Fundamental matrix of a linear differential equation ($X^{\prime }=A(t)X,\det X\ne 0$)

For a system of linear differential equations $u^{\prime }=A(t)u$, a fundamental matrix has columns forming a basis of its solution space. It solves $X^{\prime }=A(t)X$ and is invertible at every time in its interval. Every solution is $u(t)=X(t)c$ for a constant vector $c$: differentiating $X^{-1}u$ gives zero. Invertibility at one time implies invertibility throughout the interval by uniqueness of the initial value problem. This notion is distinct from a Markov-chain fundamental matrix or the projective-geometric matrix used in stereo vision.

Writing $Y=GZ$ with an invertible matrix-valued function gives $Z^{\prime }=\Lambda Z$ exactly when $\Lambda =G^{-1}AG-G^{-1}{G}^{\prime }$. For a formal power series transformation at an irregular singular point, coefficient comparison recursively removes off-diagonal terms when leading diagonal eigenvalues are distinct. This is a differential-system change of frame, not specifically an electromagnetic gauge transformation.

For $D=\mathrm{diag}(1,-1)$, $A=D{\lambda }^2+\left(\begin{array}{cc}0& U\\ V& 0\end{array}\right)\lambda +\left(\begin{array}{cc}d& P\\ R& -d\end{array}\right)$, the formal diagonal exponent has $h=d+UV/2$ and $\theta =(UR+VP)/2$. Matching the gauge equation gives the off-diagonal first coefficient $(-U/2,V/2)$ and the diagonal one $\rho D$, where $\rho =UV(d+h)/4-PR/2$. The second coefficient has diagonal entries ${\rho }^2/2-UV/8\pm h\theta /2$ and off-diagonal entries $(U\rho -P)/2,(V\rho +R)/2$. With $G_0=I$ and negative-index coefficients zero, the coefficient comparison isIts diagonal equations at $m=2,3,4,5$ give $h,\theta ,\rho$ and the second diagonal coefficient respectively. Its off-diagonal equations determine the next coefficients because the commutator with $D$ multiplies the upper and lower entries by $2$ and $-2$. This proves the expansion and supplies the recursion for all higher orders. The generally nonzero $\theta$ supplies formal monodromy.

If $\dot{\Phi }=A(t)\Phi$ and $\Phi (0)=I$, differentiation of the determinant gives $\det \Phi (t)=\exp ({\int }_0^t\mathrm{tr}A(s)ds)$. For a planar periodic orbit this gives the nontrivial Floquet multiplier, since the tangent multiplier is one.