# Does ${\rm tr}(\Pi \rho) = 1$ imply $\Pi\rho\Pi=\rho$?

Suppose I have a density matrix $$\rho$$ and an orthogonal projector $$\Pi$$. Is it true that, if $$tr(\Pi \rho) = 1$$ then it must hold that $$\Pi \rho \Pi = \rho$$?

If yes, how can I prove it?

Yes its true. Define another orthogonal projector $$\Pi_\perp$$ such that $$\Pi + \Pi_\perp = I$$ and write $$\rho$$ in terms of a spectral decomposition $$$$\rho = \sum_k \lambda_k(\rho) |\lambda_k\rangle \langle \lambda_k| \tag{1}$$$$
where we're fine to just consider components such that $$\lambda_k(\rho) >0$$. Then we have $$1 = \text{Tr}(\Pi \rho) = \text{Tr}((I - \Pi_\perp) \rho) = 1 - \text{Tr}(\Pi_\perp \rho) \tag{2}$$ implying \begin{align} 0 &= \text{Tr}(\Pi_\perp \rho)\tag{3} \\&= \text{Tr}(\Pi_\perp \rho\Pi_\perp)\tag{4} \\&= \sum_k \lambda_k(\rho)\langle \lambda_k|\Pi_\perp\Pi_\perp |\lambda_k\rangle \tag{5}\\&= \sum_k \lambda_k(\rho) \lVert\Pi_\perp |\lambda_k\rangle \rVert^2\tag{6} \end{align} where line $$(4)$$ used the cyclic property of the trace and $$\Pi_\perp \Pi_\perp = \Pi_\perp$$. Now since $$\lambda_k(\rho) >0$$ and $$\lVert \cdot \lVert \geq 0$$ the above only holds if $$\Pi_\perp |\lambda_k\rangle = 0$$ for all $$k$$. And so \begin{align} \Pi \rho \Pi &= (I - \Pi_\perp) \rho (I - \Pi_\perp)\tag{7} \\&=\rho + \sum_k \lambda_k(\rho) \Bigl(-|\lambda_k\rangle \langle \lambda_k| \Pi_\perp -\Pi_\perp |\lambda_k\rangle \langle \lambda_k| + \Pi_\perp |\lambda_k\rangle \langle \lambda_k| \Pi_\perp\Bigr)\tag{8} \\&= \rho \tag{9} \end{align}
Write the eigendecomposition of the state as $$\rho=\sum_k p_k u_k u_k^\dagger$$, where $$\{u_k\}_k$$ is a family of orthonormal vectors in the underlying space.
Suppose there is some $$u_k\notin \operatorname{supp}(\Pi)$$, that is, some $$u_k$$ such that $$\Pi u_k\neq u_k$$. Then $$\operatorname{Tr}(\Pi u_k u_k^\dagger)=\| \Pi u_k\|^2<\|u_k\|^2=1,$$ and thus $$\operatorname{Tr}(\Pi\rho)=\sum_k p_k \operatorname{Tr}(\Pi u_k u_k^\dagger)< \sum_k p_k=1.$$ It follows that if $$\operatorname{Tr}(\Pi\rho)=1$$, each $$u_k$$ must be in the support of $$\Pi$$, and thus $$\Pi\rho\Pi=\rho$$.