Let T in L(V) be an idempotent linear operator on a finite dimensional inner product space. What does it mean for T to be "the orthogonal projection onto its image"?
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Eh? What do you think a projection onto a subspace is? To me it *is* an idempotent linear map. Then nice thing about having an inner product (non-degenerat) around is that there is an obvious choice of complementary subspace
[tex]T(v)=Proj_{im(T)}(v)[/tex] for all v? And not knowing what the image of T is a priori does not matter? And how does it follow that each vector in V is a combination of some vector in its image and its nullspace?
a projection is amap onto a subspace, that sends every vector to a vector in the subspace, and leaves vectorsd that are already in the subspace where they are.
so if V is a vector space and X is a subspace we want to project on, and if
Y is any complementary subspace, i.e. X and Y together generate V, and X and Y have only the zewro vector in common, then every vector in V can be written uniquely in the form x+y where x is in X and y is in Y.
Then the map f sending x+y to x, is a projection onto X, "along" Y.
Notice that f(f(v)) = v for all v, since once v gets into X it stays put. And also Y = ker(f), since vectors in Y go to zero.
Indeed any linear map f such that f^2 = f is asuch a projection.
f is called an "orthogonal" projection if Y = ker(f) is orthogonal to X = im(f).\
at least i think so, you should of course verify everything by proving all these either trivial or false statements.
I figured it out. Of course I assumed that what is meant by "orthogonal projection onto its image" is that [tex]T(v)=Proj_{im(T)}(v)[/tex]. Thanks for the help.
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