Unitary matrix with respect to an inner product

Question

Let $A\in M_{n}({F})$ where $F$is a field. Then $\langle Ax,Ax\rangle= \langle x,x\rangle $ foer every $x\in {F}^{n}$, i.e, $A$ is unitary with respect to $\langle .,.\rangle$ if and only if there is an orthonormal basis in $F^{n}$ such that the the matrix representation of $A$ is unitary with respect to this basis i.e, $[A]_{B}[A]^{*}_{B}=I$ where $B$ is the orthonormal basis.

Can every one prove it?

Answer 1

If $A:F^n\rightarrow F^n$ fulfills $<Ax,Ax>=<x,x>$ for all $x\in F^n$ it follows by polarization that it respects the inner product i.e. $<Ax,Ay>=<x,y>$ for all $x,y\in F^n$ and this implies, that $A$ maps orthonormal bases to orthonormal bases, so if $B=\{b_1,...,b_n\}$ is an orthonormal basis then $\{Ab_1,...,Ab_n \}$ is an orthonormal basis and thus

$([A]_B[A]_B^*)_{ij}=<Ab_i,Ab_j>=<b_i,b_j>=\delta_{ij}$.

If conversely $[A]_B[A]_B^*=I$ for an orthonormal basis $B$ it follows that $<Ab_i,Ab_j>=\delta_{ij}$ i.e. $\{Ab_1,...,Ab_n \}$ is an orthonormal basis. and thus for $x=\sum_{j=1}^{n}x_jb_j\in F^n$:

$<Ax,Ax>=\sum_{i,j=1}^{n}x_i\overline{x_j}<Ab_i,Ab_j>=\sum_{j=1}^{n}x_j\overline{x_j}=<x,x>$.