# 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 $$\alpha$$?

$$c_i(x_i^TA_i(\alpha)x_i) = d_i(y_i^TB_i(\alpha)y_i), \quad i = 1,\dots k.$$

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 $$i$$.
• Thanks a lot for your answer @Carlo. What if $k> n$? – Apprentice Mar 28 at 13:53
• and you mean $n = k$ in your answer, right? – Apprentice Mar 28 at 16:32
• I tried $n=1$, $k>1$, also no solution in general. – Carlo Beenakker Mar 28 at 20:34