For $k\in \mathbb{N}$, let $\lambda_k \in \mathbb{R}^{d\times d}$ be a symetric positive definite matrix, and $\lambda_{kij}$ be it's coordinates. Suppose we have $c_k \in \mathbb{R}^d$, $c_{ki}$ its coordinates, such that $$ \sum_{k=1}^{\infty} c_k^T \lambda_k c_k < \infty $$ Can we then say that $ \sum_{k=1}^{\infty} c_{ki}^2\lambda_{kij} < \infty $, for all $i, j$?
In other words, since the converse is true (by the Cauchy-Schwarz inequality), we have that $$ \left\{[c_{k1} \, \cdots \, c_{kd} ]^T \,:\, (c_{ki}\sqrt{\lambda_{kij}}) \in \ell_2, \forall i, j \right\} \subset \left\{ c_k\in \mathbb{R}^d \,:\, \sum_{k=1}^{\infty} c_k^T \lambda_k c_k < \infty \right\}\,, $$ but do we also have equality?
No. Put $d=2$, for each $k$ put $\lambda_k=\begin{pmatrix}1 & \frac 1{k^2}-1 \\ \frac 1{k^2}-1 & 1 \end{pmatrix}$ and $c_k=\begin{pmatrix}1\\ 1\end{pmatrix}.$ Then $$c_k^T\lambda_k c_k=\frac 2{k^2},$$ so $$\sum_{k=1}^{\infty} c_k^T \lambda_k c_k=\sum_{k=1}^{\infty} \frac 2{k^2}=\frac{\pi^2}3< \infty$$ but $$c_{k1}^2\lambda_{k11}=1,$$ so $$ \sum_{k=1}^{\infty} c_{k1}^2\lambda_{k11}=\infty.$$