### 1. Coordinate maps

consider a random vector ** a**. If the values of every entry are explicitly given, one could naturally identify it as a vector under the orthonormal basis. However, it could also been understood as the

*coordinate vector*, which represents a vector under another basis by a

*coordinate map*:

### 2. Change of basis of a linear map

Now we consider a linear map (represented by a matrix) A. It could be interpreted as:

What if we want to change the basis of the objective vector space of *f* ? The following relation is obvious, if we want to find a matrix *A’* representing the map after change of basis:

Where *P* is defined as a matrix that perform this basis change.

**[Literature: Andre Lukas, Lecture note on Vectors and Matrices, University of Oxford]**

### 3. The invariant map

Suppose we have a map that is invariant under any basis change, that is:

In other words, we would like to find an operator that commutes with any other operator on the same vector space *V*. Suppose now we have a vector ** x** in

*V*. We would like to find a non-trivial linear functional on

*, so that we can define a linear map:*

**x**This is possible, for a functional in a *n *dimensional space can be express in to an *(1 x n)* matrix, so the left hand side can be expressed as *(nx1)(1xn)(nx1)* corresponding to ** v**,

*f*, and

**respectively, and the former 2 combined and form a matrix. (**

*x***)**

*My reference mentioned ” Axiom of choice” with respect to finding non-trivial functional, and yet I have not understood it perfectly*Then, the commutation relation implies:

Note that *f(T x)*, according to our pervious argument, should exist and independent on

**. Thus, what we are doing consequently is that we have constructed linear maps**

*v**P*according to our need (that is, the vector

**). In other words, we can assign every vector**

*v***in**

*v**V*a matrix

*P*and they have to satisfy (6).

Then this inplies:

*i.e.* ** T is the scalar multiple of the identity**.

**[Literature: Robert Isreal, https://math.stackexchange.com/questions/27808/a-linear-operator-commuting-with-all-such-operators-is-a-scalar-multiple-of-the]**