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Let $A:\mathbb R_+\to M_n(\mathbb R)$ be a homomorphism of the additive semi-group of positive reals into the multiplicative semi-group of real matrices. Matrix exponentials $t\to e^{tB}$ are one class of such homomorphisms. If $A$ is differentiable at $0$, then it must be the matrix exponential $e^{tA'(0)}$, since:

$$\frac{A(t+h)-A(t)}h=A(t)\frac{A(h)-I}h\xrightarrow[h\to0]{} A(t)A'(0)$$

Can this sufficient condition be relaxed further to just continuity, maybe even just continuity at zero? This answer from 2010 seems to imply that that's the case, and the theorem is certainly true when $n=1$ because of the uniqueness of positive real roots.

Note: I personally cannot understand any of the algebra in the question linked to in the last paragraph. More elementary answers would be appreciated.

Ben Grossmann
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Jack M
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1 Answers1

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Since $A$ is continuous and $A(0) = I$, $A(t)$ remains in a small disk centered at $I$ while $t$ is small. Let $L$ be a branch of log on this disk. Then $LA$ is a continuous (germ of an) additive homomorphism. It is easier to see (as in the $1$-dimensional case) that continuity applies differentiability (in fact linearity) for additive homomorphisms. $LA(t)$ must be $tB$ for some matrix $B$, and exponentiating this gives $A(t)$, both only for small $t$, but $A$ is determined by its restriction to any interval.

Bruce Evans
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  • So the only thing from the theory of matrix logarithms we need is that in some neighborhood of $I$, there is a continuous function $L$ which is (extends to) a homomorphism from the multiplicative group of matrices into the additive group of matrices? – Jack M Nov 19 '17 at 20:01
  • $L$ can be taken as the principal branch of $log$ (given by a power series), and I suppose its homomorphism property and relationship with $exp$ can be derived using brute force power series calculations as is often done for the homomorphism property of $exp$ (but it is easier to use calculus).

    Logarithms and the reduction to an additive homomorphism can be avoided using iterated square roots as in ad hoc definitions of power functions in the 1-dimensional case (define power functions on dyadic rationals and extend by continuity), but this seems to require more steps.

     – Bruce Evans Nov 20 '17 at 19:10