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matt grime

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So T(e)=e?

What does it mean in terms of projections?

What does it mean in terms of projections?

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matt grime

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

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matt grime

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Yes it is, onto the subspace itself.

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Just to be certain, is it equivalent to saying

[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?

[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?

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mathwonk

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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.

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