Nielsen & Chuang (978-1-107-00217-3) state the following on page 101:
A positive $2^n\times 2^n$ matrix $\rho$ (that can be seen as a linear mapping from $\mathbb{C}^{2^n}$ to $\mathbb{C}^{2^n}$) has a spectral decomposition
$$\rho= \sum_i \lambda_i |i\rangle\langle i|$$
where the vectors $|i\rangle\in\mathbb{C}^{2^n}$ are orthogonal and $\lambda_i$ a non negative real eigenvalue of $\rho$.
Does this mean, that there is always a orthonormal basis $|j\rangle$ of $\mathbb{C}^{2^n}$ such that $\rho$ can be written as
$$\rho= \sum_j \lambda_j |j\rangle\langle j|?$$ Is Nielsen and Chuangs way of putting it totally identical to my thought? Am I making up a difference where there is (practically) none?
For me, one still needs a proof for this - but I can't put it in precise words: Obviously one could find a $\rho$ such that the used $|j\rangle$ (the ones with $\lambda_j\neq 0$) give a basis. But to me it is necessary to say why. That's where I am stuck.
We want to show the following two statements are equivalent:
Proof: $1 \implies 2$
It is clear from the decomposition that $$ \rho |i \rangle = \lambda_i |i \rangle $$ so $|i\rangle$ are the eigenvectors with corresponding eigenvalues of $\rho$ moreover by (1) they form an orthonormal basis.
$2 \implies 1$.
In the orthonormal basis $|i\rangle$ a general matrix $\rho$ can be written as $$ \rho = \sum_{ij} c_{ij} |i \rangle \langle j |\,. $$ As $|i\rangle$ are the eigenvectors of $\rho$ we know that we must have $\rho |i\rangle = \lambda_i |i\rangle$. And because the eigenvalues form an orthonormal basis we therefore need that
$$ \delta_{kl} \lambda_l = \langle k | \rho |l \rangle = \langle k | \sum_{ij} c_{ij} |i\rangle \langle j| |l \rangle = c_{kl} $$ and therefore $$ c_{ij} = \begin{cases} 0 \qquad &\text{ if }i \neq j \\ \lambda_i \qquad &\text{ if } i = j \end{cases} $$ therefore $$ \rho = \sum_{i} \lambda_i |i \rangle \langle i|\,. $$