For a bipartite operator $M\in L(H_{AB})$, suppose $0\leq M\leq \mathbb{I}$. Prove $M^{AB}\leq M^A\otimes \mathbb{I}$

As stated in the title, let $$M$$ be a linear operator on a finite bipartite Hilbert space. Suppose $$0\leq M^{AB}\leq \mathbb{I}$$ and $$0\leq M^A,M^B\leq\mathbb{I}$$, where $$M^A=\mathrm{Tr}_B\left(M^{AB}\right)$$ and $$M^B=\mathrm{Tr}_A\left(M^{AB}\right)$$. Is it always true that $$M^{AB} \leq M^A\otimes \mathbb{I}_B?$$ It trivially holds for product operators, that is $$M^{AB} = M^A\otimes M^B$$, but the general statement is not clear to me.

Any help is appreciated.

Just take Bell state $$M^{AB} = |v\rangle\langle v|$$, where $$|v\rangle = \frac{1}{\sqrt{2}}(|00\rangle + |11\rangle)$$. It has eigenvalue 1.
But $$M^A \otimes I_2= \frac{1}{2}I_2 \otimes I_2 = \frac{1}{2}I_4$$.
• $M^{AB}$ doesn't even have to be entangled, it can be the product of two pure states and still have eigenvalue 1. Jun 2, 2021 at 20:41
• Oh, you are assuming that $M^A=\mathrm{Tr}_B (M^{AB})$? That wasn't explicit in the question but it definitely helps motivate the question. Jun 2, 2021 at 22:14
No, but it is true that $$M^{AB} \le d\, M^A \otimes \mathbb{I}_B,$$ where $$d$$ is the minimum of $$d_A$$ and $$d_B$$. This inequality is tight in the sense that entangled states saturate it, as the other answer shows. A proof can be found in Appendix A of this paper.