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Given two symmetric positive definite matrices $L_1, L_2\in\mathbb{R}^{n\times n}$ and a vector $w\in\mathbb{R}^n$, under what conditions does the following system of equations of the form $M_1(t_1,t_2)M_2(t_1,t_2)x=M_3(t_2)M_4(t_2)v$ have solutions $t_1,t_2\in\mathbb{R}$ and $x\in\mathbb{R}^{2n}$?

\begin{equation} \begin{bmatrix} t_1\mathrm{sinc}(t_1L_1) & t_2\mathrm{sinc}(t_2L_2) \\ \cos(t_1L_1) & -\cos(t_2L_2) \end{bmatrix} \begin{bmatrix} -L_1 \sin(t_1L_1) & \cos(t_1L_1) \\ L_2\sin(t_2L_2) & \cos(t_2L_2) \end{bmatrix} x = \begin{bmatrix} \cos(t_2L_2) & -t_2\operatorname{sinc}(t_2L_2) \\ L_2\sin(t_2L_2) & \cos(t_2L_2)\end{bmatrix} \begin{bmatrix} 0 & -t_2\operatorname{sinc}(t_2L_2) \\ -L_2\sin(t_2L_2) & 0 \end{bmatrix} \begin{bmatrix} w \\ 0 \end{bmatrix} \end{equation}

This equation comes from: \begin{equation} (e^{2t_1A_1}-e^{-2t_2A_2})x=e^{-t_2A_2}(e^{t_2A_2}-e^{-t_2A_2})\begin{bmatrix} w \\ 0 \end{bmatrix} \end{equation}

with $$A_i = \begin{bmatrix} 0 & I_n \\ -L_i^2 & 0\end{bmatrix}$$

When I solve this system of equations numerically (with a Newton-Raphson procedure), all the solutions I find are such that $\det(M_1(t_1,t_2))=0$. I do not see why.

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  • $\begingroup$ What does it mean for a solution to 'correspond' to a singularity? If I understand correctly which is "the very first matrix" and which is "the RHS", then the fact that the RHS is in the image just comes from the fact that the RHS equals the LHS, which is explicitly presented as a vector in the image of the first matrix. $\endgroup$
    – LSpice
    Commented May 4, 2018 at 1:54
  • $\begingroup$ @LSpice I edited to clarify. For the second part, what I meant is that generically for a square singular matrix $A$ and a vector $b$, $Ax=b$ has no solution. $\endgroup$
    – anderstood
    Commented May 4, 2018 at 11:30

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