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When a given function/equation is spanned by there respective space, i.e $\mathbb{R^n}$, $\mathbb{F^n}$, $\mathbb{T^n}$, etc. doesn't it mean that it forms a basis too?

For example:

The individual monomials $(sinx)^j(cosx)^k$ span $\mathbb{T^n}$, where $\mathbb{T^n}$ denotes the trigonometric polynomials, but it does not form a basis owing to identities stemming from the basic trigonometric formula $cos^2x + sin^2x = 1$.

Why is that? I thought if it spans it means there is a linear combination for that particular space, hence there will be a basis?

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    Not all spanning sets are basis, but certainly there will be a _subset_ of the spanning set which is linearly independent and hence forms a basis.2012-10-20
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    Iff axiom of choice holds....2012-10-20
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    @Euyu can you clarify what differentiates a basis from a span. I thought essentially they meant the same thing.2012-10-20
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    @CameronBuie what do you mean by iff axiom holds?2012-10-20
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    @diimension What Cameron Buie is referring to is the fact that _every vector space has a basis_ holds under the assumption of the axiom of choice. Perhaps we don't need to invoke that since we are working with finite dimension vector spaces.2012-10-20
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    To answer your other question, a spanning set is different from a basis. A basis if you recall, is a set which is both **spanning** _and_ **linearly independent**. Simply having the spanning property is not enough to guarantee it to be a basis.2012-10-20
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    @EuYu thank you very much, you are great! I will look it up in more detail in wiki.2012-10-20
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    "When a given function/equation is spanned by their respective fields" is the mathematical equivalent of "Colorless green ideas sleep furiously". All the parts of speech are correct, but the specific words make no sense. Fields don't span things. Things don't span functions. Things don't span equations. This is important --- you don't have a ghost of a chance of understanding these concepts as long as you throw them around so carelessly.2012-10-20
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    @GerryMyerson I think you exaggerated I bit much there but I apologize for my ignorance. I will try my best to learn the terminology correctly.2012-10-20

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In $\mathbb R^2, \{(0,1),(1,0),(1,1)\}$ is as spanning set, as every vector can be expressed as a linear combination of these, but it is not a basis as it is not linearly independent. This is an example of the theorem that all bases of finite dimensional vector spaces have the same number of components. Here we have one too many. Any two of the three are a basis.

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    Makes a lot more sense now. Especially your last two sentences! Thank you very much!2012-10-20