[proofplan]
We first define the matrix exponential by its [power series](/page/Power%20Series) and justify differentiating it term by term on compact time intervals. This gives the fundamental identity $\frac{d}{dt}e^{tA} = A e^{tA}$, and therefore $x(t)=e^{tA}x_0$ satisfies both the differential equation and the initial condition. To prove uniqueness without appealing to an external ODE theorem, we multiply an arbitrary solution by $e^{-tA}$ and show the product is constant. The same identities also give the semigroup law, which is the algebraic fact behind the uniqueness argument.
[/proofplan]
[step:Define the matrix exponential and differentiate its series]
Fix an operator norm $\|\cdot\|_{\mathrm{op}}$ on $\mathbb{R}^{n \times n}$. For each $T>0$ and each $t \in [-T,T]$, the $k$th term of the series satisfies
\begin{align*}
\left\|\frac{t^k A^k}{k!}\right\|_{\mathrm{op}} \leq \frac{T^k \|A\|_{\mathrm{op}}^k}{k!}.
\end{align*}
Since the scalar series $\sum_{k=0}^{\infty} T^k\|A\|_{\mathrm{op}}^k/k!$ converges, the matrix series defining $e^{tA}$ converges uniformly on $[-T,T]$.
For $N \in \mathbb{N}$, define the polynomial map $S_N: \mathbb{R} \to \mathbb{R}^{n \times n}$ by
\begin{align*}
S_N(t) = \sum_{k=0}^{N} \frac{t^k A^k}{k!}.
\end{align*}
Differentiating the finite polynomial sum gives
\begin{align*}
S_N'(t) = \sum_{k=1}^{N} \frac{k t^{k-1} A^k}{k!}.
\end{align*}
Reindexing with $j=k-1$ gives
\begin{align*}
S_N'(t) = A \sum_{j=0}^{N-1} \frac{t^j A^j}{j!}.
\end{align*}
The right-hand side converges uniformly on $[-T,T]$ to $A e^{tA}$. Since $S_N(0)=I_n$ for all $N\in\mathbb{N}$, the sequence $(S_N(0))_{N\in\mathbb{N}}$ converges in $\mathbb{R}^{n\times n}$. By the theorem on uniformly convergent derivatives for maps into a finite-dimensional [normed vector space](/page/Normed%20Vector%20Space), the uniformly convergent limit $t \mapsto e^{tA}$ is continuously differentiable on $[-T,T]$ and satisfies
\begin{align*}
\frac{d}{dt}e^{tA} = A e^{tA}.
\end{align*}
Because $T>0$ was arbitrary, the identity holds for every $t\in\mathbb{R}$.
[guided]
The first point is that the matrix exponential is not merely formal. We choose an operator norm $\|\cdot\|_{\mathrm{op}}$ on $\mathbb{R}^{n \times n}$, so matrix multiplication satisfies
\begin{align*}
\|BC\|_{\mathrm{op}} \leq \|B\|_{\mathrm{op}}\|C\|_{\mathrm{op}}
\end{align*}
for all $B,C \in \mathbb{R}^{n \times n}$. Therefore, for $t \in [-T,T]$,
\begin{align*}
\left\|\frac{t^k A^k}{k!}\right\|_{\mathrm{op}} \leq \frac{T^k \|A\|_{\mathrm{op}}^k}{k!}.
\end{align*}
The scalar majorant is the exponential series with parameter $T\|A\|_{\mathrm{op}}$, so it converges. This proves [uniform convergence](/page/Uniform%20Convergence) of the matrix series on every compact interval $[-T,T]$.
Now define the partial sums $S_N: \mathbb{R} \to \mathbb{R}^{n \times n}$ by
\begin{align*}
S_N(t) = \sum_{k=0}^{N} \frac{t^k A^k}{k!}.
\end{align*}
Each $S_N$ is a polynomial map, so ordinary termwise differentiation for finite sums gives
\begin{align*}
S_N'(t) = \sum_{k=1}^{N} \frac{k t^{k-1} A^k}{k!}.
\end{align*}
After setting $j=k-1$, this becomes
\begin{align*}
S_N'(t) = A \sum_{j=0}^{N-1} \frac{t^j A^j}{j!}.
\end{align*}
The same uniform convergence estimate shows that the derivative series converges uniformly on $[-T,T]$ to $A e^{tA}$. We now invoke the theorem on uniformly convergent derivatives for maps into a finite-dimensional normed [vector space](/page/Vector%20Space). Its hypotheses are satisfied: each $S_N$ is continuously differentiable, the derivatives $S_N'$ converge uniformly on $[-T,T]$, and the values $S_N(0)=I_n$ converge in $\mathbb{R}^{n\times n}$. Therefore the limit map $t \mapsto e^{tA}$ is continuously differentiable on $[-T,T]$, and its derivative is
\begin{align*}
\frac{d}{dt}e^{tA} = A e^{tA}.
