Matrix exponential of non diagonalizable matrix? I am currently self-learning about matrix exponential and found that determining the matrix of a diagonalizable matrix is pretty straight forward :).
I do not, however, know how to find the exponential matrix of a non-diagonalizable matrix.
For example, consider the matrix 
$$\begin{bmatrix}1 & 0 \\ 1 & 1\end{bmatrix}$$
Is there any process of finding the exponential matrix of a non-diagonalizable matrix? If so, can someone please show me an example of the process? :). I am not looking for an answer of the above mentioned matrix (since I just made it up), but rather I'm interested in the actual method of finding the matrix exponential to apply to other examples :)

Laplace Transforms approach is an useful for finding $e^{A t}$, whether $A$ is diagonalizable or non-diagonalizable. It is a very useful method in Electrical
Engineering.
Basically, we use the formula
$$
e^{A t} = \mathcal{L}^{-1}\left[ (s I - A)^{-1} \right]
$$
As an illustration, consider the defective matrix
$$
A = \left[ \begin{array}{cc}
1 & 0 \\
1 & 1 \\
\end{array} \right]
$$
We find
$$
s I - A = \left[ \begin{array}{cc}
s - 1  & 0 \\
-1 & s - 1 \\
\end{array} \right]
$$
where $s$ is a complex variable.
We find that
$$
(s I - A)^{-1} = \left[ \begin{array}{cc}
{1 \over s - 1} &  0 \\[2mm]
{1 \over (s - 1)^2} & {1 \over s - 1} \\[2mm]
\end{array} \right]
$$
The inverse Laplace transform of $(s I - A)^{-1}$ is the state transition matrix, which is also the matrix exponential, $e^{A t}$.
Hence, we get
$$
e^{A t} = \mathcal{L}^{-1}\left[ (s I - A)^{-1} \right]
= \left[ \begin{array}{cc}
 \mathcal{L}^{-1}\left( {1 \over s - 1} \right) & \mathcal{L}^{-1}(0) \\[2mm]
 \mathcal{L}^{-1}\left( {1 \over (s - 1)^2} \right) &  \mathcal{L}^{-1}\left( {1 \over s - 1} \right)  \\[2mm]
 \end{array} \right] = \left[ \begin{array}{cc}
 e^{t} & 0 \\[2mm]
t e^t &   e^t \\[2mm]
 \end{array} \right]
$$
