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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?

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    $\begingroup$ Your description and equation doesn't make sense. $A\in\mathbb{R}^n$ is not a psd matrix unless $n$ is a square number, but you are also claiming $\theta\in\mathbb{R}^n$ to be a vector that is compatible with $A$ with both (!) $\theta^TA$ and $\theta A$ appearing, so that doesn't happen unless $n=0$ (trivial) or $n=1$, in which case this is just a usual quadratic equation in $x$. $\endgroup$ Feb 10 at 8:57
  • $\begingroup$ Sorry there is a typo: $A\in\mathbb{R}^{n\times n}$. Updated $\endgroup$
    – Apprentice
    Feb 10 at 8:59
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    $\begingroup$ Assuming WLOG $A$ is diagonal. Then the equation is $\sum_i\theta_i^2\frac{a_i^2}{(a_i+x)^2}=a$ which you can solve numerically (I don't think the $2n$-degree polynomial is solvable by radicals for generic $a_i,\theta_i$) $\endgroup$ Feb 10 at 9:33
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    $\begingroup$ No, you can't pull $A\theta$ out, same as you can't pull $y$ out in $y^TBy\neq y^Ty\operatorname{tr}B$ $\endgroup$ Feb 10 at 9:47
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    $\begingroup$ Let $A=\operatorname{diag}(a_i)$, so $A+xI=\operatorname{diag}(a_i+x)$ and $A(A+xI)^{-1}(A+xI)^{-1}A=\operatorname{diag}((\frac{a_i}{a_i+x})^2)$ giving $\theta^TA(A+xI)^{-1}(A+xI)^{-1}A\theta=\sum_i\theta_i^2(\frac{a_i}{a_i+x})^2$. $\endgroup$ Feb 10 at 10:06
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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

$$a = \sum_{i=1}^n\frac{\eta_i^2 }{(d_i+x)^2}$$

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  • $\begingroup$ Hey, thanks a lot for your answer. Do you have any idea on how to solve the rational equation in the last line? Or at least if $a>0$ can we say something about the positiveness of the solutions? $\endgroup$
    – Apprentice
    Feb 10 at 11:57
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    $\begingroup$ @Apprentice there is no closed form expression for this equation. And you have from $2n$ real roots. The sole method you can use is numerical method. $\endgroup$
    – NN2
    Feb 10 at 12:01
  • $\begingroup$ what about the positivity of at least one of the solution for $a>0$? $\endgroup$
    – Apprentice
    Feb 10 at 12:59
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$\def\T#1{\operatorname{tr}(#1)}$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.

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