 Solution

ID: 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/i/solution

Termwise differentiation of the absolutely convergent matrix exponential series gives

\frac{d}{d t} e^{tP} = e^{tP} P = P e^{tP} .

Because

P

and

Q

P

also commutes with every power of

Q

and hence with

e^{tQ}

. The product rule therefore gives

f_{1}^{'} (t) = e^{tP} P e^{tQ} + e^{tP} e^{tQ} Q = f_{1} (t) (P + Q) .

f_{2}^{'} (t) = f_{2} (t) (P + Q)

, and both initial values are

I

f_{1} = f_{2}

. One can see uniqueness directly: the difference

D = f_{1} - f_{2}

obeys

D^{'} = D (P + Q)

and

D (0) = 0

, so

\frac{d}{d t} (D e^{- t (P + Q)}) = 0

and the product is identically zero. At

t = 1

this yields

e^{P} e^{Q} = e^{P + Q} when PQ = QP .

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