I have been going through the Halmos' Finite dimensional vector space, where in section 42 I got struck at the theorems regarding the combination of projections. I can't understand how the conditions of the linear combination of projections define the conditions for the existence of the projection. In particular,
Suppose $E_1$ and $E_2$ are projections on $U=M_1\oplus N_1$ along $M_1$, and $V=M_2\oplus N_2$ along $M_2$ respectively, and the underlying field has characteristic $\ne2$.
(i) $E_1+E_2$ is a projection iff $E_1E_2=E_2E_1=0$; if this condition is satisfied, then $E_1+E_2$ is the projection on $M$ along $N$, where $M=M_1\oplus M_2$ and $N=N_1 \cap N_2$.
(ii) $E_1-E_2$ is a projection iff $E_1E_2=E_2E_1=E_2$; if this condition is satisfied, then $E_1-E_2$ is the projection on $M$ along $N$, where $M=M_1\cap N_2$ and $N=N_1 \oplus M_2$.
I can't get why the direct sum condition is employed for $M$ and intersection for $N$. It gets more obscure when this condition for $\oplus$ and intersection swaps when we talk about $E_1-E_2$.
I would be really glad if someone could give me some explanation using examples for these theorems. Thanks in advance.
It is always useful to think of these situations in terms of 3-space. As an example think of $E_1$ as the projection on the $z$-axis ($M_1$) along the $xy$-plane ($N_1$), and $E_2$ as the projection on the $x$-axis ($M_2$) along the $yz$-plane ($N_2$). Geometrically, it is easy to see that $E_1+E_2$ is the projection on the $xz$-plane ($M_1\oplus M_2$) along the $y$-axis ($N_1 \cap N_2$).
Algebraically: $E_1(x,y,z)=(0,0,z)$ and $E_2(x,y,z)=(x,0,0)$, so that we have $(E_1+E_2)(x,y,z)=E_1(x,y,z)+E_2(x,y,z)=(x,0,z)$.
Using the same projections for $E_1$ and $E_2$ above you can verify (ii) as well...
Intuitively projections preserve certain subspaces, and "kills" the remaining subspace of the vector space (that forms a direct sum with the preserved space). Now if two projections preserve different subspaces, adding them together will preserve a larger subspace and "kill" off less. (I hope this makes some sense!). And again you can formulate a similar intuitive description of (ii).
As per OP's request: ok, you can use the same spaces as above to verify part (ii), but a more suggestive example perhaps is to let $E_3$ to the projection on the $xz$-plane along the $y$-axis. Then $E_3-E_1$ is a projection, and: $(E_3-E_1)(x,y,z)=E_3(x,y,z)-E_1(x,y,z)=(x,0,z)-(0,0,z)=(x,0,0)$. Geometrically, you are projecting onto the $xz$-plane, but then you remove the $z$-coordinate by subtracting and so in effect you are projecting onto the $x$-axis. So the nullspace of $E_3$ is augmented by the projection space of $E_1$... so you can see if the projection space of the subtracted operator does not coincide with the nullspace of of the first operator we have a direct sum of these spaces as the new nullspace. And then naturally the projected space of the combined operator must decrease...hope this helps.