Let $\rho$ be a density matrix (positive semidefinite and trace $1$), with its rank being $1$ such that \begin{equation} \rho = v v^{*}, \end{equation} where $v$ is a $n \times 1$ unit vector.

Let $M$ be a positive semidefinite matrix such that all its eigenvalues are between $0$ and $1$.

I am trying to see whether the following two inequalities are correct: \begin{equation} \text{Tr}\left(M^{2} \rho\right) \leq \text{Tr}\left(M \rho\right). \end{equation}

\begin{equation} || M v|| \leq \text{Tr}\left(M \rho\right), \end{equation} where $||\cdot||$ is the $2$-norm of a vector.

The statement is obviously true when $M$ is a projector. But what about more general $M$? Note that if the first inequality is true, it implies the second one.


1 Answer 1


Recall the eigendecomposition $M= P\Lambda P'$. Since $$\mathrm{trace}(M^2\rho) = \mathrm{trace}(v^*M^2v) = v^*M^2v$$ and $$\mathrm{trace}(M\rho) = v^*Mv,$$ we have that $$v^*(M - M^2)v = v^*(P\Lambda P' - P\Lambda^2 P')v = v^*P(\Lambda - \Lambda^2)P'v\geq 0$$ as the difference $\Lambda-\Lambda^2$ is positive semidefinite (because the eigenvalues are bounded by 0 from below and by 1 from above).

Note that your assumptions on $v$ are not necessary.


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