Orthogonal projection

Treadstone 71

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"?

matt grime

Every element in v is a combination e+f where e is in the image of T and f in the kernel and T(e+f)=e

Treadstone 71

So T(e)=e?

What does it mean in terms of projections?

matt grime

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

Treadstone 71

What about the identity transformation? It's not an orthogonal projection of anything.

matt grime

Yes it is, onto the subspace itself.

Treadstone 71

Just to be certain, is it equivalent to saying

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?

mathwonk

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.

Treadstone 71

I figured it out. Of course I assumed that what is meant by "orthogonal projection onto its image" is that . Thanks for the help.