This is from the exercise 5.7 of the book An Introduction to Quantum Stochastic Calculus by K.R. Parthasarathy. By observable we mean Hermitian Operators on a Hilbert space $H$ and by state we mean positive operator with unit trace.
Here $\rho$ be a state on $H$, let $\rho=\sum\limits_{r=1}^l p_r|u_r\rangle\langle u_r|$ be its spectral resolution where $p_r>0$, $\sum\limits_{r=1}^l p_r=1$, $u_r$'s are unit vectors and $l\le n$.
Then $\text{tr}\ \rho X_iX_j=\sum\limits_{r=1}^l p_r\langle u_r, X_iX_j u_r\rangle$. Therefore, $A_r=\sum\limits_{r=1}^l p_r A_r$ where $A_r=(\langle u_r,X_iX_ju_r\rangle)$.
Let $k_r=\text{rank}(A_r)$. Then $\text{rank}(A)\le\sum\limits_{r=1}^l k_r$.
Now if I can prove that $k_r\le n$, I have $\text{rank}(A)\le ln\le n^2$.
So I just need to prove that the $m\times m$ matrix $A_r=(\langle u_r,X_iX_ju_r\rangle)=(\langle X_iu_r,X_ju_r\rangle)$ has rank not exceeding $n$. If $m\le n$, we have nothing to prove. So assume $m>n$.
So we have observables $X_1,\ldots,X_m$ and a unit vector $u$ in a $n$ dimensional Hilbert space $H$ with $m>n$. We have to show that the matrix $(\langle X_iu,X_ju\rangle)$ has rank not exceeding $n$. I'm stuck at this point.
Can anyone help to prove the above claim? Thanks for your help in advance.
Thanks to John Doe. I'm just expressing his comments in words-
$X_1u,\ldots, X_mu$ are $m$ vectors in a $n$ dimensional space $H$. WLOG, $X_1u,\ldots,X_ku$ are the linear;y independent vectors with $k\le n$. Then for any $i>k$, $X_iu=\lambda_1 X_1u+\cdots+X_ku\implies \langle X_iu, X_iu\rangle =\lambda_1\langle X_1u,X_ju\rangle+\cdots+\lambda_k\langle X_ku,X_ju\rangle$. Hence, $i$-th row of the matrix is linear combination of the first $k$ rows. This shows that rank of the matrix$=k\le n$.