= Solution
The <matrix exponential> is the absolutely convergent power series
$$
\boxed{B(t)=e^{Lt}=\sum_{m=0}^\infty\frac{t^mL^m}{m!}.}
$$
It converges in any finite-dimensional <operator norm>, since $\|L^m\|\le\|L\|^m$. Termwise differentiation gives $B'=LB$, $B(0)=I$, 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.
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