Prove that if $\phi$ is an isometry, then $\overline \phi = \phi^{-1}$

Question

In the book of Linear Algebra by Werner Greub, at page 233, it is given that

Let $\dim E = \dim F$ and $\phi:E \to F$ be an isometric (or orthogonal) mapping, then the isometry of $\phi$ implies that $$(\phi(x), y) = (x, \phi^{-1}(y)) \quad x\in E, y\in F,$$ whence $$\overline \phi = \phi^{-1},$$ where $\overline \phi$ is the adjoint of $\phi$.

However, I can't see how does the isometry of $\phi$ implies $$(\phi(x), y) = (x, \phi^{-1}(y)) \quad x\in E, y\in F.$$

Could someone provide a proof and an explanation for that ?

Note:

The book defined a isometric mapping as a mapping that preserves the inner product, i.e $$(\phi(x), \phi(y)) = (x,y) \quad x,y \in E.$$

Answer 1

First we note that $\phi$ is an isomorphism because it’s kernel is $\{0\}$ else it won’t be isometric and by the data on the dimension we know it’s onto. So $\phi^{-1}$ is defined. $$(\phi^{-1} \circ \phi(x),\phi^{-1}(y))=(x, \phi^{-1}(y))$$ So we are left to show that the inverse is the adjoint operator but that’s easy because else we won’t have $$(\phi(x),\phi(x))=(x,x)=(x,\bar{\phi}\phi(x))$$ For every normal basis vector

Answer 2

So whenever $\phi(x)=\phi(y)$, we have by the definition that $0=(x,y)$, hence $x=y$ (by assuming inner product, not the semi one), hence $\phi$ is injective, as $E,F$ have the same dimension, $\phi$ is bijective. So $(\phi(x),y)=(\phi(x),\phi(\phi^{-1}(y)))=(x,\phi^{-1}(y))$.