Proof $\langle v,w\rangle=0\implies \langle Av,Aw\rangle=0$, given $A\in F^{m\times n}, \operatorname{rank}(A)=n$. Given two vectors $v,w$ orthogonal, and a matrix $A$ which has orthogonal columns, how to prove that $\langle Av,Aw \rangle=0$, i.e. their image is orthogonal?
 A: Consider the quantity 
$$\langle Av,Aw \rangle = w^T A^TAv $$
Zooming in on the entries of the matrix $A^TA$ ,
 $$[A^TA]_{i,j} = a_i^T a_j$$
where $a_i$ is the $i^{th}$ column of A.
Since $A^TA$ contains orthogonal columns, then $a_i^T a_j = \alpha_i \delta_{i,j}$
where $\delta_{i,j}$ is the Kronecker-delta function and $\alpha_i >0$. Finally,
$$\langle Av,Aw \rangle = w^T A^TAv = \sum_{i=1}^n \alpha_i w^T v =0$$
A: Denote th orthogonal columns of $A$ by $a_1, \ldots, a_n$, and suppose $v = (v_1, \ldots, v_n)$ and similar for $w$. Then
$$
\langle Av, Aw \rangle = \left\langle \sum_{j=1}^n v_j a_j, \sum_{k=1}^n w_k a_j \right\rangle = \sum_{j=1}^n \sum_{k=1}^n w_k v_j \langle a_j, a_k \rangle = \sum_{k=1}^n v_k w_k
$$
by othogonality of the $a_k$, but the latter is just $\langle v, w\rangle$, so that $A$ in fact preserves the scalar product on $\mathbb R^n$.
A: You have $$\langle Av,Aw\rangle=\langle v,A^TAw\rangle=\langle v,Id.w\rangle=\langle v,w\rangle=0$$by definition of the transpose of $A$ and because $A$ is supposed to be orthogonal.
