Finding the eigenvalues of a linear transformation without using the matrix of the linear transformation Define the linear transformation $L$ as 
$L: \mathbb{R}^3 \rightarrow \mathbb{R}^3: x \rightarrow x - \langle x, a \rangle b$
Let $a, b \in \mathbb{R}^3$ such that $\langle a, b \rangle = 2$ and let $\langle .,. \rangle $ be the standard inner product of $\mathbb{R}^3$. 
How would I find the eigenvalues of the linear transformation without using the matrix of the linear transformation?
 A: Hint: Assuming that $L(x)=x-\langle x,a\rangle b=\lambda x$ one obtains the equation $(1-\lambda)x=\langle x,a\rangle b$. It is easy to find the eigenvalues from this equation.  
A: It appears this all works over $\Bbb R^n$:
Given that
$L(x) = x - \langle x, a \rangle b, \tag 0$
if
$\langle a, x \rangle = 0, \tag 1$
then
$L(x) = x, \tag 2$
that is, $x$ is an eigenvector of $L$ associated with eigenvalue
$\lambda = 1; \tag 3$
thus the entire $n - 1$ dimensional subspace
$a^\bot \subset \Bbb R^n \tag 4$
is the $1$-eigenspace of $L$.  Now if 
$x = \alpha b, \; \alpha \ne 0, \tag 5$
then
$L(x) = L(\alpha b) = \alpha b - \langle \alpha b, a \rangle b = \alpha b - \alpha \langle b, a \rangle b = \alpha b - 2\alpha b = -\alpha b; \tag 6$
that is, $\alpha b$ is a $-1$-eigenvector of $L$; also
$\langle b, a \rangle = 2 \ne 0 \Longrightarrow b \notin a^\bot; \tag 7$
therefore 
$\Bbb R^n = a^\bot + \{\alpha b, \alpha \in \Bbb R\}; \tag 8$
since $a^\bot$ and $b$ generate $\Bbb R^n$, there can be no other eigenvectors/eigenvalues, so we have found all the 
eigenvectors/eigenvalues of $L$.
