is a solution to the given equation. This means that $\mathrm{Tr}(Y)=\mathrm{Tr}(AA^T)-n$.
A: $X=A^TX(I+X)^{-1}A=X=A^T(X+I-I)(I+X)^{-1}A=A^TA-A^T(I+X)^{-1}A$. 
$AXA^T = (AA^T)(AA^T)-AA^T (I+X)^{-1} AA^T$
$((AA^T))^{-1} AXA^T(AA^T)^{-1} = I-(I+X)^{-1}$ 
$ A^{-T} X A^{-1} = I-(I+X)^{-1}$ 
From this: if $X$ is a solution then $X^T$ is also a solution:
$ A^{-T} (X+X^T) A^{-1} = 2I-(I+X)^{-1}-(I+X^T)^{-1}$
Now  $Tr(A^{-T} (X+X^T) A^{-1})=Tr( (X+X^T) (A^TA)^{-1})$
Hence the problem is reduced to finding trace of product of 2 symmetric matrices.\
a) Proceeding further with assumptions with equality:

*

*Assume: $(A^TA)^{-1} = U \Sigma U^H$ and $X+X^T = U \Sigma' U^H$
$\sum_i \lambda_i((A^TA)^{-1}) 2\lambda_i(X) = 2n-2\sum_i \frac{1}{1+\lambda_i(X)}$.
Now solve for $\{\lambda_i(X)\}$ for non-trivial solutions.With this u mighe be able to get a tight bound on $Tr(X)$.


*You can use the equation (by assuming $A=U\Sigma U^H$ and $X=U\Sigma' U^H  $):
$ \frac{\lambda_i(X)}{\lambda_i(A)^2} = 1-\frac{1}{1+\lambda_i(X)}$ and solve for $\lambda_i(X)$. 
We get:
$ \frac{\lambda_i(X)}{\lambda_i(A)^2} = \frac{\lambda_i(X)}{1+\lambda_i(X)}$
Hence:
$ \lambda_i(A)^2-1 = \lambda_i(X)$
b) Now proceeding without assumptions to get an inequality: Let $||A||_F$ be frobenius norm of $A$.
$Tr(AB) \leq (||A||_F ||B||_F)^{1/2} = (\sum_i \lambda_i(A)^2 \sum_i \lambda_i(B)^2)^{1/2}$ 
$Tr( (X+X^T) (A^TA)^{-1}) \leq 2(\sum_i \lambda_i((A^TA)^{-1})^2 \sum_i \lambda_i(X)^2)^{1/2}$ 
Hence we get:
$\sqrt{\sum_i \lambda_i((A^TA)^{-1})^2 \sum_i \lambda_i(X)^2)} \geq n-\sum_i \frac{1}{1+\lambda_i(X)}$
