# On the solution of matrix equation

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.

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 (http://books.google.com/books/about/Linear_system_theory.html?id=ffNQAAAAMAAJ).

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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