Solution
= Solution
For a real <skew-symmetric matrix>, $P^T=-P$. A scalar quadratic form satisfies $y^TPy=(y^TPy)^T=y^TP^Ty=-y^TPy$, so it is zero. Therefore
$$
\frac d{dt}\lVert y(t)\rVert^2=2y^Ty'=2y^TPy=0,\qquad\boxed{\lVert y(t)\rVert=\lVert y_0\rVert\quad(t\geq0).}
$$
The finite-dimensional linear system has the global solution $e^{tP}y_0$. Equivalently, $(e^{tP})^Te^{tP}=e^{-tP}e^{tP}=I$, so its exact flow is an <orthogonal matrix> and preserves the <Euclidean norm>.