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?

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.

