Prove that $(A\vec{x})\cdot (A\vec{y})=\vec{x}\cdot \vec{y}$ if $\|A\vec{x}\|=\|\vec{x}\| \ \forall\vec{x}$

Question

How should I prove that$$(A\vec{x})\cdot (A\vec{y})=\vec{x}\cdot \vec{y}$$

if $$\|A\vec{x}\|=\|\vec{x}\|,\ \forall \vec{x}$$

could you help me to solve this step-by-step? Thank you!

Answer 1

I can’t see how this is true. If we have

$$A = \begin{bmatrix}1&1\\0&0\end{bmatrix}$$

$$x = \begin{bmatrix}1\\0\end{bmatrix}, y = \begin{bmatrix}0\\1\end{bmatrix}$$

then $||A \overrightarrow x|| = 1 = ||\overrightarrow x||$ but $(A \overrightarrow x) \cdot (A \overrightarrow y) = 1 \neq 0 = \overrightarrow x \cdot \overrightarrow y$.

Perhaps you meant $||A \overrightarrow x|| = ||\overrightarrow x||$ for all $\overrightarrow x$. In this case, your statement is indeed true, so perhaps you should clarify exactly what each of your terms mean. It’s a little bit ambiguous currently.