Having arise a question on the ambiguity on the definition of eigenvalues/eigenvectors. I try to restate a clearer version of the definition of eigenvalues and eigenvectors for matrices, which is aimed at eliminating the annoying ambigouity. I want to ask that is such definition proper, and broader, and clearer?
Let $E,~F$ be two fields, and $F\leq E$. Let $A\in M_{n}(F)$, $\lambda \in E$.
(i) If $(\exists v\in E^n\setminus\{0\},~Av=\lambda v)$, then $\lambda$ is called an eigenvalue of $A$ over $E$.
(ii) Let $v\in E^n\setminus\{0\}$. If $Av=\lambda v$, then $v$ is called an eigenvector of $A$ over $E$ corresponding to $\lambda$.
PS: Notice the use of extension field and the critical preposition "over".
Following such definition, we say:
- $\begin{bmatrix}0&-1\\1&0\end{bmatrix}$ has no eigenvalues over $\Bbb R$. It also has no eigenvectors over $\Bbb R$.
- $\begin{bmatrix}0&-1\\1&0\end{bmatrix}$ has eigenvalues $i,~-i$ over $\Bbb C$. It also has eigenvectors $\begin{cases}x=it\\y=t\end{cases}(t\in\Bbb C-\{0\})$ over $\Bbb C$ corresponding to $i$, and $\begin{cases}x=-it\\y=t\end{cases}(t\in\Bbb C-\{0\})$ corresponding to $-i$.
- $\begin{bmatrix}3&2\\2&0\end{bmatrix}$ has eigenvalues $-1,~4$ over $\Bbb Q$. It also has eigenvectors $\begin{cases}x=-t\\y=2t\end{cases}(t\in\Bbb Q)$ over $\Bbb Q$ corresponding to $-1$.
- $\begin{bmatrix}3&2\\2&0\end{bmatrix}$ has eigenvalues $-1,~4$ over $\Bbb C$. It also has eigenvectors $\begin{cases}x=-t\\y=2t\end{cases}(t\in\Bbb C)$ over $\Bbb C$ corresponding to $-1$.