5.1 The matrix exponential
The series converges absolutely for every \(M\), since \(\|M^k\| \leq \|M\|^k\) for an operator norm, so \(\|e^M\| \leq e^{\|M\|}\); for the same reason \(\sum _k(At)^k/k!\) converges uniformly for \(t\) in bounded intervals and may be differentiated term by term.
Theorem 5.1.2. For square matrices \(A\), \(B\) and an invertible \(P\):
- (i)
- \(\frac {d}{dt}e^{At} = Ae^{At} = e^{At}A\), so \(e^{At}\) is the fundamental matrix of \(x' = Ax\) with \(e^{A\cdot 0} = I\), and the solution with \(x(t_0) = x_0\) is \(e^{A(t - t_0)}x_0\);
- (ii)
- \(e^{A(t+s)} = e^{At}e^{As}\), and \((e^{At})^{-1} = e^{-At}\);
- (iii)
- if \(AB = BA\), then \(e^{A+B} = e^Ae^B\);
- (iv)
- \(e^{PBP^{-1}} = Pe^BP^{-1}\);
- (v)
- \(\det e^{At} = e^{t\tr A}\).
Proof. (i) Term-by-term differentiation gives \(\sum _{k\geq 1}kA^kt^{k-1}/k! = A\sum _{j\geq 0}(At)^j/j!\), and \(A\) commutes with every term. (ii) Both sides solve \(X' = AX\) as functions of \(t\) with value \(e^{As}\) at \(t = 0\), so they agree by uniqueness; with \(s = -t\) this gives the inverse. (iii) If \(AB = BA\), then \(X(t) = e^{At}e^{Bt}\) satisfies \(X' = Ae^{At}e^{Bt} + e^{At}Be^{Bt} = (A + B)X\), since \(B\) commutes with \(e^{At}\), and \(X(0) = I\); so \(X(t) = e^{(A+B)t}\), and \(t = 1\) gives the claim. (iv) \((PBP^{-1})^k = PB^kP^{-1}\) for every \(k\). (v) Liouville’s formula, Theorem 4.2.1, with \(\Phi = e^{At}\) and \(\det \Phi (0) = 1\). □
Property (iii) fails without commutativity (Problem 5.7.3), which is the constant-coefficient shadow of Proposition 4.5.1.
Questions on this section
Stuck on something here? Ask below and it stays attached to this topic.