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

$$T(v)=Proj_{im(T)}(v)$$ 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 $$T(v)=Proj_{im(T)}(v)$$. Thanks for the help.