Matrix equation with diagonal matrix Let $A \in\mathbb{R}^{n\times n}$ be a given Positive-Semi Definite matrix ,$\theta\in\mathbb{R}^n$ a vector, $x\in\mathbb{R}$ an unknown variable, and $a\in\mathbb{R}^+$ a positive constant. I have the following equation
$$
\theta^TA(A+xI)^{-1}(A+xI)^{-1}A\theta  = a
$$
where $I$ is the identity matrix.
Is it possible to solve this equation for $x?$
What I have tried:
$$
\text{trace}(\theta^TA(A+xI)^{-1}(A+xI)^{-1}A\theta)  = \text{trace}(a)
\\\theta^TAA\theta  \cdot\text{trace}((A+xI)^{-1}(A+xI)^{-1})  = a
\\ \sum_{i = 1}^n\frac{1}{(x+s_i)^2}=\frac{a}{\|\theta\|_{A^2}^2}
$$
where $s_i$ are the singular values of the $A$. Is it correct? How can i proceed from here?
 A: Because $A$ is symetric psd matrix, we can diagonalize $A$ as: $A = UDU^T$ where $U$ is a orthogonal matrix ($U^T = U^{-1}$) and $D$ is a diagonal matrix.
Hence, we have
\begin{align}
(A+xI)^{-1} &= (UDU^T+xUU^T)^{-1} \\
&= (U(D+xI)U^{-1})^{-1} \\
&= U(D+xI)^{-1}U^{-1} \\
\end{align}
Then
\begin{align}
a & =\theta^TA(A+xI)^{-1}(A+xI)^{-1}A\theta  \\
& =\theta^TUDU^T \left(U(D+xI)^{-1}U^{-1}\right) \left(U(D+xI)^{-1}U^{-1}\right)UDU^T\theta  \\
& =\theta^TUD \left((D+xI)^{-1}\right)^2DU^T\theta  \\
\end{align}
The matrix $ \left((D+xI)^{-1}\right)^2$ is a diagonal matrix of $\frac{1}{(d_i+x)^2}$ with $(d_1,...,d_n)$ is the diagonal of the matrix $D$.
Let's denote $\eta =(\eta_1,...,\eta_n) = DU^T\theta \in \mathbb{R}^n$, we have
$$a = \sum_{i=1}^n\frac{\eta_i^2 }{(d_i+x)^2}$$
A: $\def\T#1{\operatorname{tr}(#1)}$I would suggest using Newton's Method, i.e.
$$\eqalign{
f(x) &= 0, \quad
f_k &= f(x_k), \quad
f'_k &= f'(x_k) \quad\implies\quad
x_{k+1} &= x_k - \frac{f_k}{f'_k} \\
}$$
For typing convenience, define the matrices
$$\eqalign{
B &= A+Ix \quad\implies\quad
\frac{dB^n}{dx} = nB^{n-1} \\
M &= A\,\theta\theta^T\!A \\
}$$
Set up an appropriate function and apply Newton's Method
$$\eqalign{
f  &= a - \T{MB^{-2}} \\
f' &= +\T{2MB^{-3}} \\
B_k &\doteq A + Ix_k \\
x_0 &= 1, \qquad
x_{k+1} 
  = x_k - \frac{a-\T{MB_k^{-2}}}{\T{2MB_k^{-3}}} \\
}$$
NB:  The iteration will converge to different roots depending upon the choice of $x_0$
Also note that if $A$ is diagonal, then $B$ is also diagonal and the matrix inversion required by the iteration can be computed very efficiently.
