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A projection $P$ on a real vector space is defined to be a linear mapping such that $P^2 = P$. For projections on complex vector spaces why does one require the extra condition that $P^* = P$, where $P*$ is the Hermitian of $P$?

edited Nov 20 2009 at 8:53

asked Nov 20 2009 at 8:34

Aston Symthe
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You don;t mean Hermitian; you mean "adjoint". See also Darsh's answer below. – Yemon Choi Nov 20 2009 at 9:11

Darsh Ranjan · Answer 1 · 2009-11-20 09:09:04Z UTC

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You're getting your concepts confused. $P^2=P$ defines projections in any vector space; the additional constraint $P^*=P$ only makes sense in inner product spaces and defines orthogonal projections. (Projections don't have to be orthogonal.)

answered Nov 20 2009 at 9:09

Darsh Ranjan
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Don't you mean unitary if we're talking about complex vector spaces? – John McCarthy Nov 20 2009 at 9:16

You could be right, but I've only heard the phrase "orthogonal projection," even for complex vector spaces. I've only heard the word "unitary" used in the context of unitary transformations, i. e., ones that satisfy UU* = U*U = I. If the word applies here, then I'm happy to edit my answer... – Darsh Ranjan Nov 20 2009 at 9:34

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@John: In the context of inner product spaces and operators on them, unitary almost always means what Darsh has said. At least, this is in my experience. (The word orthogonal is being used in the sense of "mutually orthogonal vector", not in the O(n) sense) – Yemon Choi Nov 20 2009 at 9:38

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Over a real vector space, one sometimes requires that P^T=P. These are called the orthogonal projections. This is precisely analogous to requiring that P^*=P in a complex vector space. – David Speyer Nov 20 2009 at 13:05

Real and Complex Projections

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