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If $F(x) = \exp(x A) = \sum_{i = 0}^\infty \frac{1}{i!} x^i A^i$ where $F(x), A \in \mathbb{R}^{n \times n}$, then \begin{equation} F(x + y) = F(x) F(y) = F(y) F(x) \end{equation} holds.

When is the converse true, i.e. if $F(x + y) = F(x) F(y) = F(y) F(x)$, then there exists a matrix $A$ such that $F(x) = \exp(x A)$?

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    $\begingroup$ I think you need (at least) the additional hypothesis that $F$ is continuous. $\endgroup$ – Omnomnomnom Apr 27 '15 at 16:29
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    $\begingroup$ What about $F(x) = 0$? $\endgroup$ – AlexR Apr 27 '15 at 16:30
  • $\begingroup$ what if $F(x)=a^{(xA)}$ where $a$ is not equal to $e$ ? $\endgroup$ – sashas Apr 27 '15 at 16:30
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    $\begingroup$ @sasha $a^{xA} = e^{x[\ln(a) A]}$ $\endgroup$ – Omnomnomnom Apr 27 '15 at 16:32
  • $\begingroup$ If $P $ is an idempotent, you can take $F (x)=P $. $\endgroup$ – Martin Argerami Apr 27 '15 at 16:34
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"When is the converse true?"

One suitable hypothesis on $F(x)$, other than

$F(x + y) = F(x)F(y) = F(y)F(x), \tag{1}$

is that $F(x)$ be differentiable at $x = 0$, that is, that

$F'(0) = \lim_{h \to 0} \dfrac{1}{h} (F(h) - F(0)) \tag{2}$

exist as a well-defined matrix; another is that

$F(0) \;\; \text{is invertible}. \tag{3}$

Indeed, both (2) and (3) are in fact necessary, as we shall see. In the event that (1) and (2) apply, we have for any $x$ and $h \ne 0$

$\dfrac{1}{h}(F(x + h) - F(x)) = \dfrac{1}{h}(F(x)F(h) - F(x)) = \dfrac{1}{h}(F(h)F(x) - F(x)); \tag{4}$

we further note that

$F(x) = F(x + 0) = F(0 + x) = F(0)F(x) \tag{5}$

via (1); thus (4) becomes

$\dfrac{1}{h}(F(x + h) - F(x)) = \dfrac{1}{h}(F(h)F(x) - F(0)F(x)) = \dfrac{1}{h}(F(h) - F(0))F(x); \tag{6}$

now taking the limit as $h \to 0$ yields, using (2),

$F'(x) = \lim_{h \to 0} \dfrac{1}{h}(F(x + h) - F(x)) = \lim_{h \to 0} \dfrac{1}{h} (F(h) - F(0))F(x) = F'(0) F(x); \tag{7}$

we see that $F(x)$ must satisfy the ordinary differential equation

$F'(x) = F'(0) F(x); \tag{8}$

the unique solution to (8) is

$F(x) = F(0) e^{F'(0) x}, \tag{9}$

which may be easily checked by direct differentiation:

$F'(x) = (F(0) e^{F'(0) x})' = F(0)(e^{F'(0) x})' = F(0) F'(0) e^{F'0) x} = F'(0) (F(0)e^{F'(0) x}) = F'(0) F(x); \tag{10}$

in performing the validating computation (10), we have (tacitly) used the fact that

$F'(0) F(0) = F(0) F'(0)\tag{11}$

which follows from (1), (2):

$F(0) F'(0) = F(0) \lim_{h \to 0}\dfrac{1}{h}(F(h) -F(0)) = \lim_{h \to 0} F(0)\dfrac{1}{h}(F(h) - F(0)) = \lim_{h \to 0}\dfrac{1}{h}(F(h) - F(0)) F(0) = F'(0)F(0). \tag{12}$

In (9), we are almost there, needing only to deal with the factor $F(0)$; we have

$(F(0))^2 = F(0) F(0) = F(0 + 0) = F(0); \tag{13}$

here (3) at last comes into play; we further have

$F(0) = F^{-1}(0)(F(0))^2 = F^{-1}(0) F(0) = I, \tag{14}$

whence (9) becomes

$F(x) = e^{F'(0) x}, \tag{15}$

the desired form with $A = F'(0)$.

We have thus seen that conditions (1)-(3) are sufficient for (15) to bind; they are in fact also necessary; (3) follows from (15) by setting $x = 0$:

$F(0) = e^{F'(0) 0} = e^0 = I, \;\; \text{invertible}; \tag{16}$

(2) is such an easy consequence of the (given) series expansion of $F(x) = e^{Ax} = e^{F'(0)x}$ that I leave the details to my readers; finally (1), which given (15) may be written

$e^{A(x + y)} = e^{Ax} e^{Ay} \tag{17}$

is a well-known property of exponential matrices; see my answer to $M,N\in \Bbb R ^{n\times n}$, show that $e^{(M+N)} = e^{M}e^N$ given $MN=NM$.

This question and answer are related to Show solution to character identity.

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