Let $S : \mathbb{R}^n \rightarrow \mathbb{R}^n$ be the transformation induced by the $3 \times 3$ matrix $\left(I-u\,u^T\right)$. Show that for $x \in \mathbb{R}^3$, $S(x)$ is perpendicular to $u$, where $u = [b , c, d]^T.$

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  • $\begingroup$ Show that $u^\top S(x) = 0$ by plugging in the definition of $S(x)$. I think you also need the condition that $u^\top u = 1$ which isn't mentioned in the problem. $\endgroup$
    – angryavian
    Oct 13 at 22:54

$S(x) = (I-u\,u^T)\,x = x - u\,u^T\,x$

$\langle u,S(x)\rangle = u^T\,S(x) = u^T\,x - u^T\,u\,u^T\,x = (u^T\,x)\,(1-u^T\,u)$

If we include the condition: $b^2 + c^2 + d^2 = 1 \Leftrightarrow \|u\| = 1$

$1-u^T\,u = 0$, which means that $\langle u,S(x)\rangle = 0$, i.e., $S(x)$ is perpendicular to $u$.

If $u$ is a unit vector, then $(u\,u^T\,x) = (u^T\,x)\,u$ is the projection of $x$ in the $u$-direction. Thus, $S(x)$ subtracts from $x$ its projection in the $u$-direction, and only the perpendicular part remains.


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