Which matrices can be embedded in flows? I've been thinking about matrices, recently, and wondering about a fairly simple but maybe hard-to-answer question. A real $n\times n$ matrix $A$ yields a discrete dynamical system on $\mathbb{R}^n$, where from time $n$ to time $n + 1$ the vector $\mathbf{x}$ jumps to the vector $A\mathbf{x}$. Under what conditions on $A$ can we "interpolate" this to a continuous-time dynamical system, i.e., find some vector field $\mathbf{X}$ such that the time-one flow of $\mathbf{X}$ takes any vector $\mathbf{x}$ to $A\mathbf{x}$?
Here is what I think I know (but please correct me if I'm wrong). A smooth homomorphism $\phi\colon \mathbb{R} \to \text{GL}_n(\mathbb{R})$ always takes the form $t \mapsto e^{tB}$ for some $B \in \text{M}_{n\times n}(\mathbb{R})$. Thus, there exists a flow through matrices whose time-one map equals $A$ if and only if $A = e^B$ for some matrix $B \in \text{M}_{n\times n}(\mathbb{R})$. There are known conditions specifying when this is true; for instance, a paper by Walter Culver titled "On the Existence and Uniqueness of the Real Logarithm of a Matrix" says that such a $B$ exists if and only if $A$ is nonsingular, and each Jordan block of $A$ belonging to a negative eigenvalue occurs an even number of times.
But do there exist matrices $A$ which are not the exponential of some other real matrix, but nevertheless are the time-one map of some more complicated, non-linear, flow? Thus, we are looking for a smooth homomorphism $\phi\colon \mathbb{R} \to \text{Diff}(\mathbb{R}^n)$ such that $\phi(1) = A$, or equivalently (I think), a not-necessarily-linear vector field $\mathbf{X}$ whose time-one flow equals $A$.
By the way, I don't need this for any work I'm doing; I just got curious. In terms of my level, I went to grad school for math, but it was years ago and I don't remember things as well as I should. Thanks in advance for your help!
Edit: As Alp Uzman points out in his answer, if $A$ embeds in a nonlinear flow, then clearly $A$ must commute with nonlinear diffeomorphisms. By Kopell’s Theorem (again, see Uzman’s answer), this is not possible if $A$ is a contraction and certain conditions on its eigenvalues hold.
In dimension 1, the possibilities for diffeomorphisms commuting with any smooth contraction $f$ (not necessarily a linear map) are quite restricted, thanks to Kopell’s Lemma. It’s natural to ask whether this result can be generalized to higher dimensions. I posed this question on Mathoverflow years ago, and have not yet gotten an answer: https://mathoverflow.net/questions/168550/kopells-lemma-in-higher-dimensions
 A: There are certain obstructions to this, and associated partial answers. Below are two separate arguments (for the latter I drop the "not an exponential" condition; I added it for further context, as even without the "not an exponential" condition the question is interesting). The two arguments essentially exemplify the two dynamical paradigms.
(To give further context, this question is closely related to the so-called "centralizer problem" in dynamics. See Smale's 12th problem in https://en.wikipedia.org/wiki/Smale%27s_problems as well as the announcement "The centralizer of a $C^1$ generic diffeomorphism is trivial" by Bonatti, Crovisier and Wilkinson (https://arxiv.org/abs/0705.0225v2) and the references therein; especially Kopell's paper "Commuting Diffeomorphisms" and Palis' paper/announcement "Vector fields generate few diffeomorphisms" seem to be among the important older papers.)

First Argument: Since we don't want $A$ to be an exponential, it does not have a root in $\text{GL}(d,\mathbb{R})$ (see Image of Matrix Exponential Map); but if it were in addition the time-$1$ map of some anonymous flow $\psi_\bullet$; it would have a root in $\text{Diff}^r_+(\mathbb{R}^d)$, say $\psi_{1/2}=f$:
$$f\circ f = A.$$
Taking derivatives we have for any $x\in \mathbb{R}^d$, $T_{f(x)}f \circ T_x f = T_xA=A$. This means that $f$ has no fixed points (by "no linear roots"), and in particular, $f(0)\neq 0 $ and $f (f(0))= 0$, so that $0$ is a periodic point of $f$ of period $2$. Then $A\circ f (0) = f\circ A(0) = f(0)$ gives that $A$ must have $1$ as an eigenvalue.
Hyperbolic matrices (i.e. those that have no eigenvalue of modulus $1$) are open and dense in $\text{GL}(d,\mathbb{R})$ (see my answer in Space of linear, continuous, hyperbolic functions is open, dense in the set of invertible functions for an outline and references), thus we have:
Claim: In $\text{GL}(d,\mathbb{R})$, the collection of matrices $A$ that is not the exponential of some matrix but still are the time-$t$ map of a $C^1$ flow for some $t\in\mathbb{R}$ is nowhere dense.
(Palis in the above mentioned article stated/proved a much more general version of this for compact manifolds: the collection of $C^1$ diffeomorphisms that embed in a $C^1$ flow is meager in the $C^1$ compact-open topology; the Bonatti-Crovisier-Wilkinson work says that even the collection of $C^1$ diffeomorphisms that commute with a diffeomorphism that is not an iterate is meager in the $C^1$ compact-open topology.)
Still it might be the case that there are nonhyperbolic examples, my estimation is that if there are constructing them would require some subtle diophantine analysis.

