- #1

fatpotato

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**Summary::**Problem interpreting a vector space of functions f such that f: S={1} -> R

Hello,

Another question related to Jim Hefferon' Linear Algebra free book. Before explaining what I don't understand, here is the problem :

I have trouble understanding how the dimension of resulting space for part (a) has to be 1 and not infinity. Generally speaking, I can't wrap my head around how this can represent a vector space.

If I understand correctly, this definition means that the only tolerated argument for any function ##f: S=\{1\} \rightarrow \mathbb{R}## has to be the value 1. However, if we can find a function ##f## such that ##f(1) = g(2)##, wouldn't we be able to span a space of dimension 2 without violating our rule? Thus, we could span any space as long as we can construct a function that behaves like previous equation, but for any argument. What am I misunderstanding here?

For part (b), I tried to come up with an analogy using my understanding of basis. As an example, if ##B = <\vec{e_1},\vec{e_2}> ##, namely ##B## is the basis containing the set of natural basis vectors in ##\mathbb{R}^2##, and any linear combination can be described as ##c_1\vec{e_1} + c_2 \vec{e_2}##, could we see some sort of similar behaviour where two functions ##f,g## are analogous to the coefficients ##c_1, c_2## and the arguments ##(1),(2)## are analogous to the vectors ##\vec{e_1},\vec{e_2}##? So, any two functions ##f,g## would constitute a basis of this space as long as we evaluate them with arguments ##(1)## and ##(2)## respectively?

To go on with the analogy in ## \mathbb{R}^2##, this would mean that for any function ##f: S=\{1,2\} \rightarrow \mathbb{R}##, the resulting spanned space would be some kind of 2D plane where each coordinate is defined by taking the sum of any two functions (possibly the same one) and evaluating them with the arguments ##(1)## and ##(2)##?

I know that my analogy might not be the best, but I don't know how to grasp this concept better.

Any help is more than welcome. Thanks in advance.

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