Elementary geometrical interpretation of $2\langle a,x\rangle a -x$ Let $V$ be an euclidean vector space with $a \in V$ of length 1.
I showed that $$f: V \rightarrow V, x \rightarrow 2\langle a,x\rangle a -x $$ is an orthogonal transformation and hence either a rotation, reflection or a combination of both. 
How excatly am I supposed to figure out, what kind of transformation is taking place?
 A: Hint:
By the definition of orthogonal transformation you have to prove that:
$\langle f(x),f(y)\rangle=\langle x,y\rangle \quad \forall x,y \in V$.
Calculate:
$$
\langle (2\langle a,x\rangle a -x),(2\langle a,y\rangle a -y) \rangle
$$
using the properties of the inner product, and remember that $\langle a,a\rangle=1$

This transformation can be interpreted as a reflection in the plane ( $2-$dimensional subspace) spanned by $a$ and $x$.
A sketch of this interpretation:
without loss of generality, we can find a basis in $span\{a,x\}$ such that $a=(1,0)$ and $x=(x_1,x_2)$
So we have $f(x)=(2x_1,0)-(x_1,x_2)=(x_1,-x_2)$ 
That is a reflection.
A: If $f$ is a linear transformation, write $f$ in matricial form. This can be helpful.
A: $\newcommand{\Brak}[1]{\left\langle #1\right\rangle}$A natural approach (whether you know the answer or not) is to try to compute
$$
f(x) = 2\Brak{a, x} a - x
$$
for as many vectors as you can. Here, the primary impediment is our (lack of) ability to evaluate $\Brak{a, x}$.


*

*Since $\|a\| = 1$, we have $\Brak{a, a} = \|a\|^{2} = 1$. Consequently,
$$
f(a) = 2\Brak{a, a} a - a = a.
$$
Success!

*If $\Brak{a, x} = 0$, i.e., $x$ is orthogonal to $a$, then
$$
f(x) = -x.
$$
Again, success!
But this answers the question: $a$ and the set of vectors orthogonal to $a$ span $V$, so we've found the eigenspace decomposition of $f$.