\end{align*}
Since $T>0$ was arbitrary, the same derivative identity holds on all of $\mathbb{R}$.
[/guided]
[/step]
[step:Verify that $e^{tA}x_0$ satisfies the initial value problem]
Define $x: \mathbb{R} \to \mathbb{R}^n$ by
\begin{align*}
x(t) = e^{tA}x_0.
\end{align*}
Since $t \mapsto e^{tA}$ is continuously differentiable, the map $x$ is continuously differentiable. Using the derivative identity from the previous step,
\begin{align*}
x'(t) = \left(\frac{d}{dt}e^{tA}\right)x_0.
\end{align*}
Therefore
\begin{align*}
x'(t) = A e^{tA}x_0.
\end{align*}
By the definition of $x(t)$, this is
\begin{align*}
x'(t) = A x(t).
\end{align*}
At $t=0$, the series gives
\begin{align*}
e^{0A} = I_n,
\end{align*}
where $I_n \in \mathbb{R}^{n \times n}$ is the identity matrix. Hence
\begin{align*}
x(0) = I_n x_0 = x_0.
\end{align*}
So $x(t)=e^{tA}x_0$ is a solution of the initial value problem.
[/step]
[step:Multiply the two absolutely convergent exponential series]
Fix $s,t\in\mathbb{R}$. The two matrix series defining $e^{tA}$ and $e^{sA}$ converge absolutely in the operator norm because
\begin{align*}
\sum_{k=0}^{\infty}\left\|\frac{t^kA^k}{k!}\right\|_{\mathrm{op}} \leq \sum_{k=0}^{\infty}\frac{|t|^k\|A\|_{\mathrm{op}}^k}{k!}
\end{align*}
and
\begin{align*}
\sum_{l=0}^{\infty}\left\|\frac{s^lA^l}{l!}\right\|_{\mathrm{op}} \leq \sum_{l=0}^{\infty}\frac{|s|^l\|A\|_{\mathrm{op}}^l}{l!}.
\end{align*}
The absolute convergence of these two series in the finite-dimensional normed algebra $\mathbb{R}^{n\times n}$ justifies the Cauchy product. Since every power of $A$ commutes with every other power of $A$, we obtain
\begin{align*}
e^{tA}e^{sA} = \sum_{m=0}^{\infty}\sum_{k=0}^{m}\frac{t^kA^k}{k!}\frac{s^{m-k}A^{m-k}}{(m-k)!}.
\end{align*}
For each $m\in\mathbb{N}\cup\{0\}$, the inner sum equals
\begin{align*}
\sum_{k=0}^{m}\frac{t^k s^{m-k} A^m}{k!(m-k)!} = \frac{(t+s)^m A^m}{m!},
\end{align*}
where the last equality is the [binomial theorem](/theorems/750) applied to the scalar polynomial $(t+s)^m$. Therefore
\begin{align*}
e^{tA}e^{sA} = \sum_{m=0}^{\infty}\frac{(t+s)^mA^m}{m!} = e^{(t+s)A}.
\end{align*}
Thus
\begin{align*}
e^{(t+s)A} = e^{tA}e^{sA}
\end{align*}
for all $s,t\in\mathbb{R}$.
[/step]
[step:Show that every solution agrees with the matrix exponential solution]
Let $y: \mathbb{R} \to \mathbb{R}^n$ be any continuously differentiable solution satisfying
\begin{align*}
y'(t) = A y(t)
\end{align*}
for every $t \in \mathbb{R}$ and
\begin{align*}
y(0) = x_0.
\end{align*}
Define $z: \mathbb{R} \to \mathbb{R}^n$ by
\begin{align*}
z(t) = e^{-tA}y(t).
\end{align*}
The derivative identity for $e^{-tA}$ follows by applying the already-proved formula to the matrix $-A$. Using this identity and the product rule,
\begin{align*}
z'(t) = -A e^{-tA}y(t) + e^{-tA}y'(t).
\end{align*}
Since $y'(t)=Ay(t)$ and $A$ commutes with every power of $A$, hence with $e^{-tA}$, we obtain
\begin{align*}
z'(t) = -A e^{-tA}y(t) + e^{-tA}A y(t) = -A e^{-tA}y(t) + A e^{-tA}y(t) = 0.
\end{align*}
Therefore $z$ is constant on $\mathbb{R}$. Since
\begin{align*}
z(0) = e^{0A}y(0) = I_n x_0 = x_0,
\end{align*}
we have
\begin{align*}
e^{-tA}y(t) = x_0.
\end{align*}
Multiplying by $e^{tA}$ and using the semigroup identity with $s=-t$ gives
\begin{align*}
y(t) = e^{tA}x_0.
\end{align*}
Thus every solution equals the solution constructed above, so the solution is unique.
[/step]
