A linear transfomation $h:\Bbb R^n\rightarrow\Bbb R^n$ is orthogonal iff carries a base $\mathcal V$ in an orthonormal base $\mathcal V'$.

Question

What shown below is a reference form Analysis on Manifolds by James Munkres.

So I want understand better the final statement. Indeed if the matrix $A$ is calculated only respcect the canonical base then clearly the transformation $h$ carries the canonical base in an orthonormal base but I don't understand if $A$ is calculated respect the bases $\mathcal V:=\{v_1,...,v_n\}$ and $\mathcal W:=\{w_1,...,w_n\}$ then the statement holds again. Then if we use another dot product the statement is generally false, right? So could someone help me, please?

Answer 1

The statement says that if $A \in \mathbb R^{n \times n}$ is an orthogonal matrix, the columns of $A$, namely$Ae_1,\dots,Ae_n$, forms an orthonormal basis for $\mathbb R^n$. I don't understand what do you mean by if $A$ is calculated respect to the basis..., but it is true that if $v_1,\dots,v_n$ is any other orthonormal basis for $\mathbb R^n$ (not necessarily the standard basis) then $Av_1,\dots,Av_n$ is again an orthonormal basis for $\mathbb R^n$. This is because if $B$ is the matrix whose $j$-th column is $v_j$, then $B$ and $AB$ are orthogonal matrices, and then note that $Av_j = (AB)e_j$. Is this that you are looking for?