Prove that $\left|\left|I-\frac{ss^t}{s^ts}\right|\right|=1$

Question

Let $0\neq s\in \mathbb{R}^n$. Prove that $$\left|\left|I-\frac{ss^t}{s^ts}\right|\right|=1$$

I truly don't have idea on how to solve this. Any hints?

Answer 1

Let $$M=I-\frac 1{s^t s} ss^t$$

The matrix being symmetric, let's look at its eigenvalues and eigenvectors.

First, let $u\in\mathbb R^n$ be a vector orthogonal to $s$. Then $s^t u=0$, and $$Mu=u$$ So $u$ is an eigenvector of $M$ with eigenvalue $1$. The space of vectors orthogonal to $s$ being of dimension $n-1$, we know that $1$ is an eigenvalue of multiplicity $n-1$. We now need to find the last eigenvalue of $M$. Now note that $$Ms=s-s=0$$ so $M$ has $0$ as an eigenvalue.

To summarize, $M$ has two eigenvalues: $1$ (multiplicity $n-1$) and $0$ (multiplicity 1).

It follows that $$\|M\|=1$$

Answer 2

This is a less elegant, more 'analysis' approach. For any $v\in\mathbb R^n$, we have

$$ss^tv = s \cdot \langle s, v\rangle,$$

so $ss^tv$ is a multiple of $s$. On the other hand, $s^ts = \langle s,s\rangle = \lVert s \rVert^2$. Hence, for $v \in\mathbb R^n$ we have

$$\left(I - \frac{ss^t}{s^ts} \right)v = v - \frac{\langle s, v\rangle}{\lVert s \rVert^2}\, s,$$

and thus

\begin{align} \left\lVert \left(I - \frac{ss^t}{s^ts}\right)v \right\Vert^2 &= \left\langle v - \frac{\langle s, v\rangle}{\lVert s \rVert^2}\, s , v - \frac{\langle s, v\rangle}{\lVert s \rVert^2}\, s \right\rangle \\&= \langle v, v \rangle - 2 \left\langle v, \frac{\langle s, v\rangle}{\lVert s \rVert^2}\,s\right\rangle +\left\langle \frac{\langle s, v\rangle}{\lVert s \rVert^2}\, s , \frac{\langle s, v\rangle}{\lVert s \rVert^2}\, s \right\rangle \\&= \langle v, v \rangle - \frac{2\,{\langle v, s\rangle}^2}{\lVert s \rVert^2} +\frac{{\langle v, s\rangle}^2}{\lVert s \rVert^2} \\&= \langle v, v \rangle - \frac{\,{\langle v, s\rangle}^2}{\lVert s \rVert^2}. \end{align}

Now, if $\lVert v \rVert = 1$, we have

$$\left\lVert \left(I - \frac{ss^t}{s^ts}\right)v \right\Vert^2 = 1 - \frac{\,{\langle v, s\rangle}^2}{\lVert s \rVert^2}$$

so that $\left\lVert I - \frac{ss^t}{s^ts} \right\Vert \leqslant 1$. To see that $1$ is achieved, it suffices to take $v$ with $\langle v, s\rangle = 0$, that is, any $v$ orthogonal to $s$.