Suppose $Q$ is orthogonal $n$ by $n$ matrix and $v,u$ are vectors in $\mathbb R^n$. Show that $$Qu \cdot Qv = u \cdot v$$
My attempt:
We will denote arbitrary element of $Q$ as $q_{ij}$
So
$$Qv = \begin{pmatrix} \sum_{i}^{n}q_{1i}v_i\\ \cdots\\ \sum_{i}^{n}q_{ni}v_i \end{pmatrix} $$
and
$$Qu = \begin{pmatrix} \sum_{i}^{n}q_{1i}u_i\\ \cdots\\ \sum_{i}^{n}q_{ni}u_i \end{pmatrix} $$
Hence
$$\begin{align}Qu \cdot Qv & = \sum_{j=1}^{n}\biggr(\sum_{i=1}^{n}q_{ji}u_i \sum_{i=1}^{n}q_{ji}v_i\biggl ) \\ & = \sum_{j=1}^{n}\sum_{i=1}^{n}q_{ji}^2v_iu_i \\ & = \sum_{i=1}^{n}\sum_{j=1}^{n}q_{ji}^2v_iu_i = \sum_{i=1}^{n}\underbrace{\sum_{j=1}^{n}q_{ji}^2}_{= 1}v_iu_i = u \cdot v \\ \end{align}$$
$\blacksquare$
Is it correct?
Is there a better way to prove it?
Since $Q$ is orthogonal, $QQ^T=Q^TQ=I$. Therefore,
$$Qu\cdot Qv:= (Qu)^T(Qv)=u^TQ^TQv=u^Tv=u\cdot v$$