A basis to a orthogonal set Let $(E, \langle \cdot, \cdot \rangle)$ be an $n$-dimensional Hilbert space and $A,B \colon E \to E$ linear isomorphisms. 
Does there exist a basis $\{e_{1},...,e_{n}\}$ of $E$ such that $\mathcal{B}=\{A(e_{1}),...,A(e_{n}),B(e_{1}),...,B(e_{n})\}$  are a orthogonal set?
Hints or solutions are greatly appreciated.
Related: Find a basis for two operators

No, it doesn't (for $n\neq 0$). Because a orthogonal set of non-zero vectors is linearly independend. As your space is $n$ dimensional there are at most $n$ linearly independend vectors.
Indeed, let $\{v_1, \dots, v_m \}\subseteq E$ be a orthogonal set of non-zero vectors. Assume they are linearly dependend
$$ \sum_{j=1}^m a_j v_j=0,$$
then we get
$$ 0 = \langle 0, v_i\rangle = \langle \sum_{j=1}^m a_j v_j, v_i\rangle 
= \sum_{j=1}^m a_j \langle v_j , v_i \rangle = a_i \langle v_i, v_i \rangle.$$
As $v_i\neq 0$, we have $\langle v_i, v_i \rangle \neq 0$ and thus $a_i=0$. Hence, the vectors are linearly independend.
Thus $2n\leq n$, which immediately implies $n=0$. In that case it is true (the empty set is a basis of the trivial vector space and the empty set is orthogonal).
