Principal motivation behind eigenvectors is, of course, eigenspaces. And eigenspaces, in turn, are an important case of invariant subspaces and one of few constructs that we can make of a linear operator using pure linear algebra (without Euclidean norm, coordinates etc.). In some cases eigenspaces classify invariant subspaces. Suppose we don’t know anything about an abstract vector space. What is the only way to specify a linear operator? Multiplication by a fixed scalar, i.e.
$$ v\mapsto \lambda v\,, $$
of course. And finding subspaces where some not-so-simple operator has such form shows how to describe an operator using subspaces and scalars. When linear operator is diagonalizable (i.e. hasn’t Jordan blocks, that is a general position), it provides complete description of the operator, albeit over complex numbers (or another algebraically closed field).
Students should know that:
- An eigenvector multiplied by a non-zero scalar is an eigenvector again.
- A linear combination of eigenvectors with the same eigenvalue is an eigenvector again (if non-zero), with the same eigenvalue – that’s how eigenspaces arise.
- Linear span of some set of eigenvectors (or direct sum of eigenspaces) produces an invariant subspace, but not necessarily an eigenspace.
This is an archetypical situation that reproduces, to some extent, many times in algebra: decomposition to elementary structural blocks.
Some examples requested by the original poster.
Let, over some F,
$$ A = \begin{pmatrix}
0 && 1 && 0 \\
1 && 0 && 0 \\
0 && 0 && 1
\end{pmatrix}. $$
The simplest choice of eigenvectors of $A$ is:
$$
\begin{pmatrix}1\\1\\0\end{pmatrix},
\begin{pmatrix}0\\0\\1\end{pmatrix}\ \text{for }\lambda=1;
\quad
\begin{pmatrix}1\\-1\\0\end{pmatrix}\ \text{for }\lambda=-1.
$$
Then an arbitrary linear combination of two $\lambda=1$ eigenvectors, namely
$$\begin{pmatrix}α\\α\\β\end{pmatrix},$$
describes the $\lambda=1$ eigenspace of $A$, a two-dimensional subspace of F3. Its non-zero elements are all $\lambda=1$ eigenvectors of $A$.
An arbitrary linear combination of the first and the third eigenvectors, namely
$$\begin{pmatrix}α+γ\\α-γ\\0\end{pmatrix},$$
describes an $A$-invariant subspace, but not an eigenspace of $A$ (assuming 2 ≠ 0) because “1” and “−1” eigenvalues are mixed.