# Prove that $P_{M_1}P_{M_2}= P_{M_2}P_{M_1}$ implies $\text{Pr}(\text{span}[M_1, M_2]) = \text{Pr}(M_1)+\text{Pr}(M_2)−\text{Pr}(M_1\cap M_2)$

Prove that if

$$\text{Proj}_{M_1}\text{Proj}_{M_2}= \text{Proj}_{M_2}\text{Proj}_{M_1}$$ then

$$\text{Pr}(\text{span}[M_1, M_2]) = \text{Pr}(M_1) + \text{Pr}(M_2) − \text{Pr}(M_1 \cap M_2)$$.

In the case where the projection operators are non-commutative, I understand how to show that the above formula is actually false. However, I am unsure of how the commutativity of the projectors implies the above equation. I have included an image from the text to provide the definitions of $$M_1$$ and $$M_2$$.

1. It follows that $$M_1$$ is invariant subspace of operator $$\text{Proj}_{M_2}$$. Indeed, if $$v \in M_1$$ then $$\text{Proj}_{M_1}\text{Proj}_{M_2}v=\text{Proj}_{M_2}\text{Proj}_{M_1}v = \text{Proj}_{M_2}v,$$ so $$\text{Proj}_{M_1}(\text{Proj}_{M_2}v) = \text{Proj}_{M_2}v,$$ but this can happen only if $$\text{Proj}_{M_2}v \in M_1$$.
2. Similarly, it can be proved that $$M_1^{\perp}$$ is invariant for $$\text{Proj}_{M_2}$$, and also $$M_2,M_2^{\perp}$$ are invariant for $$\text{Proj}_{M_1}$$ by symmetry.
3. Since $$M_1$$ is invariant for $$\text{Proj}_{M_2}$$ and $$\text{Proj}_{M_2}$$ has two eigenspaces $$M_2,M_2^{\perp}$$, then $$M_1$$ can be split into $$M_1 = (M_1 \cap M_2) \oplus (M_1 \cap M_2^{\perp})$$ Similarly, $$M_2 = (M_2 \cap M_1) \oplus (M_2 \cap M_1^{\perp})$$
4. Now, clearly, $$(M_1 \cap M_2^{\perp}) \perp (M_2 \cap M_1^{\perp})$$, so $$M_1 + M_2 = (M_1 \cap M_2) \oplus (M_1 \cap M_2^{\perp}) \oplus (M_2 \cap M_1^{\perp}),$$ hence $$\text{Pr}(M_1+M_2) = \text{Pr}(M_1 \cap M_2) + \text{Pr}(M_1 \cap M_2^{\perp}) + \text{Pr}(M_2 \cap M_1^{\perp}) =$$ $$= \text{Pr}(M_1) + \text{Pr}(M_2) - \text{Pr}(M_1 \cap M_2)$$
• Thank you! Is $M_1$ invariant under $Proj_{M_2}$ because the operator projects onto a space with basis vectors $|0>$ and $|1>$ and $M_1$ is a space that has only $|0>$ as a basis? Apr 12 '20 at 18:22
• No, my answer is general. Also for $M_1, M_2$ from that exercise there will be no commutativity of projectors. Apr 12 '20 at 19:01