2 dimensional plane of a 3 dimensional space

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In a 3-dimensional space, the projection operator onto a 2-dimensional subspace acts as an identity operator for that subspace, confirming that the identity operator is also applicable. The projection operator is generally unique, as it is defined by its image. However, it was clarified that the projection is not uniquely determined solely by the image due to the existence of complementary subspaces. This distinction highlights the complexity of projection operators in linear algebra. The discussion emphasizes the need for careful consideration of definitions in mathematical contexts.
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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?
 
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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.
 
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Thanks for the reply. Before asking, I'll give the proofs a try.
 


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

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