Finding the Exponential of a Matrix that is not Diagonalizable Consider the $3 \times 3$ matrix
$$A =
\begin{pmatrix} 
1 & 1 & 2 \\ 
0 & 1 & -4 \\ 
0 & 0 & 1  
\end{pmatrix}.$$
I am trying to find $e^{At}$.
The only tool I have to find the exponential of a matrix is to diagonalize it. $A$'s eigenvalue is 1. Therefore, $A$ is not diagonalizable.
How does one find the exponential of a non-diagonalizable matrix?
My attempt:
Write
$\begin{pmatrix} 
1 & 1 & 2 \\ 
0 & 1 & -4 \\ 
0 & 0 & 1  
\end{pmatrix} = M  + N$,
with $M = \begin{pmatrix} 
1 & 0 & 0 \\ 
0 & 1 & 0 \\ 
0 & 0 & 1  
\end{pmatrix}$ and $N = \begin{pmatrix} 
0 & 1 & 2 \\ 
0 & 0 & -4 \\ 
0 & 0 & 0  
\end{pmatrix}$.
We have $N^3 = 0$, and therefore $\forall x > 3$, $N^x = 0$.  Thus:
$$\begin{aligned}
e^{At}
&= e^{(M+N)t} = e^{Mt} e^{Nt} \\
&= \begin{pmatrix} 
e^t & 0 & 0 \\ 
0 & e^t & 0 \\ 
0 & 0 & e^t  
\end{pmatrix} \left(I + \begin{pmatrix} 
0 & t & 2t \\ 
0 & 0 & -4t \\ 
0 & 0 & 0  
\end{pmatrix}+\begin{pmatrix} 
0 & 0 & -2t^2 \\ 
0 & 0 & 0 \\ 
0 & 0 & 0  
\end{pmatrix}\right) \\
&= e^t \begin{pmatrix} 
1 & t & 2t \\ 
0 & 1 & -4t \\ 
0 & 0 & 1  
\end{pmatrix} \\
&= \begin{pmatrix} 
e^t & te^t & 2t(1-t)e^t \\ 
0 & e^t & -4te^t \\ 
0 & 0 & e^t  
\end{pmatrix}.
\end{aligned}$$
Is that the right answer?

In general if $f$ is a matrix function and $J$ a Jordan block with eigenvalue $\lambda_{0}$ then
\begin{equation}
f(J)=\left(\begin{array}{ccccc}
f(\lambda_{0}) & \frac{f'(\lambda_{0})}{1!} & \frac{f''(\lambda_{0})}{2!} & \ldots & \frac{f^{(n-1)}(\lambda_{0})}{(n-1)!}\\
0 & f(\lambda_{0}) & \frac{f'(\lambda_{0})}{1!} &  & \vdots\\
0 & 0 & f(\lambda_{0}) & \ddots & \frac{f''(\lambda_{0})}{2!}\\
\vdots & \vdots & \vdots & \ddots & \frac{f'(\lambda_{0})}{1!}\\
0 & 0 & 0 & \ldots & f(\lambda_{0})
\end{array}\right),
\end{equation}
This means that in your case since A has just one Jordan block $J$, i.e. $A=SJS^{-1}$  with eigenvalue $\lambda_{0}=1$, since your matrix function is $f(J)=e^{tJ}$, then you have
\begin{equation}
e^{At}=Se^{Jt}S^{-1}=S\left(\begin{array}{ccccc}
e^{t} &  te^{t}  & \frac{t^2e^{t}}{2} \\
0 & e^{t} & te^{t}  \\
0 & 0 & e^{t} & \\
\end{array}\right)S^{-1}.
\end{equation}
