Orthogonal matrix

Question

I am given that the vectors $x$ and $x'$ have the same Euclidean length and $Qx=x'$ where $Q=I-\frac{2uu^T}{\|u\|^2}$ and $u=x-x'$.

I need to show that $Q$ is orthogonal but I don't know how to do this.

Any help would be appreciated.

Answer 1

Taking the transpose, we have $$ Q^t = I^t - \frac{2uu^t}{\def\norm#1{\left\|#1\right\|}\norm u^2} = Q, $$ hence, $Q$ is symmetric. As, moreover \begin{align*} Q^2 &= I - 4\frac{uu^t}{\norm u^2} + 4\frac{uu^tuu^t}{\norm u^4}\\ &= I - 4\frac{uu^t}{\norm u^2} + 4 \frac{u\norm u^2 u^t}{\norm u^4}\\ &= I - 4 \frac{uu^t}{\norm u^2} + 4 \frac{uu^t}{\norm u^2}\\ &= I, \end{align*} $Q$ fulfills $QQ^t = Q^tQ = Q^2 = I$, hence, $Q$ is orthogonal.