Exercises
[1N0]Prerequisites:[1T1].Difficulty:**.In the general case (even when we do not know if \(A,B\) commute), we can express \(\exp (A+sB)\) using a power series. Define
\[ C(t)=\exp (-tA)B \exp (tA) \]and (recursively) set \(Q_ 0={\mathbb {I}}\) (the identity matrix) and then
\[ Q_{n+1}(t)=β«_ 0^ t C(π) Q_ n(π)\, {\mathbb {d}}π \]then
\begin{equation} \exp (-A)\exp (A+sB)=β_{n=0}^β s^ n Q_ n(1)~ ~ ;\label{eq:exp_ A+sB} \end{equation}24this series converges for every \(s\).
In particular, the directional derivative of \(\exp \) at the point \(A\) in the direction \(B\) is
\[ \frac{d\hskip5.5pt}{d{s}} \exp (A+sB)|_{s=0}=\exp (A) Q_ 1(1) = β«_ 0^ 1 \exp ((1-π) A)B\exp (π A) \, {\mathbb {d}}π~ ~ . \]( Hint: Use the exercise [1T1] with \(Y(t,s) = \exp (-tA)\exp (t(A+sB))\) and then set \(t=1\). )
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