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 $1$?


1 Answer 1


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.


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