Matrix equation involving quadratic form Let $X,Y\in\mathbb{R}^{n\times k}$, $\Lambda(\alpha) = \text{diag}(\alpha)$, with $\alpha\in\mathbb{R}^k$, and let $c,d\in\mathbb{R}^+$ be positive constants. Let
$$A_i(\alpha) = (X\Lambda(\alpha)  X^T)^{-1}x_ix_i^T(X\Lambda(\alpha) X^T)^{-1},$$ $$B_i(\alpha) = (Y\Lambda(\alpha)  Y^T)^{-1}y_iy_i^T(Y\Lambda(\alpha) Y^T)^{-1},$$ where $x_i, y_i\in\mathbb{R}^n$ are the $i$-th column of matrix $X$ and $Y$ respectively.
Is there any efficient way to solve the following system of equations for $\alpha$?
$$
c_i(x_i^TA_i(\alpha)x_i) = d_i(y_i^TB_i(\alpha)y_i), \quad i = 1,\dots k.
$$
 A: Let me try first the simple case $n=k$ of square symmetric matrices. Define the vector $v^{(i)}$ with elements $v^{(i)}_{n}=\delta_{ni}$ and denote $X'=X^T$. Then
$$A^{(i)}=(X'^{-1}\Lambda^{-1}X^{-1}Xv^{(i)})(X'^{-1}\Lambda^{-1}X^{-1}Xv^{(i)})^T=(X'^{-1}\Lambda^{-1}v^{(i)})(X'^{-1}\Lambda^{-1}v^{(i)})^T$$
$$\qquad=\frac{1}{\alpha_i^2}(X'^{-1}v^{(i)})(X'^{-1}v^{(i)})^T\Rightarrow x_i^TA^{(i)}x_i=\frac{1}{\alpha_i^2}.$$
Similarly, $x_i^TB^{(i)}x_i=\frac{1}{\alpha_i^2}$, and so the equations
$$c_i(x_i^TA_i(\alpha)x_i) = d_i(y_i^TB_i(\alpha)y_i), \quad i = 1,\dots k$$
have no solution unless $c_i=d_i$ for all $i$.

At the other extreme, let me try $n=1$ and arbitrary $k$. Then $X_{1i}=x_i$ and
$$A^{(i)}=\left(\sum_{j=1}^k x_j^2\alpha_j\right)^{-2} x_i^2,$$
$$B^{(i)}=\left(\sum_{j=1}^k y_j^2\alpha_j\right)^{-2} y_i^2,$$
so the equation to solve is
$$c_i\left(\sum_{j=1}^k x_j^2\alpha_j\right)^{-2} x_i^4=d_i\left(\sum_{j=1}^k y_j^2\alpha_j\right)^{-2} y_i^4.$$
There is no solution for $k>1$, unless $(c_i/d_i)(x_i/y_i)^4$ is independent of $i$.
