Page 136 - Matrices theory and applications

This calculation also shows that if M is Hermitian, then
1

exp M =exp M.
2
We shall use the following more precise statement:
Proposition 7.2.3 The map exp : H n → HPD n is a homeomorphism
(that is, a bicontinuous bijection). 7.2. Exponential of a Matrix 119
Proof
Injectivity: Let A, B ∈ H n with exp A =exp B =: H.Then
1 √ 1
exp A = H =exp B.
2 2
By induction, we have
exp 2 −m A =exp 2 −m B, m ∈ ZZ.
m
Substracting I n , multiplying by 2 , and passing to the limit as m →
+∞,we obtain

d d
exp tA = exp tB;

dt dt
t=0 t=0
that is, A = B.
Surjectivity: Let H ∈ HPD n be given. Then H = U diag(d 1 ,... ,d n )U,
∗
where U is unitary and d j ∈ (0, +∞). From above, we know that
H =exp M for
∗
M := U diag(log d 1 ,... , log d n )U,
which is Hermitian.
Continuity: The continuity of exp has already been proved. Let us in-
l
vestigate the continuity of the reciprocal map. Let (H ) l∈IN be a
sequence in HPD n that converges to H ∈ HPD n .We denoteby
l
l
M ,M ∈ H n , the Hermitian matrices whose exponentials are H and
H. The continuity of the spectral radius gives
l
l −1

lim ρ(H )= ρ(H), lim ρ (H ) = ρ (H) −1 . (7.5)
l→+∞ l→+∞
l
l
Since Sp(M ) = log Sp(M ), we have
l −1
l
l

ρ(M )= log max ρ(H ),ρ (H ) . (7.6)
Keeping in mind that the restriction to H n of the induced norm · 2
coincides with that of the spectral radius ρ, we deduce from (7.5, 7.6)
l
that the sequence (M ) l∈IN is bounded. If N is a cluster point of the
sequence, the continuity of the exponential implies exp N = H.But
l
the injectivity shown above implies N = M. The sequence (M ) l∈IN ,
bounded with a unique cluster point, is convergent.