1' The map X↦eX is a continuous map of L(E) into the group of automorphisms of E (invertible elements of* L(E)).
2 The map t↦eXt of R into L(E) is differentiable, and
dtd(eXt)=XeXt=eXtX. (24)
3 For any X∈L(E) one has
eX=n=0∑∞n!Xn (25)
the right-hand side being absolutely and uniformly convergent on every bounded subset of L(E); in particular, eIt=etI for t∈R.
4) If X and Y commute then Y and eY commute with eX, and
eX+Y=eXeY. (26)
The relation (24) follows from the expression (23) for C(t,0) and from the fact that this function is an integral of (20); by recursion on n one deduces from (24) that t↦eXt is indefinitely differentiable on R and that
Dn(eXt)=XneXt.
By Taylor’s formula one thus can write
eX=I+1!X+2!X2+⋯+n!Xn+Xn+1∫01n!(1−t)neXtdt. (27)
On the other hand, cor. 3 of IV, p. 181 shows that ∥eXt∥≤exp(∥X∥∣t∣). Thus the remainder rn(X)=Xn+1∫01n!(1−t)neXtdt in formula (27) satisfies the inequality
∥rn(X)∥≤(n+1)!∥X∥n+1e∥X∥
whence one deduces the formula (25), the series on the right-hand side being absolutely and uniformly convergent on every bounded subset of L(E). For every pair of elements X, T of L(E) one thus has
eX+T−eX=n=1∑∞n!1((X+T)n−Xn).
Now, one can write (X+T)n−Xn=∑(Vi)V1V2…Vn, where the sum is taken over the 2n−1 sequences (Vi) of elements of L(E) such that Vi=X or Vi=T for 1≤i≤n, and at least one of the Vi is equal to T; the inequality
∥(X+T)n−Xn∥≤(∥X∥+∥T∥)n−∥X∥n follows immediately, whence
∥exp(X+T)−expX∥≤exp(∥X∥+∥T∥)−exp∥X∥ which establishes the continuity of the map
X↦expX.
Finally, if X and Y commute, then Y commutes with eXt (IV, p. 181, prop. 2), so
dtd(eXteYt)=XeXteYt+eXt(YeYt)=(X+Y)eXteYt. Since, on the other hand,
eXteYt is equal to
I for
t=0, one has
eXteYt=e(X+Y)t, whence formula (26). From this latter, one deduces in particular that for arbitrary real
s and
t one has
eX(s+t)=eXseXt and also that
e−X=(eX)−1.
On the contrary one notes that (26) need not remain valid if X and Y are no longer assumed to commute: indeed it would imply that expX and expY always commute, which is not the case, as is shown by simple examples (IV, p. 204, exerc. 3).
Now let us assume that E is a vector space of finite dimension over the field C, and A an endomorphism of E (for its vector space structure over C) which one can identify with its matrix with respect to a basis of E; then, for all t∈R, eAt is an automorphism of E for the same structure (IV, p. 181, prop. 2). Let rk (1≤k≤q) be the distinct roots (in C) of the characteristic polynomial φ(r)=det(A−rI) of the endomorphism A (the "characteristic roots" of A); if nk is the order of multiplicity of rk then ∑k=1qnk=n. One knows that (Alg., VII, 31, n° 3) to each root rk there corresponds a subspace Ek of E, of dimension nk, such that Ek is stable under A, and that E is the direct sum of the Ek: Ek can be defined as the subspace of vectors x such that
(A−rkI)nk.x=0. Let
a be any vector in
E; one can write
a=∑k=1qak, where
ak∈Ek; the integral of the equation (19) of IV, p. 188, taking the value
a at the point
t=0 is thus given by
u(t)=eAt.a=k=1∑qeAt.ak=k=1∑qerkte(A−rkI)t.ak.
But since ak∈Ek one has
e(A−rkI)t⋅ak=ak+1!t(A−rkI)⋅ak+2!t2(A−rkI)2⋅ak+⋯
+(nk−1)!tnk−1(A−rkI)nk−1⋅ak.
Thus every integral of the equation (19) of IV, p. 188, can be written as
u(t)=k=1∑qerktpk(t)
where pk(t) is a polynomial in t, with coefficients in the vector space Ek, and of degree ≤nk−1. In particular, if all the roots of the characteristic equation of A are simple, the spaces Ek (1≤k≤n) are all of dimension 1 over the field C, so there exist n vectors ck such that the n functions erktck (1≤k≤n) form a fundamental system of integrals of the equation (19) of IV, p. 188.
The characteristic roots of the endomorphism A are also called characteristic roots of the linear equation (19) of IV, p. 188. One may observe that one obtains the characteristic equation of A by writing that the function cert is an integral of (19) for a vector c=0.
When one has determined the roots rk (1≤k≤q) explicitly, and thus the order of multiplicity nk of rk, in practice one obtains the integrals of (19) by writing that this equation is satisfied by the expression (32) of IV, p. 191, where pk is an arbitrary polynomial of degree ≤nk−1, with coefficients in E; on identifying the coefficients of erkt (for 1≤k≤q) in the two sides of the equation so obtained one obtains linear equations for the coefficients of the polynomials pk: one establishes easily that these equations determine the terms of degree >0 of pk as functions of the constant term, and that the latter is a solution of the equation (A−rkI)nk⋅x=0, which defines the subspace Ek (method of "undetermined coefficients").
Bourbaki puts the proof directly after the statement, so what is above is the statement and its proof together.
Nothing.