# Understanding canonical basis

1. Sep 23, 2014

### rkaminski

In many places the canonical basis is defined as a set of vectors with coordinates as:
$$\boldsymbol{e}_i=(0,...,1,...0)$$
where "1" is on the i-th place. In my undestanding of such definicion every basis is canonical basis. If we write coordinates of basis vectors in the same basis we will get such sets of coordinates for any possible basis we choose in n-dimensional vector space. Is there any inconsistency in my thinking and understanding of what a canonical basis is?

2. Sep 23, 2014

### FactChecker

No. A basis for the (x,y) plane is (1,0) and (0,2). That is not canonical. There are infinitely many more (1,-1) and (1,1) is another non canonical basis.

3. Sep 29, 2014

### rkaminski

OK. So there is my problem, maybe it arises from my bad understanding of the definition. According to the above example, consider a Cartesian coordinate system. It obviously has two basis vectors: (1,0) and (0,1), and this set of vectors fulfills this a 'definition' of the canonical basis. Now, we form two vectors, as suggested above: (1,0) and (0,2). Now, we can find a transformation from one Cartesian basis to the new one. In this new basis, the two new two basis vectors will have new components, namely (1,0) and (0,1). So where is the problem?

4. Sep 29, 2014

### FactChecker

I see what you are saying. Canonical basis vectors are defined in a coordinate system. So (0,2) is not canonical in the original coordinate system. When you talk about transforming that vector to another coordinate system, that opens a can of worms that would only be confusing at this point IMO.

5. Sep 29, 2014

### rkaminski

But is my original 'definition' of the canonical basis wrong or incomplete? Is this definition even needed at all? Because still, my understanding is that all possible set of basis vectors for a given vector space (for simplicity in 2D) can be expressed as (1,0) and (0,1).

6. Sep 29, 2014

### FactChecker

I think that the valid statement is that (1,0) is canonical in the given coordinate system and (0,2) is not.

The mathematical definition of "vector" is too broad for you to say what you want to say and the physics definition is too strict for you to want to say that.

In mathematics, I can legitimately define a "vector" that is (0,2) in any coordinate system. That is a well defined vector and it is (0,2) in every coordinate system. It can not be transformed into anything else and it can never be canonical in any coordinate system.

In physics, I would want a "vector" and transformations to have physical properties that are independent of coordinate system. Your transformation above (sending (1,0) => (1,0) and (0,2) => (0,1) ) has to distort one coordinate without distorting the other. In physics, you might not want to say that that is a valid transformation.

For now, it's probably better to just leave the definition of a canonical basis as (1,0), (0,1) in the given coordinate system and to say that other vectors are not canonical in that coordinate system.

Last edited: Sep 29, 2014
7. Oct 10, 2014

### platetheduke

Math perspective:
The canonical basis of $\mathbb{R}^n$ does not have any special properties, it is just a very useful convention. Given any ordered basis of an $n$-dimensional real vector space we can write our vectors using the coordinates given by our basis, and it will be formally equivalent to working in $\mathbb{R}^n$ under the canonical basis. This is a handy convention in the same way that we write our functions in terms of polynomials and transcendental functions when possible.

Before you say that "the canonical basis interacts well with inner products and other vector operations" I'll mention that all of the operations on vectors can be defined in a coordinate free manner and once reduced to coordinates in the way I mentioned above, they will look like the coordinate specified definitions in the canonical basis, regardless of which basis you are actually using.