Here is the specification again:
Let $ w \in \mathbb{R}^{n} $ and $||w||=1$. The $n \times n$- matrix.
$$ P:=I-2ww^t $$
is called the Householder transform and $w$ is called the Householder vector.
Determine the eigenvalues of P and its eigenvectors.
Here is my way of determining the eigenvalues and eigenvectors. Does this make any sense or is there something missing?
A Householder transform is orthogonal ($P^{-1}=P^t$) and symmetric ($P^t=P$).
I.e. it holds:
$$ (I - 2ww^t)^t = I - 2ww^t $$
$$ (I - 2ww^T)^2 = I - 4ww^T + 4w(w^Tw)w^T = I $$
Here $w^t w = 1$, since according to the above $w$ is a unit vector.
Eigenvalues of orthogonal matrices have the absolute value $1$, since multiplication by an orthogonal matrix is isometry (length-preserving).
Show here in a short proof that holds: Let A be an orthogonal matrix and $v \in \mathbb{R}^{n}$ be an eigenvector of A to eigenvalue $\lambda \in \mathbb{R}$. Then $\lambda \in \{ \pm 1\}$ holds.
Proof. Let A be an orthogonal matrix and $v \in \mathbb{R}^{n}$ be an eigenvector of A at eigenvalue $\lambda \in \mathbb{R}$. As an orthogonal matrix, A is length-faithful:
$$ ||v||=||Av||=||\lambda v||=| \lambda | ||v||$$. The last equality follows from the bilinearity of the scalar product.
As an eigenvector $v \ne 0$ and thus $| \lambda |=1$.
By assumption, $\lambda \in \mathbb{R}$, so $ \lambda \in \{ \pm 1 \} $. $\square $\\
Since the Householder matrix $P = I - 2ww^T$ is real and symmetric, its eigenvalues are real. The only real numbers with absolute value $1$ are then $\pm 1$.\\
Since $Pw = w - 2w(w^t w) = -w$, $-1$ is eigenvalue of $P$ and $vw$ for any $v \in \mathbb{R}, v \ne 0$, is an associated eigenvector.
For $x \in w^t$, i.e. $x^tw = 0$, holds: $$Px = x-2ww^tx=x$$, i.e. 1 is eigenvalue of P and every $x \in \mathbb{R}^{n},x \ne 0$ with $x^tw=0$ is eigenvector.
By decomposing \begin{equation} y = \langle y, w \rangle w + (\ y - \langle y, w \rangle w )\ \end{equation}
into its projection on span$ \{ w \} $ and a perpendicular part, one sees that there are no other eigenvectors of P.
P therefore has the eigenvalues $ \lambda_1 = 1 (\ (\ n-1 )\ -fold )\ $ with eigenspace $E_{\lambda_1} = \{ x \in \mathbb{R}^{n} : x^tw = 0 \}$ and $\lambda_2 = -1 $ (single) with eigenspace $E_{\lambda_2} = \{ vw : v \in \mathbb{R} \}$.
The eigenspace at eigenvalue 1 is the mirror plane, i.e. the orthogonal complement of the $v$ generated one-dimensional subspace.