Eigenvalues and eigenvectors for 0
I can find the solutions mathematically for p = 0 or p = 1, but not for 0 < p <1. I also don't know how to justify geometrically. Any help/ relevant links would be greatly appreciated. 
 A: HINT
Eigenvalues are the roots of
$$0 = \det (\Sigma - \lambda I) = (1-\lambda)^2 - \rho^2,$$
which should be trivial to find for $\rho \in (0,1)$. Once you know the eigenvalues, can you solve for the eigenvectors?
UPDATE
To work with inequalities, you just solve what you need in a generic way and use the information in the inequality when you need to. For example, for eigenvalues, you are essentially solving $$(1-\lambda)^2 = \rho^2$$ which means
$$
1-\lambda = \pm \rho,
$$
Can you solve for $\lambda$ now? Can you find the eigenvectors?
UPDATE 2
So the eigenvalues must be $\lambda = 1 \pm \rho$. The eigenvectors will satisfy $\Sigma\vec{v} = \lambda \vec{v}$, so for $\lambda = 1+\rho$ we must have
$$
0 = (\Sigma-\lambda I) \vec{x} =
\begin{pmatrix}
1 - (1+\rho) & \rho\\
\rho & 1-(1+\rho)
\end{pmatrix}
\begin{pmatrix}x \\y\end{pmatrix}
=
\begin{pmatrix}
- \rho & \rho\\
\rho & -\rho
\end{pmatrix}
\begin{pmatrix}x \\y\end{pmatrix}
$$
which is equivalent to the equation $0 = -\rho x + \rho y$, i.e. $x=y$. Hence the eigenvector indeed looks like $$\vec{v}_+ = \left(\frac{1}{\sqrt2}, \frac{1}{\sqrt2} \right)^T.$$
Similarly, using $\lambda_- = 1 - \rho$ yields
$$
0 = (\Sigma-\lambda I) \vec{x} =
\begin{pmatrix}
1 - (1-\rho) & \rho\\
\rho & 1-(1-\rho)
\end{pmatrix}
\begin{pmatrix}x \\y\end{pmatrix}
=
\begin{pmatrix}
\rho & \rho\\
\rho & \rho
\end{pmatrix}
\begin{pmatrix}x \\y\end{pmatrix}
$$
which is equivalent to the equation $0 = \rho x + \rho y$, i.e. $x=-y$. Hence the eigenvector looks like $$\vec{v}_- = \left(\frac{1}{\sqrt2}, \frac{-1}{\sqrt2} \right)^T.$$
