# Two projection operators may not be commutative, with an explicit example.

A projection operator is a linear mapping from a linear space X to X that then satisfies $p^2=p$

Then in a geometrical point of view, looking at the linear space, any element in $X$ mapping under $p$ only has two possibilities, one is $px=x$, and under the action of $p$ image of $x$ is himself. The second possibility, x under the action of $p$ is not itself but another element $y$, then $p$ in action, on $y$ should also be equal to $y$ because we know that $p^2x=px = y$ so $py=p^2x=px = y$.

Well, then this way, if we only look at it from the perspective of mapping between sets, we have to the following figure. from the perspective of mapping between sets, figure of mapping $p$

But we are not from the set because the projection $p$ is a linear space of an operator, so we should look at a set of bases in this linear space $X$, and see what happens to $p$ on this set of bases, and then we know that there is a part of it that is not moving, and then another part will be mapped to this part that is not moving.

Now if there are two projections p1 p2, of course they are not necessarily exchanged, because if I am not moving, because I am I am certainly not necessarily exchanged, because I two projection operators, for example OK, for example, even where they are not moving, is the same place they are not necessarily exchanged, because OK for example, I first, for example, for the two of them to move together That piece, they are certainly the same role, but for both of them is. For both of them that is to say there is an X, and then I I send this student this P with P2, X in px = Y1px = 2 ~ X does not = Y1X does not = 2, that this px1OK, then in case must = yy = Y, so that the place where they two holes or where they hit the past is the same place, must be immobile to accept, if that. But if say two projection arithmetic amount, on but if say two projection arithmetic.