Choose a proper basis An exponent $e^A$ of a diagonal matrix $A=\begin{pmatrix}
                             \lambda_{1} & 0\\
                             0 & \lambda_{2} \\
                             \end{pmatrix}$
can be found directly by the rule
$e^A=\begin{pmatrix}
                             e^{\lambda_{1}} & 0\\
                             0 & e^{\lambda_{2}} \\
                             \end{pmatrix}$
.
Use a proper basis in the 2-dimensional space to calculate $e^A$, given $A= \begin{pmatrix}
                             0 & \sqrt{2}\\
                             \sqrt{2} & 1 \\
                             \end{pmatrix}$.

If you diagonalize $A$ as $A=PDP^{-1}$, then you can compute the exponent as
$$e^A = Pe^DP^{-1}$$.
