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Suppose $N(t)$ be a family of bounded $n$-by$n$ matrices with $t>0$ such that

$N(s+h)=N(s)N(h)$ holds for all $s,h>0$

What kind of structure of the solution could enjoy?

Can one always have that $N(t)=e^{At}$ holds for some $A$?

This assumption seems too strong, for instance, $N(s)=0$ is some solution.

So what structure does the whole solutions enjoy?

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Where do your $s, t$ and $h$ range? Where do the coefficients live? Presumably this will have an impact on what the answer should be. In particular because I don't really see why $M$ would have to be non-singular unless your $t$ is allowed to take negative values. – Thierry Zell Aug 10 '11 at 12:15
Non-singular isn't enough; how do you define $M^t$ if $M$ isn't diagonalizable? If its eigenvalues are not all positive reals? – Qiaochu Yuan Aug 10 '11 at 14:37
Also, what do you mean by a "bounded" matrix? – Tom Leinster Aug 10 '11 at 15:42

Maybe you are already aware of this, but in response to the setup that you have, the following comes to mind: One parameter semigroups

The linked PDF discusses evolution equations of the form $f(s+t)=f(s)f(t)$ in great detail, and might prove helpful.

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As I seen, it assumed that $f(x)$ is continuous and $f(0)=I$ , then the result is valid. But does it still work for only bounded condition? – gondolf Aug 11 '11 at 5:08
I think you can adapt the proof for the 1-dimensional case. Sketch: (1) fix any time $T$, e.g. $T=1$, and call $N(1)=e^A$ (2) prove that $N(tT)=e^{tA}$ for $t\in\mathbb{N}_+$ and then for each $t\in\mathbb{Q}_+$ (3) if the partial solutions for two incommensurable $T_1$ and $T_2$ do not agree, use irrationality to find rational $a$, $b$ such that $t=aT_1-bT_2$ is small and $\left\Vert N(t) \right\Vert>M$ is large (4) depending on what you mean by "bounded", play with this $t$ and $M$ to find a contradiction. – Federico Poloni Aug 11 '11 at 7:32
This seems not true, because $0$ is also a solution – gondolf Aug 18 '11 at 4:44
In the scalar case the solution $0$ is usually excluded, but I think you can easily add it back in if you wish: $0$ must be a simple eigenvalue (as $N(t)=N(t/2)^2$), thus simply change $e^A$ to $\begin{bmatrix}e^A & 0\\\\ 0 & 0\end{bmatrix}$. By the way, since you're checking this discussion, do you mind explaining us what "bounded" means? – Federico Poloni Aug 18 '11 at 7:43

As Thierry and Qiaochu have pointed out, it's not clear what exactly the question means. Even so, maybe it's possible to say something helpful.

The question asks

Can one always have that $N(t) = M^t$ holds for some non-singular $M$?

This doesn't quite make sense, so I'll interpret it as

Must there exist a non-singular matrix $M$ such that $N(t) = M^t$ for all $t$?

I'm guessing this was the intention.

The answer is probably no. (I say "probably" because the hypotheses on $N$ weren't stated precisely.) It's even false for $n = 1$. That is, there exist functions $N: \mathbb{R} \to (0, \infty)$ satisfying $$ N(s + t) = N(s) N(t) $$ for all $s, t \in \mathbb{R}$, but not of the form $N(t) = M^t$ for any $M > 0$. The existence of such functions requires the axiom of choice.

This is a problem going back to Cauchy. It's equivalent to consider the equation $f(s + t) = f(s) + f(t)$ (by putting $f(s) = \log N(s)$), and this is what's usually called "Cauchy's functional equation". You can find information about it here and here, for example.

Maybe you're happy to assume that $N(t)$ is continuous in $t$, in which case there's a genuine question about matrices to be answered. But in the absence of precise hypotheses, I don't know what you're happy to assume.

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True, but with any reasonable interpretation of the word "bounded" in the text, these solutions may be excluded: the graph of all non-linear Cauchy solutions is dense in $\mathbb{R}^2$. – Federico Poloni 0 secs ago – Federico Poloni Aug 11 '11 at 7:28
What do you reckon the OP means by bounded? The phrase s/he uses is "family of bounded n by n matrices". Any n by n matrix, as a linear operator, is bounded. So perhaps s/he actually means "bounded family of n by n matrices", i.e. that there's a uniform bound on the entries of the matrices N(t). (In that case, not everything of the form $M^t$ is a solution, e.g. if n = 1 and the single entry of M is greater than 1.) I suppose that's the most plausible interpretation, but I'd prefer not to have to guess. – Tom Leinster Aug 11 '11 at 20:46
Yes, I agree that uniformly bounded seems the most reasonable interpretation. The other one that comes to my mind is essentially "continuous at 0 as a function of $t$", but his comment rules it out. – Federico Poloni Aug 14 '11 at 8:59

Let $N$ be the solution to a linear differential equation $\dot{N}(t) = A(t) N(t)$, with appropriate initial conditions. Then $N$ has the property you desire, but it cannot be represented as an exponential unless $A$ is independent of $t$. You can find this and more in books on linear system theory such as Jack Rugh's (

I am not sure but I believe it is true that the converse also holds - under suitable technical conditions, if $N$ has the one-parameter semigroup property then it can be represented as the solution to a linear differential equation.

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I shall assume $N(s)$ depends continuously on $s$; otherwise there are many discontinuous solutions even in the one-dimensional case. After a change of basis, we may assume that $$N(1)=\pmatrix{M&0\cr 0&Q},$$ where $M$ is nonsingular and $Q$ is nilpotent. Since all the matrices commute, we have the same block structure for all of them: $$N(s)=\pmatrix{M(s)&0\cr 0&Q(s)}.$$

Since $Q(1)$ is nilpotent, and $Q(1)=Q(1/n)^n$, it is easy to see that $Q$ is nilpotent for every rational. This is possible only if $Q(s)$ is identically zero. Moreover, $M$ is always nonsingular: if $s<1$, then $M(1)=M(s)M(1-s)$, so $M(s)$ is nonsingular for $s<1$, and for $s>1$, we can use that $M(s)=M(s/n)^n$. Now continuity implies that $M(\epsilon)=M(1)^{-1}M(1+\epsilon)$ approaches the identity as $\epsilon\to 0$, and it follows that $M(s)$ must be of the form $\exp(As)$.

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