Eigenvalues and eigenvectors for 0<p<1

Question

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.

Answer 1

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 $$\vec{v}_- = \left(\frac{1}{\sqrt2}, \frac{-1}{\sqrt2} \right)^T.$$