Dimension of the space of all linear functions

[Zermelo]

Hi,
I have a question about the following:

let V be a n-dimensional vector space, and L(V) the set (space) of all linear functions on the space V. Then [MATH]\dim L(V) = n\cdot n = n^2[/MATH]I have no idea where this statement comes from. It’s really strange even thinking of L(V) as a vector space, and even stranger to thing about it’s basis vectors and dimension. I have a feeling that it might be n*n because there are n dimensions in the domain of any function [MATH]f \in L(V)[/MATH], and also n dimensions in the codomain, aka [MATH]\forall f \in L(V),\ f:L(V) \rightarrow L(V)[/MATH]. What even could be the basis vectors of this space? I know that [MATH]\{ 1, x, x^2, ... , x^n \} [/MATH] is the basis of the space of all polynomials on order n, maybe we could construct a base in L(V) in a similar way?

 

[Steven G]

Pick some finite dimensional vector space V and look at what L(V) would be. Please think about this some more. You can do it!!

 

[Zermelo]

Pick some finite dimensional vector space V and look at what L(V) would be. Please think about this some more. You can do it!!

Okay thanks for the push, maybe this:
Let’s consider [MATH]R^2[/MATH], and the L space of linear functions on that space.
Let’s consider 4 functions, aka the set
[MATH]B = \{(x,y) \rightarrow (x,0), (x,y) \rightarrow (0,y), (x,y) \rightarrow (0,x), (x,y) \rightarrow (y,0).\}[/MATH]Now let’s consider an arbitrary function [MATH]f \in L, f(a (x,y) + b (z,t)) = af(x,y) + bf(z,t). [/MATH] This is a very general condition, and I don’t really see how can I express an arbitrary function f from L. We need to show that any function from this set can be expressed as a linear combination of functions from B. Let’s observe the span of these functions, the span is the set of all functions [MATH](x,y) \rightarrow (a x + by, cx + dy)a,b,c,d \in R[/MATH]. It is trivial to show that every one of these functions is linear, aka [MATH]span(B) \subseteq L[/MATH], but I don’t know how to prove the opposite.
This can be easily generalised to any vector space, and we get that for any n standard basis vectors in the domain ( aka (1, 0,...,0), (0,1,0,...,0) ... ) there are n “image” vectors in the codomain, thus dim(L) = n*n, if we follow the rule of permutations with repetition. Not completely proven but close enough, I think at least ?