What is the limit of $\mathrm{e}^{At}$ when $t\to \infty$, for $A$ a square matrix? I was doing a matrix calculation and need to find
$$\lim_{t\to \infty} \mathrm{e}^{At}=?$$
What is the limit of $\mathrm{e}^{At}$ when $t\to \infty$, for $A$ a matrix?
 A: This limit depends on the eigenvalues of $A$.
Let $A$ be diagonalizable, i.e. $A=U^{-1}DU$, $D=\mathrm{diag}(d_1,\ldots,d_n)$. Then 
$$
\mathrm{e}^{tA}=
U^{-1}\mathrm{e}^{tD}U=U^{-1}\mathrm{diag}(\mathrm{e}^{d_1t},\ldots,\mathrm{e}^{d_nt})U.
$$
So, the limit of $\mathrm{e}^{tA}$ depends on the $d_1,\ldots,d_n$. If all their real parts are negative, then $\mathrm{e}^{tA}\to 0$.
A: The limit is $0$ if and only if $A$ is Hurwitz (ie. real part of any eigenvalue of $A$ is negative). To see why this is true, apply the Spectral Mapping Theorem, which basically says the eigenvalues of $e^{At}$ is of the form $e^{\lambda t}$ where $\lambda$ is an eigenvalue of $A$.
A: Recall the definition of exponential function of one variable in terms of power series:
$$
e^x := \sum_{n=0}^{\infty} \frac{x^n}{n!} = 1 + x + \frac{x^2}{2!} +  \frac{x^3}{3!} + \dots
$$
Similarly, we can define the matrix exponent  of $n\times n$ matrix $A$:
$$
e^A : = I + A + \frac{1}{2!}A^2 +  \frac{1}{3!}A^3 + \dots = \sum_{n=0}^{\infty}\frac{A^n}{n!},
$$
where $I$ is the identity matrix of the size $n\times n$. 
Assume now that $A$ is diagonalizable, i.e. it can be represented as $A = P^{-1}DP$, where 
$$
D = 
\begin{bmatrix}
d_{11} & 0 &  \cdots& 0\\
0 & d_{22} & \cdots & 0\\
\vdots& \vdots & \ddots & 0\\
0 & 0& \cdots & d_{nn}
 \end{bmatrix}
$$ 
is a diagonal matrix, i.e. it has non-zero entries only on its diagonal. Note that 
positive integer $n$, the power $D^n$ is easy to compute:
$$
D^n = \underbrace{D \cdot D \cdot \dots \cdot D}_{n \text{ times }} =  \left(\ 
\begin{bmatrix}
d_{11} & 0 &  \cdots& 0\\
0 & d_{22} & \cdots & 0\\
\vdots& \vdots & \ddots & 0\\
0 & 0& \cdots & d_{nn}
 \end{bmatrix}\ \right) ^n
=
\begin{bmatrix}
\big(d_{11}\big)^n & 0 &  \cdots& 0\\
0 & \big(d_{22}\big)^n & \cdots & 0\\
\vdots& \vdots & \ddots & 0\\
0 & 0& \cdots & \big(d_{nn}\big)^n
 \end{bmatrix}
$$
Then for any positive integer $n$ we can compute $A^n$ as
$$
A^n = \left(P^{-1} D P \right) ^n  = \underbrace{P^{-1} D P  \cdot P^{-1} D P  \cdot \dots \cdot P^{-1} D P }_{n\text{ times }} 
=P^{-1} D^n P 
$$
Therefore the exponent
$$
e^A = I + \left(P^{-1} D P\right) + \left(\frac{1}{2!} P^{-1} D^2 P\right) + \left(\frac{1}{3!} P^{-1} D^3 P\right)  + \dots = 
P^{-1}\left(\sum_{n=0}^{\infty}\frac{1}{n!}D^n\right)P=
\\
=
P^{-1}\sum_{n=0}^{\infty}\frac{1}{n!}\left(\ \begin{bmatrix}
\big(d_{11}\big)^n & 0 &  \cdots& 0\\
0 & \big(d_{22}\big)^n & \cdots & 0\\
\vdots& \vdots & \ddots & 0\\
0 & 0& \cdots & \big(d_{nn}\big)^n
 \end{bmatrix}\ \right)P=\\
=
P^{-1}\left(
\begin{bmatrix}
\displaystyle\sum_{n=0}^{\infty}\dfrac{\left(d_{11}\right)^n}{n!} & 0 &  \cdots& 0\\
0 & \displaystyle\sum_{n=0}^{\infty}\dfrac{\left(d_{22}\right)^n}{n!}  & \cdots & 0\\
\vdots& \vdots & \ddots & 0\\
0 & 0& \cdots & \displaystyle\sum_{n=0}^{\infty}\dfrac{\left(d_{nn}\right)^n}{n!} 
\end{bmatrix}\right)P
=
P^{-1}
\left(
\begin{bmatrix}
e^{d_{11}}  & 0 &  \cdots& 0\\
0 & e^{d_{22}} & \cdots & 0\\
\vdots& \vdots & \ddots & 0\\
0 & 0& \cdots & e^{d_{nn}}
\end{bmatrix}
\right)P
=
P^{-1} e^D P
$$
Finally, multiplying $A$ by $t$, you get
$$
e^{At} = P^{-1} e^{Dt} P = 
P^{-1}
\left(
\begin{bmatrix}
e^{d_{11}t}  & 0 &  \cdots& 0\\
0 & e^{d_{22}t} & \cdots & 0\\
\vdots& \vdots & \ddots & 0\\
0 & 0& \cdots & e^{d_{nn}t}
\end{bmatrix}
\right)P,
$$
where $P^{-1}DtP = A$ is diagonalization of $At$.
