Past exam of the mathematics course of the year 2 of the natural sciences course of the University of Cambridge / 2025 / Paper 1 / 6 / d

Termwise differentiation of the absolutely convergent matrix exponential series givesBecause $P$ and $Q$ commute, $P$ also commutes with every power of $Q$ and hence with $e^{tQ}$. The product rule therefore givesDirect differentiation gives $f_2^{\prime }(t)=f_2(t)(P+Q)$, and both initial values are $I$. The uniqueness theorem for ordinary differential equations, applied to the finitely many matrix entries, implies $f_1=f_2$. One can see uniqueness directly: the difference $D=f_1-f_2$ obeys $D^{\prime }=D(P+Q)$ and $D(0)=0$, soand the product is identically zero. At $t=1$ this yields

Let $K(t)=e^{tR}S{e}^{-tR}$. The product rule and differentiation of the matrix exponential givewhere brackets are the operator commutator. Evaluating at zero gives $K(0)=S$, $K^{\prime }(0)=[R,S]$ and $K^{\prime \prime }(0)=[R,[R,S]]$. The Taylor series at zero is consequentlyIn particular $[R,[R,S]]=R^{2}S-2RSR+S{R}^2$, which also follows by multiplying the two matrix exponential series through quadratic order. For fixed finite matrices the remainder has the stated Big O notation in any matrix norm.