Not sure why eigenvalue is proposed for linking twice and vector once.
I have selected the option to "Only link once". The "Include headers" options is also selected.
If $T$ has a one-dimensional invariant subspace $U$, it is generated by some non-zero [[vector]] $u \in V$, that is, $U = \{ au \mid a \in \mathbb{F} \}$. This means that $Tu = \lambda u$ for some $\lambda \in \mathbb{F}$.
By definition, $\lambda$ is an [[eigenvalue]] of $T$ and $u$ is the [[eigenvector]] corresponding to $\lambda$.
# Example: Eigenvalues of a rotation
Consider the operator $R(x, y) = (-y, x)$, so $R: \mathbb{F}^2 \to \mathbb{F}^2$ .
Any eigenvalue of $R$ satisfies the equation $(-y, x) = \lambda(x, y$), with the solutions $y = -\lambda x$, $x = \lambda y$, which gives $y = -\lambda^2 y$ and so $\lambda = \pm i$. The two eigenvectors are
- $(1, -i$) with eigenvalue $i$
- $(1, i)$ with eigenvalue $-i$.
This means there are no eigenvalues when $\mathbb{F} = \mathbb{R}$ which makes sense, since the operator $R$ can be interpreted as a counterclockwise rotation of $\pi/2$. No vector in $\mathbb{R}^2$ is left invariant under this transformation.
In $\mathbb{C}^2$, we get the eigenvalues as calculated above.