<s>[INST] 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?
 [/INST] this is my first answer on this site so if anyone can help to improve the quality of this answer, thanks in advance.
That said, let us get to business.


*

*Compute the Jordan form of this matrix, you can do it by hand or check this link. (or both). 

*Now, we have the following case: $$ A = S J S^{-1}.$$ You will find $S$ and $S^{-1}$ on the previous link. For the sake of simplicity, $J$ is what actually matters, 
$$
J = 
\begin{pmatrix}
1 & 1 & 0\\
0 & 1 & 1\\
0 & 0 & 1\\
\end{pmatrix}
$$
because:

*$e^A = e^{SJS^{-1}} = e^J$ And the matrix $J$ can be written as: $J = \lambda I + N$, where $I$ is the identity matrix and $N$ a nilpotent matrix.

*So, $e^J = e^{\lambda I + N} = \mathbf{e^{\lambda} \cdot  e^N}$ 
By simple inspection, we get that:
$$
J = \lambda I + N = 
1 \cdot
\begin{pmatrix}
1 & 0 & 0\\
0 & 1 & 0\\
0 & 0 & 1\\
\end{pmatrix}
 + \begin{pmatrix}
0 & 1 & 0\\
0 & 0 & 1\\
0 & 0 & 0\\
\end{pmatrix}
$$
where you can check that $\lambda =1$ and N is 
$$
\begin{pmatrix}
0 & 1 & 0\\
0 & 0 & 1\\
0 & 0 & 0\\
\end{pmatrix}
$$

*So, 
$e^A = e \cdot e^N$
we just apply the definition $ e^N \equiv \sum^{\infty}_{k=0} \frac{1}{k!} N^k$. And, of course, it converges fast: $N^2 \neq 0$ but $N^3=0$.

*Finally:
$$
e^A = e \cdot \left[
1 \cdot I + 
1 \cdot N^1 + 
\frac{1}{2} N^2
\right]
$$
where $$
N^2 = 
\begin{pmatrix}
0 & 0 & 1\\
0 & 0 & 0\\
0 & 0 & 0\\
\end{pmatrix}
$$ then:
$$
\mathbf{
e^A = e \cdot
\begin{pmatrix}
1 & 1 & 1/2\\
0 & 1 & 1\\
0 & 0 & 1\\
\end{pmatrix}
}
$$
Last but not least, $$ e^{At} = e^{A \cdot t} = e^{\lambda \cdot t} \cdot e^{N \cdot t}  = 
e^t \cdot
\begin{pmatrix}
1 & t & 1/2 t^2\\
0 & 1 & t\\
0 & 0 & 1\\
\end{pmatrix}
$$ 
You replace $N$ by $At$ in the exp definition and that's it.

</s>