Show that a Hilbert space with two inner products has a basis that is orthogonal with respect to both inner products Let $\mathcal{H}$ be a complex, $n$-dimensional Hilbert space with two inner products $\langle \cdot, \cdot  \rangle_1$, $\langle \cdot, \cdot  \rangle_2$. 
Show that there exists a basis $ X = x_1, \dots, x_n$ of $\mathcal{H}$ such that for $i \neq j$, $0 = \langle x_i, x_j  \rangle_1 = \langle x_i,   x_j \rangle_2$. That is, we want to find a basis that is orthogonal with respect to both inner products.
Using Gram Schmidt, I know that we can find an orthonormal basis for each inner product separately. The question is how we then construct a basis that is orthogonal with respect to both $\langle \cdot, \cdot  \rangle_1$ and $\langle \cdot, \cdot  \rangle_2$ at the same time.
I have thought about trying to use an induction argument on $n$. The $n=1$ case is trivial. For $n > 1$, we can write $\mathcal{H}$ as a direct sum $\mathcal{H} = V \bigoplus \mathbb{C} x $, where $V$ is an $n-1$ dimensional Hilbert space. We can then find a basis $Y$ of $V$ that is orthogonal with respect to both $\langle \cdot, \cdot  \rangle_1$ and $\langle \cdot, \cdot  \rangle_2$. The question would then be how we make $x$ orthogonal to the rest of the basis elements in $Y$. 
Hints or solutions are greatly appreciated.
 A: Let us write ${\cal H}_i$ to denote ${\cal H}$ equipped with $\langle\cdot,\cdot\rangle_i$, for $i=1,2$.
Consider an orthonormal basis $(e_1,\ldots,e_n)$ in ${\cal H}_1$, and consider the operator
$$
G:{\cal H}\longrightarrow{\cal H},G(x)=\sum_{i=1}^n \langle e_i,x\rangle_2 e_i
$$
we have
$$ 
\langle y, G(x)\rangle_1 =\sum_{i=1}^n \langle e_i,x\rangle_2 \langle y, e_i\rangle_1
=\sum_{i=1}^n \overline{\langle e_i,y\rangle_1}\langle e_i,x\rangle_2 =
\left\langle \sum_{i=1}^n\langle e_i,y\rangle_1e_i,x\right\rangle_2 =\left\langle y,x\right\rangle_2 
$$
Thus
$$
\langle y, G(x)\rangle_1=\left\langle y,x\right\rangle_2 =\overline{\left\langle x,y\right\rangle_2}=\overline{\left\langle x,G(y)\right\rangle_1}=
\langle G(x),y\rangle_1
$$
So $G$ is hermitian, and since $\langle G(x),x\rangle_1=||x||_2^2>0$ for nonzero $x$, we conclude that $G$ is definite positive. Thus, there is an orthonormal basis $(f_1,\ldots,f_n)$ of  ${\cal H}_1$ that diagonalizes $G$ that is $G(f_k)=\lambda_k f_k$,  with $\lambda_k>0$, for $k=1,\ldots,n$. For $k\ne m$, we have by construction,$\langle f_k, f_m\rangle_1=0$ , and
$$
\langle f_k, f_m\rangle_2= \langle f_k, G(f_m)\rangle_1=\lambda_m\langle f_k,  f_m\rangle_1
=0$$
Thus $(f_1,\ldots,f_n)$ does the job.$\qquad\square$
${\bf Remark.}$ $(f_1,\ldots,f_n)$ is orthonormal in ${\cal H}_1$ and orthogonal in ${\cal H}_2$.
