Let $d = 12$ and $m = 6$, and denote by $0_n$ and $I_n$ the zero matrix and the identity matrix of size $n \times n$.

Let $D_+ \in \mathbb{R}^{m \times m}$ be a diagonal matrix with positive diagonal entries and $D \in \mathbb{R}^{d \times d}$ be defined by \begin{align*} D = \begin{bmatrix} -D_+ & 0_m\\ 0_m & D_+ \end{bmatrix} \end{align*} Let $B = B(x) \in \mathbb{R}^{d \times d}$ have the form \begin{align*} B = \frac{1}{2} (D \bar{P}+ \bar{J}^{-1}\bar{P}^\intercal\bar{J}D), \qquad \text{with }\bar{P} = \begin{bmatrix} P & P \\ P & P \end{bmatrix}, \quad \bar{J} = \begin{bmatrix} \hat{J} & 0_m \\ 0_m & \hat{J} \end{bmatrix}, \end{align*} where $P,J \in \mathbb{R}^{m \times m}$ are defined by \begin{align*} P = \begin{bmatrix} \Upsilon & \Gamma \\ 0_3 & \Upsilon \end{bmatrix}, \quad \hat{J} = \begin{bmatrix} I_3 & 0_3 \\ 0_3 & J \end{bmatrix} \end{align*} for $\Upsilon, \Gamma \in \mathbb{R}^{3 \times 3}$ skew-symmetric matrices, $J \in \mathbb{R}^{3 \times 3}$ a diagonal matrix with positive diagonal entries.

My question is:

Is it possible to find a diagonal matrix $Q = Q(x) \in \mathbb{R}^{d \times d}$ (for $x \in [0,L]$), such that the matrix $M = M(x)\in \mathbb{R}^{d \times d}$ \begin{align*} M(x) = Q'(x)D + Q(x)B + B^\intercal Q(x) \end{align*} is negative definite for any $x \in [0,L]$ ?

(Here $B^\intercal$ denotes the transpose of $B$.)

Remark: the diagonal of $B$ contains only zeros (since $\Upsilon$ is skew-symmetric), hence the diagonal of $QB+B^\intercal Q$ contains only zeros.

Any suggestion or reference is welcome. Thank you.


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