Solving Complex Eigenvalues The problem I am struggling with is this:
Solve the system
\begin{equation}
x' = \begin{pmatrix}2&-5\\5&2\end{pmatrix}x
\end{equation}
With $x(0)$ = \begin{pmatrix}-2\\-2\end{pmatrix}
Give your solution in real form.
So I tried to follow my notes and find the eigenvalue. Solving for $\lambda$ yielded (through the quadratic equation) $2\pm50i$. 
From here I am completely lost and cannot seem to follow my notes. Where do I go from here? Thank you for your help in advance.
 A: the eigenvalues of $\pmatrix{2&-5\\5&2}$ are $2\pm 5i.$ the eigenvector corresponding to the eigenvalues is given by $$\pmatrix{-5i&-5\\5&-i}\pmatrix{x\\y} = \pmatrix{0\\0.}$$  ann eigenvector is $\pmatrix{1\\-i}$ and a solution to $x' = Ax$ is $$\pmatrix{1\\-i}e^{2t+5it}=e^{2t}\{\left(\pmatrix{1\\0}+i\pmatrix{0\\-1}\right)\left(\cos 5t+i\sin 5t\right)\}$$ whose real and imaginer parts are $$e^{2t}\pmatrix{\cos 5t \\\sin 5t}, e^{2t}\pmatrix{\sin 5t\\-\cos 5t}$$ these satisfy the initial conditions $$\pmatrix{1\\0},\pmatrix{0\\-1} $$ therefore the solution satisfying the initial condition $\pmatrix{-2\\-2}$ is 
$$-2e^{2t}\pmatrix{\cos 5t \\\sin 5t}+2 e^{2t}\pmatrix{\sin 5t\\-\cos 5t} =2e^{2t}\pmatrix{\sin 5t - \cos 5t\\-\sin 5t - \cos 5t}.$$
A: So, you found the eigenvalues (following Graydad's comment, these are rather $2\pm5i$), it means that the matrix can be written as
$$A:=\ \pmatrix{2&-5\\5&2}=B\cdot\pmatrix{2-5i&0\\0&2+5i}\cdot B^{-1}$$
where the columns of $B$ are the corresponding eigenvectors (with complex coefficients).
Then, we have the solution
$$x(t)=e^{At}\cdot c= B\pmatrix{e^{(2-5i)t} &0\\0&e^{(2+5i)t}}B^{-1}\cdot c $$
with $c=\pmatrix{-2\\-2}$ from the initial condition.
For the required real form, calculate $B$, $B^{-1}$ and use the following identities and see if the complex parts score out each other.
$$ e^{z+w}=e^z\cdot e^w \\
e^{\alpha i}=\cos\alpha+i\,\sin\alpha\,.$$
