Prove that at most $d^2$ Kraus operators are sufficient to represent any quantum operation

Question

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"?

Answer 1

The missing step that you ask about in your last paragraph is to use the following fact (labeled Theorem 8.2 in my edition of Nielsen and Chuang): If $\{E_1,\ldots,E_n\}$ and $\{F_1,\ldots,F_n\}$ are two sets of operators defining operations $\mathcal{E}$ and $\mathcal{F}$ respectively, then $\mathcal{E}=\mathcal{F}$ iff there is an $n$-by-$n$ unitary matrix $u$ such that $E_i = \sum_j u_{ij} F_j$.

Use the $u$ that diagonalizes $W$ to define new operators $F_i = \sum_j u^*_{ij}E_j$. Then for all but $d^2$ of the $F_i$ we have $\operatorname{tr}(F_i^\dagger F_i)=0$ (and hence $F_i=0$). By the above theorem the $F_i$ describe the same operation as the original $E_i$.