# show that two vector are perpendicular to other

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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• 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. 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.