Why does $(Av)\cdot (Aw)=v\cdot w$ hold?

Question

Let $A\in M_n(\mathbb{R})$ be orthogonal and $U\leq_A V$, where $V=\mathbb{R}^n$

Then we have that $U$ $\ A$-invariant, this means that $Au \in U$ for all $u \in U$.

How can we show that $(Av)\cdot (Aw)=v\cdot w$ for all $v,w\in V$ ?

Answer 1

We have $A$ is orthogonal.

Hence $$(Av)\cdot (Aw)=v\cdot (A^TA)w = v \cdot w$$

Since $A^TA=I$.

Answer 2

That is because

$$(Av)\cdot (Aw)=(Av)^T(Aw)=v^TA^TAw=v^Tw=v\cdot w.$$

This uses the definition of dot product, a property of transpose, associativity of matrix product and the fact that $A$ is orthogonal.

Answer 3

If you want to stay with the scalar product, you can use the notion of adjoint operator: $$\langle Av,Aw\rangle=\langle A^*Av,w\rangle,$$ where $A^*=A^\top$ if you work with reals. Since $A^\top A=I$, you get the requested result.