Let $E$ $n$-dimensional space with $\langle \cdot, \cdot \rangle$ inner product. Consider $A:E \to E$ isomorphism linear.
Is there exist a orthonormal basis $\{f_{1},...,f_{n}\}$ such that $\{A(f_{1}),...,A(f_{n})\}$ is a orthonormal basis too?
Using the Gram–Schmidt process I can obtain a basis $\{g_{1},...,g_{n}\}$ such that $\{A(g_{1}),...,A(g_{n})\}$ is a orthonormal basis.
Hints or solutions are greatly appreciated.
No. In fact, $A$ sends some orthonormal basis to an orthonormal basis if and only if $A$ is unitary. To see that this is necessarily, assume that $A$ sends the orthonormal basis $(f_1, \dots, f_n)$ to $(g_1, \dots, g_n)$ (so that $Af_i = g_i$). But then
$$ \| Av \|^2 = \big\| A \left( \sum_{i=1}^n \left< v, f_i \right> f_i \right) \big\|^2 = \big\| \sum_{i=1}^n \left< v, f_i \right> g_i \big\|^2= \sum_{i=1}^n |\left< v, f_i \right>|^2 = \big| \sum_{i=1}^n \left
which shows that $A$ is unitary. On the other hand, it is clear that a unitary map sends (any) orthonormal basis to an orthonormal basis since it preserves orthogonality and length.