1
$\begingroup$

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?

  • 0
    @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

1 Answers 1

2

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.

  • 0
    Makes a lot more sense now. Especially your last two sentences! Thank you very much!2012-10-20