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!
 A: Note that $A^k = \begin{pmatrix} 1 & k \\ 0 & 1 \end{pmatrix}$.
So $e^A = \sum_{k=0}^{\infty} \frac{1}{k!}A^k = \sum_{k=0}^{\infty} \frac{1}{k!} \begin{pmatrix} 1 & k \\ 0 & 1 \end{pmatrix} =  \begin{pmatrix} \sum_{k=0}^{\infty} \frac{1}{k!} & \sum_{k=0}^{\infty} \frac{k}{k!} \\ 0 & \sum_{k=0}^{\infty} \frac{1}{k!} \end{pmatrix} = \begin{pmatrix} e & e \\ 0 & e \end{pmatrix}$
A: You can write : $A = I_2 + N$ with $N = \begin{pmatrix} 0 & 1 \\ 0 & 0 \end{pmatrix}$ nilpotent. Then, $N^2 = 0$ and $N$ commutes with $I_2$. So :
$$ A^n = \sum_{k = 0}^n \binom{n}{k} I_2^{n-k} N^k = I_2 + n N $$
Therefore :
$$ e^A = \sum_{k = 0}^{\infty} \frac{A^k}{k!} = I_2 \sum_{k = 0}^{\infty} \frac{1}{k!} + N \sum_{k = 0}^{\infty} \frac{k}{k!} = I_2 e + N e $$
This is a general strategy to compute the exponential even when the matrix is not diagonal.
A: I offer a third method for finding matrix exponentials, which is a popular technique in Electrical Engineering.
Given matrix is
$$
A = \left[ \begin{array}{cc}
1 & 1 \\
0 & 1 \\
\end{array} \right]
$$
We find
$$
s I - A = \left[ \begin{array}{cc}
s - 1  & -1 \\
0 & s - 1 \\
\end{array} \right]
$$
where $s$ is a complex variable.
We find that
$$
(s I - A)^{-1} = \left[ \begin{array}{cc}
{1 \over s - 1} & {1 \over (s - 1)^2} \\[2mm]
0 & {1 \over s - 1} \\[2mm]
\end{array} \right]
$$
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
$$
e^{A t} = \mathcal{L}^{-1}\left[ (s I - A)^{-1} \right]
= \left[ \begin{array}{cc}
 \mathcal{L}^{-1}\left( {1 \over s - 1} \right) &  \mathcal{L}^{-1}\left( {1 \over (s - 1)^2} \right) \\[2mm]
\mathcal{L}^{-1}(0) &  \mathcal{L}^{-1}\left( {1 \over s - 1} \right)  \\[2mm]
 \end{array} \right] = \left[ \begin{array}{cc}
 e^{t} & t e^t \\[2mm]
0 &   e^t \\[2mm]
 \end{array} \right]
$$
In particular, we get
$$
e^{A} = \left[ \begin{array}{cc}
 e^{1} & 1 e^1 \\[2mm]
0 &   e^1 \\[2mm]
 \end{array} \right] = \left[ \begin{array}{cc}
 e  & e \\[2mm]
0 &   e\\[2mm]
 \end{array} \right]
$$
