Show that a finite D-dimensional Hilbert Space has $D^2$ operators

Question

I'm studying Group Theory and i've just arrived at this chapter about Hilbert Spaces and the author states that a this finite Hilbert Space has $D^2$ operators, one being the trivial one, by completeness: \begin{equation} \sum_{j=0}^D\,|\,j\,\rangle\,\langle\,j\,|=1 \quad,\quad\langle\,j\,|\,k\,\rangle=\delta_{jk} \end{equation} So, it leaves $(D^2-1)$ non-trivial linear operators.

My questions are:

1. How do I prove that for this space, it has $D^2$ operators?
2. The author also states that it has $(D-1)$ diagonal combinations such that: \begin{equation} (|1\rangle\langle1|-|2\rangle\langle2|),...,(|1\rangle\langle1|+|2\rangle\langle2|+...+|D-1\rangle\langle D-1|-(D-1)|D\rangle\langle D|) \end{equation} How can I show this?

Thanks in advance for any help, and by the way, it's not a homework. It's a study of my own to my undergraduate thesis.

Answer 1

Hopefully you are misquoting: what the author means is that there are $D^2$ linearly independent linear operators, forming a basis of the linear operators on this $D$-dimensional Hilbert space. For example, consider the operators $| i \rangle \langle j|$ for $i, j = 1 \ldots D$.

Answer 2

A Hilbert space has many operators. It seems that the author is suggesting that the operators on a finite dimensional Hilbert space (often denoted $B(\mathcal{H})$) are spanned by $D^2$ linearly independent operators. A canonical choice for these operators are $|i\rangle \langle j|$, sometimes written $E_{ij}$.