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**BASIS**to a

**FINITE DIMENSIONAL**Vector Space, it is enough to show that it is Linearly Independent.

No Need to prove that it spans the whole vector space?

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- Thread starter Bachelier
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No Need to prove that it spans the whole vector space?

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Fredrik

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Hint: What is the definition of finite-dimensional? A vector space is n-dimensional, if...

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Hint: What is the definition of finite-dimensional? A vector space is n-dimensional, if...

...if it has a basis consisting of a

This is my point, we still have to show that the number of elements in the set is equal to the dimension of the vector space V. For instance if the dim is unknown, then we must show the set spans. Otherwise using just the fact that V is finite dimensional is not enough.

Am I reading this correctly?

Thank you.

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If you are given a dimension for a vector space V, then by the replacement theorem , any linearly independent subset of V that's of size n will form a basis for V

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Fredrik

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OK, with that definition things get circular. The one I had in mind is that dim V=n if there exists a linearly independent subset of V with n members but no linearly independent subset of V with n+1 members. Suppose that dim V=n, and that B is a linearly independent subset of V with n members. Let x be an arbitrary member of V. If x is in B, then x is obviously a linear combination of the members of B. If x is in V-B, then [itex]B\cup\{x\}[/itex] is linearly dependent (since it has n+1 members), so x is a linear combination of the members of B in this case too....if it has a basis consisting of afinite number of vectors. In this casenvectors.

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Landau

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The set

[tex]\{(1,0,0),(0,1,0)\}\subset\mathbb{R}^3[/tex]

is linearly independent, but not a basis.

The set

[tex]\{(1,0,0),(0,1,0),(0,0,1)\}\subset\mathbb{R}^3[/tex]

is also linearly independent, but moreover it spans R^3, hence it is a basis.

Of course, if you know the dimension of the vector space, say n, then in order to prove that some set consisting of precisely n vectors is a basis, it IS enough only to show linear independence. But that was not included in the question.

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Fredrik

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Great answers everbody. Thank you very much.

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