<s>[INST] Calculating a matrix-exponential Let A be the following matrix.
$$\begin{pmatrix}
1 & 1 \\
0 & 1 \\
\end{pmatrix}
$$
I have to calculate $e^A$.
My idea was to diagonalize A because then $e^A = Pe^DP^-1$ if $A = PDP^-1$.
But A cannot be diagonalized since 1 is a double eigenvalue and therefore A does not have 2 linearly independent eigenvectors.
How else can I calculate $e^A$?
Thank you!
 [/INST] Note that $A^k = \begin{pmatrix} 1 & k \\ 0 & 1 \end{pmatrix}$.
So $e^A = \sum_{k=0}^{\infty} \frac{1}{k!}A^k = \sum_{k=0}^{\infty} \frac{1}{k!} \begin{pmatrix} 1 & k \\ 0 & 1 \end{pmatrix} =  \begin{pmatrix} \sum_{k=0}^{\infty} \frac{1}{k!} & \sum_{k=0}^{\infty} \frac{k}{k!} \\ 0 & \sum_{k=0}^{\infty} \frac{1}{k!} \end{pmatrix} = \begin{pmatrix} e & e \\ 0 & e \end{pmatrix}$
</s>