Projections: Self-square Automorphisms

A morphism is described as a projection if it is self-composable and equal to its self-composite (ie it is self-square). Thus any identity morphism is a projection: indeed, any projection can be thought of as the identity morphism on a `subset' of the domain on which its identity acts. In practice, the main area in which projections arise is in linear algebra (particularly when used as a tool of quantum mechanics).

If two projections P, Q are composable (in both orders, though this usually follows from their self-composability) and commute with composite Q, we say that P subsumes Q and that Q refines P. One may think of this as saying that Q is the identity on a subset of that on which P acts as identity. We say two morphisms commute if they are composable in either order and the two composites are equal (in which case the two morphisms must be self-composable and composable with their product).

For projectors P, Q, R, if P refines Q and Q refines R we have PQ=P=QP, QR=Q=RQ, whence PR=PQR=PQ=P=QP=RQP=RP, so P refines R: `refines' is a transitive relation.

Projections in Linear Algebra

Linear algebra concerns itself with vector spaces: its morphisms are linear automorphisms. These support scalar magnification, addition and subtraction: they have an additive identity, 0 (which maps any vector to the vector space's additive identity) whose composite with any other automorphism is 0. In particular, it is a projection: and it refines any other projection. We describe a linear projection as irreducible precisely if 0 is its only refinement. We say that a linear morphism, P, annihilates another, Q, precisely if PQ=0; note that this need not imply that Q annihilates P. We call the identity automorphism 1.

The presence of subtraction allows us to consider what happens when we subtract a linear projection, P, from 1: we call the resulting linear automorphism the complement of P. We obtain, using PP=P, P(1-P) = P-PP = 0 = (1-P)P, whence (1-P)(1-P) = (1-P) - P(1-P) = 1-P. Thus the complement of a projection is itself a projection; any projection commutes with an annihilates its complement.

We define the range of a linear morphism, (V|P:U), to be the collection of values (in U) it takes, (V|P|); and its kernel to be the collection of values (in V) which it maps to the additive identity of U, (|P|{0}). For a projection, V=U and P acts as the identity on its range because: any vector in its range can be written as Pv for some v; and PPv=Pv. The range of a projection is the subset (a vector subspace) as whose identity we think of P.

Now, any vector, v, in V can be expressed, in terms of some projection (V|P:V), as v = Pv + (1-P)v, a sum of one vector in range(P) and another in range(1-P). Each of these is preserved by one of P and 1-P but annihilated by the other. Consequently, we are able to write V as the Cartesian product of range(P) and range(1-P).

We obtain (range(P)|1-P|) = {(1-P)Px: x in V} = {0}, whence range(P) is a subset of kernel(1-P). But for v in kernel(1-P) we have 0=(1-P)v=v-Pv, so v=Pv and v is in range(P): so kernel(1-P) is a subset of range(P), whence the two must be equal. Likewise, kernel(P) = range(1-P) and our decomposition of V into range(P) and range(1-P) becomes a decomposition into range(P) and kernel(P). Note that, whereas any linear (V|P:U) has V isomorphic to the Cartesian product of P's range and kernel, this provides a specific natural isomorphism between V and this Cartesian product.

Suppose we have two projections, Q and P, on V. Let us look, first, at what happens if they commute, that is PQ = QP (eg when one subsumes the other). Then v in kernel(P) implies Pv = 0 = Q0 = QPv = PQv, whence Qv in kernel(P). Now, since (1-P)Q = Q - PQ = Q - QP = Q(1-P), we can apply the same to kernel(1-P), ie range(P), so that v in range(P) implies Qv in range(P). Thus Q respects the decomposition of V into range(P) and kernel(P). The other consequence of PQ = QP is that (PQ)(PQ) = P(QP)Q = P(PQ)Q = (PP)(QQ) = PQ, so this product is also a projection. Thus, when projections commute each respects the range/kernel decomposition of the other and their product is also a projection.

It is worth noting that the product of two projections may be a projection without the original projections commuting: for instance, in the simple world of two dimensions,

P = (1 0), Q = (1 0) => PQ = P, QP = Q
    (1 0)      (0 0)

so P and Q don't commute, but both their products are projections. It is notable that, with respect to any metric, at least one of these lengthens at least some vectors: ie at least one of P and Q is not normal projection onto some subspace.

Maintained by Eddy.
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