Inequality in a Hilbert space This was a homework question a couple weeks ago that I couldn't solve.  I'd appreciate a solution.

Let $\mathcal{H}$ be a Hilbert space and $\{\eta_n \}_{n \in \mathbb{Z}}$ a set of not necessarily orthogonal elements of $\mathcal{H}$.  Let $A$ be the matrix, defined on $\ell_2(\mathbb{Z})$, such that $a_{nm} = |(\eta_n,\eta_m)|$.  Show that for all $f \in \mathcal{H}$,
  $$\sum_{n \in \mathbb{Z}} |(f,\eta_n)|^2 \le ||A|| \, ||f||^2$$

 A: For notational simplicity I will take the index set to be $\mathbb N$ instead of $\mathbb Z$ (it doesn't change the substance of the proof).
Fix $f\in\mathcal H$. Consider the rank-one operators $E_k=\langle\cdot,\eta_k\rangle\,\eta_k$. Then
$$\tag{1}
\left\|\sum_{k=1}^nE_kf\right\|^2=\left\langle\sum_{k=1}^nE_kf,\sum_{j=1}^nE_jf\right\rangle=\sum_{j,k=1}^n\langle E_jE_kf,f\rangle=\sum_{j,k=1}^n\langle f,\eta_j\rangle\,\langle\eta_k,\eta_j\rangle\,\langle\eta_k,f\rangle.
$$
Let $g_n\in\ell^2(\mathbb N)$ be given by 
$$
g_n(k)=\begin{cases}\langle f,\eta_k\rangle,&\text{ if }k\leq n \\ 0,&\text{ otherwise} \end{cases}
$$
Then, from (1), 
$$
\left\|\sum_{k=1}^nE_kf\right\|^2=\langle Ag_n,g_n\rangle\leq \|A\|\,\|g_n\|^2.
$$
So
$$
\sum_{k=1}^n|\langle f,\eta_k\rangle|^2=\sum_{k=1}^n\langle f,\eta_k\rangle\,\langle \eta_k,f\rangle=\sum_{k=1}^n\langle E_kf,f\rangle=\langle\sum_{k=1}^n E_kf,f\rangle\leq\left\|\sum_{k=1}^nE_kf\right\|\,\|f\|\\ \leq \|A\|^{1/2}\,\|g_n\|\,\|f\| =\|A\|^{1/2}\,\|f\|\,\left(\sum_{k=1}^n|\langle f,\eta_k\rangle|^2\right)^{1/2}.
$$
After dividing by $\left(\sum_{k=1}^n|\langle f,\eta_k\rangle|^2\right)^{1/2}$ and squaring, we get
$$
\sum_{k=1}^n|\langle f,\eta_k\rangle|^2\leq\|A\|\,\|f\|^2.
$$
As this holds for any $n$,
$$
\sum_{k=1}^\infty|\langle f,\eta_k\rangle|^2\leq\|A\|\,\|f\|^2.
$$
