Projection operators and their subspaces (of Hilbert space)

I've been watching Susskind's lectures on Quantum Entanglement, and something he said regarding (non-)commuting projection operators confused me.

Consider two subspaces {$|a>$} and {$|b>$} of Hilbert space, with operators $K$ and $L$ for which:

• $K |a> = \lambda |a> (1)$
• $L |b> = \mu |b> (2)$

Now considers operators $P_K$ and $P_L$ that project any vector in Hilbert space onto their respective subspaces, that is:

• $K (P_K |\psi>) = \lambda (P_K |\psi>)$
• $L (P_L |\psi>) = \mu (P_L |\psi>)$

We want to find simultaneous eigenstates of both $K$ and $L$. If $P_K$ and $P_L$ commute: $P_K (P_L |\psi>) = P_L (P_K |\psi>)$. Now the left-hand satiesfies $(1)$, and the right-hand side satisfies $(2)$, so these are the required states.

In fact, if $P_K$ and $P_L$ operators commute, they share a complete set of eigenstates. The eigenstates of projection operators are those that span the subspace they project onto, so apparently $P_K$ and $P_L$ project onto the same subspace, which means they're the same operator? Then, is the statement: "projection operators commute $\rightarrow$ they're the same" correct, or do they somehow project states onto the same subspace in a different way?

Furthermore, we can imagine the subspaces geometrically as 'planes', and where these planes intersect we can find states that satisfy both $(1)$ and $(2)$. Now, according to Susskind, if $P_K$ and $P_L$ do not commute, finding such states is impossible. If the previous paragraph holds (does it?), then them commuting implies the intersection of their subspaces is the entire subspace. I don't know what non-commuting means geometrically, but shouldn't there be a case where the intersection of their subspaces isn't the entire subspace (for example, imagine two 2D perpendicular planes intersecting eachother on a 1D line)? Susskind's comment seems to contradict that, and can't see exactly where I'm going wrongly.

• Think of projection operators onto spin eigenstates; $\frac{1}{2} ( 1 + \sigma_i)$ – taupunkt Apr 12 '14 at 23:16