# Given that for every valid density matrix $\rho$, $\text{Tr}(M\rho) = 1$; what can we conclude about matrix $M$?

My intuition says that $$M$$ has to be the identity matrix, but I am not able to show it rigorously. I tried playing around using spectral decomposition. If $$\rho = \sum_i \lambda_i |\lambda_i \rangle \langle \lambda_i|\,,$$ then we get the condition $$\sum_i \lambda_i M_{ii} = 1$$, where $$\lambda_i$$s are the eigenvalues of the density matrix and $$M_{ii}$$ are the diagonal matrix elements of $$M$$ in the eigenbasis of $$\rho$$.

This should be true for any $$\{ \lambda_i \}$$ which sums to 1 with diagonal matrix elements $$M_{ii}$$ in the corresponding eigenbasis.

TL;DR: Yes, if $$\mathrm{tr}(M\rho)=1$$ for every density matrix $$\rho$$, then $$M=I$$.
We need three facts. First, the inner products of any element $$A$$ of a Hilbert space with the elements of a basis $$\mathcal{B}$$ determine $$A$$ uniquely$$^1$$. Second, $$\langle A,B\rangle_{HS}:=\mathrm{tr}(A^\dagger B)$$ is an inner product$$^2$$. Third, the set of density matrices $$\mathcal{D}$$ contains$$^3$$ a basis$$^4$$ $$\mathcal{Q}$$. Thus, just the finite set of equations $$\mathrm{tr}(M\rho)=1$$ for $$\rho\in\mathcal{Q}\subset\mathcal{D}$$ fixes $$M$$ uniquely, let alone the infinite set $$\mathrm{tr}(M\rho)=1$$ for $$\rho\in\mathcal{D}$$.
$$^1$$ This follows from the fact that these inner products are coefficients in the representation of $$A$$ as a linear combination of the elements of the basis dual to $$\mathcal{B}$$.
$$^2$$ Also known as the Hilbert-Schmidt inner product.
$$^3$$ Exercise. Hint: Construct the one-hot basis from projectors.
$$^4$$ Unfortunately, none of the bases in $$\mathcal{D}$$ are orthogonal with respect to $$\langle.,.\rangle_{HS}$$, which is why we don't assume it in $$^1$$ and instead resort to the dual basis.
• There exist a basis $\mathcal{Q}$ of the complex vector space of all linear operators on the Hilbert space in which every element is a valid density matrix. For example, in the single-qubit case we can take $\mathcal{Q}=\{|0\rangle\langle 0|, |1\rangle\langle 1|, |+\rangle\langle +|, |{+i}\rangle\langle{+i}|\}$. Sep 21, 2023 at 23:55