Second Argument: Say $n=1$, so that we have a $C^1$ flow $\psi_\bullet$ on $\mathbb{R}$ such that $\psi_1(x)=cx$ for some $c\in\mathbb{R}_{>0}$. By reversing time if needed we may assume that $c<1$; clearly when $c=1$ there is no obstruction. Then for any $t$, $n$ and $x$ we have $\psi_t(c^nx)=c^n\psi_t(x)$ and plugging in $x=0$ we have $\psi_t(0)=0$, whence
$$\psi_t(x)=\dfrac{\psi_t(c^nx)}{c^n}=\dfrac{\psi_t(c^nx)-\psi_t(0)}{c^nx}x \xrightarrow{n\to\infty}\psi_t'(0)x,$$
so that $\psi_\bullet$ factors through a group homomorphism $\mathbb{R}\to\mathbb{R}_{>0}$; since $\psi_1=c$, we have that $\psi_t(x)=e^{t\log(c)}(x)$.
Consequently in dimension $1$, even without the "not an exponential" condition (which is satisfied automatically) there is no embedding of a linear map into a nonlinear flow.
One can do similar things if the time-$t$ map is a hyperbolic matrix (in arbitrary dimension); just the property of a diffeomorphism (or worse) being able to commute with a hyperbolic matrix is quite strong.
For instance here is one theorem from Kopell's paper I mentioned above (Thm.6 on p.167). Before stating it let me define the (multiplicative) resonator to be the function $\mathfrak{M}:\{1,2,...,d\}\times \mathbb{Z}_{\geq0}^d\times \mathbb{C}^d\to \mathbb{C}$, $(i,\alpha;z)\mapsto z_i-z_1^{\alpha_1}z_2^{\alpha_2}\cdots z_d^{\alpha_d}$ (these are infinite dimensional analogs of root systems; see https://www.desmos.com/calculator/qxfuw4ehsy for a humble interactive graph).
Theorem (Kopell): Let $A\in\text{GL}(d,\mathbb{R})$ have eigenvalues (with multiplicity) $\lambda=(\lambda_1,...,\lambda_d)$. Put $L=\min_i|\lambda_i|$, $R=\max_i|\lambda_i|$. Suppose $0< L\leq R< 1$ (so that $A$ is contracting), and let $r\in\mathbb{Z}_{\geq1}$ be the least integer such that $R^r<L$. Then for any $g\in\text{Diff}^r(\mathbb{R}^d)$,

*

*$g\circ A = A\circ g$ implies that $g$ is a polynomial of degree strictly less than $r$. In particular the centralizer of $A$ in $\text{Diff}^r(\mathbb{R}^d)$ is a (finite dimensional) Lie group.

*If in addition for any $i$ and $\alpha$, $\mathfrak{M}(i,\alpha;\lambda)\neq0$ (i.e. "there are no resonances"), then $g\circ A=A\circ g$ implies that $g$ is linear.

The idea of the proof is fairly straightforward; the starting point is the Taylor approximation of $g$; comparing the homogeneous terms ($A$ is linear, hence preserves degree of homogeneity) one realizes the higher order terms vanish via an argument using the spectral gap condition. Similarly comparing coefficients of the monomials, together with "no resonances", allows one to realize all but linear monomials vanish.
(Resonators (and more broadly "normal form theory") is also used in the nonhyperbolic case, but they are of a different nature, roughly speaking.)
A: It is not an answer, but some observations which are too long for a comment.
The main issue here should has to do with complex numbers. In a complex set up, every nonsingular matrix is the exponential of a matrix, so your particular real nonsingular matrix (and its powers) belongs to the flow generated by this complex matrix. There is a beautiful visualization of this fact. Consider a very simple discrete dynamical system: the points in $\mathbb{R}$ skipping from $x$ to $-x$ alternatively. That is, $n=1$ and $A=(-1)$ with your notation. If you want to "embed" this discrete dynamical system into a differentiable one, you need to go through 0 with two different velocities. So you need a second-order system (the harmonic oscillator, indeed).

Or you can "extend the space", by considering complex numbers, and then you can embed this discrete dynamical system into a first order linear continuous dynamical system: $\dot{x}=i\cdot x$

Interestingly enough, in your setup, although it is not sure that $A^n$ can be embedded into a linear flow, it is the case for $A^{2n}$, because the matrix $A^2$ does have a logarithm (see here). So you can get a linear continuous dynamical system meeting your discrete system half the time. I think that this can also be related to the intuition given by the example above.
Indeed, I think that this example could answer partially the question. Because it should be easy to formalize that if you had a (possibly nonlinear) dynamical system such that $\phi(1,x)=-x$ and $\phi(2,x)=x$ then it cannot be an autonomous system, because of the two different velocities at $0$.
