*1.Vectors*

It is important to formalised the definition of a **vector** and a** vector space**. Here I shall pay no more attention on the this issue. But they are important as they characterised the condition that, for example, the scalar field accompanying with V, must satisfy.

*1.Functionals*

One should be very familiar with the definition of “**linear**” and “linear maps”(And from the definition we see linear maps can from a linear space: space of linear functions, as well). Very interestingly, we can show the following two important relations: for *f: V→W:*(See reference)

and that if *ker= { 0}*,

*f: V→V*is

**isomorphism**.

We can get the following observation: consider a linear map with *dim(imf)* = 1 (Which, since vector spaces are defined base on a scalar field, this 1 dimensional space is the field itself. We name this special type of map linear functionals. Since we know that linear functions themselves form a vector space, we call it dual space* V *(n,K)* , this apply to linear functionals as well. Form (1) we know that one particular functional actually only act on one specific dimension, we conclude that there are only* dim(V)* types of linearly independent functionals. i.e.* dim(V)=dim(V*)*. It follows that for basis {** e**} in

*V*and basis {

*} in*

**g***V**, we naturally pair them together:

So that *V* and *V* *are putted in an equal status (symmetry).

Now, suppose we have *f: V→W *and* g: W→F *where *g* is a functional, it is easy to find the a functional in *V**, namely, *g*( *f ). *Then, we could also define another space of functionals *h: W*→V* *as *h: g→f = h(g). *

We can also define a map *i: V→V* *and we can show the condition for it to be isomorphism. Since dim*V* = dim *V** if we also know Ker(*i*) = 0 this should do it. Remember that for defining a dual space, it is necessary to recall that it represents a space of linear maps. For any non-zero ** v**, if

*w = f(v)*is non zero (Note that the image of zero is always zero) the corresponding dual vector is (by our definition) non zero. (In other word, this map is not degenerate) However, we shall note that the proportionality between

**and image of**

*e***by**

*e***is not restricted. We could try to more specify it:**

*g*Consider the effect on w, we also use it to define the inner product *<,>: V *x *V→F:*

[Reference: Geometry, Topology and Physics. M Nakahara(2003)]