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 $A= \begin{pmatrix}
                             0 & \sqrt{2}\\
                             \sqrt{2} & 1 \\
                             \end{pmatrix}$.

$$Ax=\lambda x \text{, where } x \neq \theta_n.$$
$$p(A)x=p(\lambda)x, \ p \text{ is a polynomial}.$$
Any function can be written as a polynomial using Taylor's series. So,
$$f(A)x=f(\lambda)x.$$
Choose $x$ such a way that it is the eigen vector of $A$, so if $A$ is diagonalizable then we will have $n$ linearly independent eigen vectors, and this will be your best choice of basis vector for $n$-dimensional space to evaluate the function. (In your case, $n=2$.)
