 Solution

The matrix exponential is the absolutely convergent power seriesIt converges in any finite-dimensional operator norm, since $\Vert L^m\Vert \le \Vert L{\Vert }^m$. Termwise differentiation gives $B^{\prime }=LB$, $B(0)=I$, so the solution is $q(t)=B(t)q_0$. Also $e^{Lt}{e}^{-Lt}=I$, proving invertibility without requiring $L$ itself to be invertible.

The question's expansion in eigenvectors tacitly needs an eigenbasis: an invertible matrix alone need not be a diagonalizable matrix. The exponential solution and the singular-value arguments remain valid without that assumption.