Solution EE.T20 Robert Beezer

This problem asks you to prove that two sets are equal, so use Definition SE.

First show that $\set{\zerovector}\subseteq\eigenspace{A}{\lambda}\cap\eigenspace{A}{\rho}$. Choose $\vect{x}\in\set{\zerovector}$. Then $\vect{x}=\zerovector$. Eigenspaces are subspaces (Theorem EMS), so both $\eigenspace{A}{\lambda}$ and $\eigenspace{A}{\rho}$ contain the zero vector, and therefore $\vect{x}\in\eigenspace{A}{\lambda}\cap\eigenspace{A}{\rho}$ (Definition SI).

To show that $\eigenspace{A}{\lambda}\cap\eigenspace{A}{\rho}\subseteq\set{\zerovector}$, suppose that $\vect{x}\in\eigenspace{A}{\lambda}\cap\eigenspace{A}{\rho}$. Then $\vect{x}$ is an eigenvector of $A$ for both $\lambda$ and $\rho$ (Definition SI) and so \begin{align*} \vect{x} &=1\vect{x} &&\knowl{./knowls/property.O.knowl}{\text{Property O}}\\ &=\frac{1}{\lambda-\rho}\left(\lambda-\rho\right)\vect{x} &&\lambda\neq\rho,\ \lambda-\rho\neq 0\\ &=\frac{1}{\lambda-\rho}\left(\lambda\vect{x}-\rho\vect{x}\right) &&\knowl{./knowls/property.DSAC.knowl}{\text{Property DSAC}}\\ &=\frac{1}{\lambda-\rho}\left(A\vect{x}-A\vect{x}\right) &&\text{$\vect{x}$ eigenvector of $A$ for $\lambda$, $\rho$}\\ &=\frac{1}{\lambda-\rho}\left(\zerovector\right)\\ &=\zerovector &&\knowl{./knowls/theorem.ZVSM.knowl}{\text{Theorem ZVSM}}\\ \end{align*} So $\vect{x}=\zerovector$, and trivially, $\vect{x}\in\set{\zerovector}$.