2 dimensional plane of a 3 dimensional space

[Gear300]

For a subspace of Rⁿ, such as a 2 dimensional plane of a 3 dimensional space, wouldn't the projection operator onto the subspace act as an identity operator for that subspace, while at the same time, (I) would also be a (trivial) identity operator?

Furthermore, is a projection operator for a subspace unique?

 

[losiu99]

Yes, by the definition of projection operator, it's an identity on its image. Undoubtedly, I is also an identity on this subspace.
[STRIKE]Yes, a projection operator is uniquely determined by its image. It follows (easily) from the definition as well.[/STRIKE]
Hope that helps. Should you need proofs, just say.

 

[Gear300]

Thanks for the reply. Before asking, I'll give the proofs a try.

 

[losiu99]

I apologise, the projection is NOT uniquely determined by the image, because complementary subspace is not. Sorry if I wasted your time