Determine geometrically the eigen-values and eigen-vectors of $F$ 
Let $F=I-2ww^*$ where $w\in\mathbb{R}^2$ and $\Vert w\Vert_2=1$. Determine geometrically the eigen-values and eigen-vectors of $F$.

Pick $w=[1/\sqrt{2}, 1/\sqrt{2}]^T$ which its two-norm is $1$. So
$$F=I-2ww^*=\begin{bmatrix}0 & -1\\-1 & 0\end{bmatrix}$$ I can find the correspond eigen-vectors and eigen-values for this example. How to do it for all $w\in\mathbb{R}^2$ with two-norm is $1$?
 A: A natural guess for an eigenvector of $F$ is $w$ itself, and indeed
\begin{align*}
F w &= (I - 2 w w^{*})w \\
&= Iw - 2w||w||_2^2 \\
&= w - 2w \\
F w &= -w \rm{,}
\end{align*}
since $w^{*}w = ||w||_2^2 = 1$. Let 
\begin{align*}
R = \begin{pmatrix}
0 & -1 \\
1 & 0
\end{pmatrix} \rm{.}
\end{align*}
Letting $w = \left(w_1 \ w_2 \right)^{\mathrm{T}}$, it is straightforward to see that
\begin{align*}
w^{*} R  w = w_1 w_2 - w_2 w_1 = 0 \rm{,}
\end{align*}
so $R w$ is orthogonal to $w$ and is also an eigenvector of $F$, as
\begin{align*}
F\left(Rw\right) &= (I - 2 w w^{*})\left(Rw\right) \\
&= I\left(Rw\right) - 2 w \left(w^{*}Rw \right) \\
&= Rw - 0 \\
F\left(Rw\right) &= Rw \rm{.}
\end{align*}
Therefore, $w$ and $Rw$ are the eigenvectors of $F$, with eigenvalues $-1$ and $+1$, respectively. We can thus interpret $F$ as a reflection about the line through the origin defined by the vector $Rw$, since $Rw$ is left unchanged by $F$, but the sign of $w$, the orthogonal vector to $Rw$, is flipped by $F$.
EDIT: Mixed up eigenvalues with respect to their corresponding eigenvectors in the last paragraph.
