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?
 A: Hint: 
Write your matrix $A$ as $I+N$ where $I$ is the identity matrix and $N$ is a nilpotent matrix. Then use the definition of $e^{At}$ as a power series, noting that $N^k=0$ for some $k$.
A: If you know about the Jordan Canonical Form you can use that.
Another method, probably more elementary, was mentioned in a comment. The comment was deleted; I don't know why. Note that $A=I+N$, where $N^3=0$. It follows that $$A^k=I+kN+\frac{k(k-1)}{2}N^2.$$You can use that to calculate $e^{At}=\sum t^kA^k/k!$.
Edit: Oh, that comment was converted to an answer. I'll leave this here anyway, being more detailed (at least regarding one approach).
A: (This question was edited a lot, I'm referring to this revision.)
Yes, this is correct. Note however that:


*

*You've used $e^{(M+N)t}=e^{Mt}e^{Nt}$. Note that this is only valid if $M$ and $N$ commute (that is, $MN=NM$). In this case it's ok because $M$ is scalar and commutes with everything, but you should mention it explicitly.

*In general, it may be easier to find the Jordan form of $A$ and use that. You can calculate the exponential in blocks, and there is an elegant expansion for each block.

A: 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.

A: 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}
A: Hint:
Find the Jordan matrix, by which the exponential can always be found.
A: It's actually possible (though somewhat tedious) to calculate $\exp(tA)$ using the series definition.  The definition states that,
$$\exp(tA) = \sum_{n=0}^\infty \frac{t^n}{n!}A^n.$$
Now, since $\exp(tA)$ is a matrix, it is enough to find the values of $\exp(tA)e_i$ for $i=1,2,3$, where the $e_i$ are the standard basis vectors.
For $e_1$, we have
$$\exp(tA)e_1 = \sum_{n=0}^\infty \frac{t^n}{n!}A^ne_1 = \sum_{n=0}^\infty \frac{t^n}{n!} e_1 = \exp(t)e_1$$
so the first column of $\exp(tA)$ is
$$\left[ \exp(t)\quad 0\quad 0\right]^T$$
For $e_2$, we have that $Ae_2 = e_1 + e_2$, so
$$A^n e_2 = A^{n-1}e_1 + a^{n-1}e_2 = e_1 + A^{n-1}e_n = \cdots = ne_1 + e_2$$
and consequently,
$$\exp(tA)e_2 = \sum_{n=0}^\infty \frac{t^n}{n!}A^ne_2 = \sum_{n=0}^\infty \frac{t^n}{n!} ne_1 + \sum_{n=0}^\infty \frac{t^n}{n!} ne_2 = t\sum_{n=1}^\infty \frac{t^n}{(n-1)!} e_1 + \exp(t)e_2 = t\exp(t)e_1 + \exp(t)e_2$$
and the second row of the matrix is
$$\left[t\exp(t)\quad \exp(t)\quad 0\right]^T$$
Finally,
$$A e_3 = 2e_1 + -4e_2 + e_3$$
so
$$A^n e_3 = 2A^{n-1}e_1 - 4A^{n-1}e_2 + A^{n-1}e_3 = 2e_1 - 4((n-1)e_1 + e_2) + A^{n-1} e_3=\cdots$$ $$ \cdots = 2n e_1 - 4\left(\frac{n(n-1)}{2}e_1 + ne_2\right) + e_3 = -(2n^2+4n) e_1 - 4ne_2 + e_3.$$
Thus,
$$\exp(tA)e_3 = \sum_{n=0}^\infty \frac{t^n}{n!}A^ne_1 = \sum_{n=0}^\infty \frac{t^n}{n!}\left(-(2n^2-4n) e_1 - 4ne_2 + e_3\right)$$
$$ = -2\sum_{n=0}^\infty \frac{t^n}{n!}n(n-2)e_1 - 4\sum_{n=0}^\infty \frac{t^n}{n!}ne_2 + \sum_{n=0}^\infty \frac{t^n}{n!} e_3$$
$$= -2t\sum_{n=1}^\infty \frac{t^n}{(n-1)!}(n-2)e_1 -4 t\exp(t)e_2 + \exp(t)e_3$$
$$= -2t\sum_{n=0}^\infty \frac{t^n}{n!}(n-1)e_1 -4 t\exp(t)e_2 + \exp(t)e_3 $$
$$ = -2t\sum_{n=0}^\infty \frac{t^n}{n!}ne_1 + 2t\sum_{n=0}^\infty \frac{t^n}{n!}e_3 -4 t\exp(t)e_2 + \exp(t)e_3$$
$$= -2t \sum_{n=1}^\infty \frac{t^n}{(n-1)!} e_1 + 2t\exp(t)e_1 - 4t\exp(t)e_2 + \exp(t)e_3$$
$$= -2t^2 \sum_{n=0}^\infty \frac{t^n}{n!} e_1 + 2t\exp(t)e_1 - 4t\exp(t)e_2 + \exp(t)e_3$$
$$= -2t(t-1)\exp(t)e_1 -4t\exp(t)e_2 + \exp(t)e_3$$
so the final column of $\exp(tA)$ is
$$\left[-2t(t-1) \exp(t) \quad -4t\exp(t) \quad \exp(t)\right]^T$$
Thus,
$$\exp(tA) = \exp(t)\begin{bmatrix}
  1 & t & -2t(t-1)\\
  0 & 1 & -4t\\
  0 & 0 & 1\\
\end{bmatrix}$$
