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The statement of the theorem does not involve dimension, except in the innocuous looking qualification that V should be finite-dimensional, so one might think that there was a way of doing things more canonically, and avoiding the unpleasantly arbitrary choice of basis. For example, let V be the space of all infinite real sequences with only finitely many non-zero terms. A sketch of this is as follows. If you have some infinite real sequences v 1 ,v 2 , The resulting sequence is not a finite linear combination of the v i.

Unfortunately, we must now rely on a further piece of theory: every vector space has a basis. If the vector space V is infinite-dimensional, then this means that V contains a subset B such that every vector v in V is a linear combination of finitely many elements of B, and any finite subset of B is linearly independent in the usual sense.

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This statement requires the axiom of choice for its proof. It is also true that every linearly independent subset of V can be extended to a basis. This shows that the natural embedding is not an isomorphism, and that is enough to indicate that there will not be an easy proof for finite-dimensional vector spaces since such a proof would use the natural embedding. This justifies the use of Step 7 in the proof.

One might still ask for Step 10 to be simplified. To do this we took a basis of V and used an element of the dual basis. However, the choice of basis was non-canonical and the statement looks pretty obvious.

• Definition:Isomorphism (Hilbert Spaces).
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