Proof of Theorem EMS

We will check the three conditions of Theorem TSS. First, Definition EM explicitly includes the zero vector in $\eigenspace{A}{\lambda}$, so the set is nonempty.

Suppose that $\vect{x},\,\vect{y}\in\eigenspace{A}{\lambda}$, that is, $\vect{x}$ and $\vect{y}$ are two eigenvectors of $A$ for $\lambda$. Then \begin{align*} A\left(\vect{x}+\vect{y}\right)&=A\vect{x}+A\vect{y}&&\knowl{./knowls/theorem.MMDAA.knowl}{\text{Theorem MMDAA}}\\ &=\lambda\vect{x}+\lambda\vect{y}&&\knowl{./knowls/definition.EEM.knowl}{\text{Definition EEM}}\\ &=\lambda\left(\vect{x}+\vect{y}\right)&&\knowl{./knowls/property.DVAC.knowl}{\text{Property DVAC}} \end{align*}

So either $\vect{x}+\vect{y}=\zerovector$, or $\vect{x}+\vect{y}$ is an eigenvector of $A$ for $\lambda$ (Definition EEM). So, in either event, $\vect{x}+\vect{y}\in\eigenspace{A}{\lambda}$, and we have additive closure.

Suppose that $\alpha\in\complex{\null}$, and that $\vect{x}\in\eigenspace{A}{\lambda}$, that is, $\vect{x}$ is an eigenvector of $A$ for $\lambda$. Then \begin{align*} A\left(\alpha\vect{x}\right)&=\alpha\left(A\vect{x}\right)&&\knowl{./knowls/theorem.MMSMM.knowl}{\text{Theorem MMSMM}}\\ &=\alpha\lambda\vect{x}&&\knowl{./knowls/definition.EEM.knowl}{\text{Definition EEM}}\\ &=\lambda\left(\alpha\vect{x}\right)&&\knowl{./knowls/property.SMAC.knowl}{\text{Property SMAC}} \end{align*}

So either $\alpha\vect{x}=\zerovector$, or $\alpha\vect{x}$ is an eigenvector of $A$ for $\lambda$ (Definition EEM). So, in either event, $\alpha\vect{x}\in\eigenspace{A}{\lambda}$, and we have scalar closure.

With the three conditions of Theorem TSS met, we know $\eigenspace{A}{\lambda}$ is a subspace.