All quantum operations $\mathcal{E}$ on a system of Hilbert space dimension $d$ can be generated by an operator-sum representation containing at most $d^2$ elements, $$ \mathcal{E}(\rho)=\sum_{k=1}^M E_k\rho E_k^\dagger $$ where $1\leq M\leq d^2$.
A possible attempt to prove this is given in Exercise 8.10, Page 373, Chapter 8, Quantum Computation and Quantum Information by Nielsen and Chuang, as
Let $\{E_j\}$ be a set of operation elements for $\mathcal{E}$. Define a matrix $W_{jk}\equiv tr(E_j^\dagger E_k )$. Show that the matrix $W$ is Hermitian and of rank at most $d^2$, and thus there is unitary matrix $u$ such that $uWu^\dagger$ is diagonal with at most $d^2$ non-zero entries. Use $u$ to define a new set of at most $d^2$ non-zero operation elements $\{F_j\}$ for $\mathcal{E}$.
Proof of this is given in Kraus operator rank, as
$E_j=(I\otimes \langle e_j|)U(I\otimes |e_0\rangle)$
Given that the quantum operation $\mathcal{E}$ acts on a Hilbert space of dimension $d$, i.e., $\rho\in\mathbb{C}^{d\times d}$ and $E_k\in\mathbb{C}^{d\times d}$
Therefore, a maximum of $d^2$ of the $E_j$ can be linearly independent.
$$ W=\begin{bmatrix} tr(E_1^\dagger E_1)&tr(E_1^\dagger E_2)&\cdots&tr(E_1^\dagger E_M)\\ tr(E_1^\dagger E_1)&tr(E_1^\dagger E_2)&\cdots&tr(E_1^\dagger E_M)\\ \vdots&\vdots&\ddots&\vdots\\ tr(E_M^\dagger E_1)&tr(E_M^\dagger E_2)&\cdots&tr(E_M^\dagger E_M)\\ \end{bmatrix} $$ $W_{jk}=tr(E_j^\dagger E_k)=tr((E_k^\dagger E_j)^\dagger)=tr(E_k^\dagger E_j)^*=W_{kj}^*$
$\therefore W$ is hermitian.
Let the $k^{th}$ column of $W$ is, $|w_k\rangle=\sum_{j=1}^M W_{jk}|j\rangle=\sum_{j=1}^M tr(E_j^\dagger E_k)|j\rangle$
Let $E_j$ are arranged such that $E_1,\cdots,E_{d^2}$ are linearly independent, then for any $k>d^2$, we have $E_k=\sum_{l=1}^{d^2} c_{kl}E_l$ where $c_{kl}\in\mathbb{C}$.
\begin{align} |w_k\rangle=\sum_{j=1}^M tr(E_j^\dagger E_k)|j\rangle=\sum_{j=1}^M tr(E_j^\dagger\sum_{l=1}^{d^2} c_{kl} E_l)|j\rangle=\sum_{j=1}^M tr(\sum_{l=1}^{d^2} c_{kl} E_j^\dagger E_l)|j\rangle\\ =\sum_{j=1}^M \sum_{l=1}^{d^2} c_{kl} tr( E_j^\dagger E_l)|j\rangle=\sum_{l=1}^{d^2} c_{kl}\bigg[\sum_{j=1}^M tr( E_j^\dagger E_l)|j\rangle\bigg]=\sum_{l=1}^{d^2} c_{kl}|w_l\rangle \end{align} $\implies$ for every $k>d^2$, the $k^{th}$ column of $W$ is a linear combination of the first $d^2$ linearly independent columns of $W$.
$\implies$ the number of linearly independent columns of $W$ is therefore at most $d^2$
$\implies rank(W)\leq d^2$
And since $W$ is a hermitian matrix, there exists a decomposition $uWu^\dagger$ which is a diagonal matrix with at most $d^2$ nonzero diagonal entries.
This seems fine, but
How does showing "there is unitary matrix $u$ such that $uWu^\dagger$ is diagonal with at most $d^2$ non-zero entries" proves "all quantum operations $\mathcal{E}$ on a system of Hilbert space dimension $d$ can be generated by an operator-sum representation containing at most $d^2$ elements"?
