Taking vectors from one basis to another [Byron and Fuller]

In summary: So the above notation is just a way of writing ##\delta_{i j}##, and it's not necessary to write out the transpose operation every time.However, if the matrix A is not orthogonal, then the right-hand side of the first equality will not be equal to ##\delta_{ij}##. In that case, we need to use the transpose operation to get the right-hand side of the first equality.
  • #1
Elwin.Martin
207
0
B&F have the following:
[itex]\delta_{i j} = e'_i \cdot e'_j = a_{i k} \left( e_k \cdot e'_j \right) = a_{i k} a_{j k} [/itex]

and they ask the reader to show that
[itex] a_{k i} a_{k j} = \delta_{i j} [/itex]

Does it suffice to show the following? :

[itex] \delta_{i j} = a_{i k} a_{j k} \to \delta_{j i} = \left( a_{i k} a_{j k} \right)^T = a_{k i} a_{k j} [/itex]
and note that

[itex] \delta_{j i} = \delta_{i j} [/itex]
?

I believe this is sufficient, but I have a feeling I short-cut this.
 
Physics news on Phys.org
  • #2
Elwin.Martin said:
B&F have the following:
[itex]\delta_{i j} = e'_i \cdot e'_j = a_{i k} \left( e_k \cdot e'_j \right) = a_{i k} a_{j k} [/itex]

and they ask the reader to show that
[itex] a_{k i} a_{k j} = \delta_{i j} [/itex]

Does it suffice to show the following? :

[itex] \delta_{i j} = a_{i k} a_{j k} \to \delta_{j i} = \left( a_{i k} a_{j k} \right)^T = a_{k i} a_{k j} [/itex]
and note that

[itex] \delta_{j i} = \delta_{i j} [/itex]
?

I believe this is sufficient, but I have a feeling I short-cut this.

Hey Elwin.Martin.

One suggestion for you is to use the fact that the inner product is commutative. So swap the e^k and the e'^j around, to show the result.

One note is that you can't just take things like this for granted: it's a pain in the arse but its just a necessary habit in mathematics. However in saying this, things will become more intuitive as you get more experience with them.
 
  • #3
I'm not a fan of notation like ##\left( a_{i k} a_{j k} \right)^T##. The transpose operation acts on matrices, not numbers. I'd write ##a_{ki}a_{kj}=(A^T)_{ik}A_{kj}=(A^TA)_{ij}##. If the matrix A is known to be orthogonal, then ##A^TA=I##, and the right-hand side of the first equality is ##\delta_{ij}##.
 

Related to Taking vectors from one basis to another [Byron and Fuller]

1. How do you take a vector from one basis to another?

To take a vector from one basis to another, you first need to determine the transformation matrix between the two bases. This can be done by setting up a system of equations using the basis vectors as the columns of the matrix. Once the transformation matrix is determined, you can multiply it by the vector to obtain the coordinates of the vector in the new basis.

2. What is a basis?

A basis is a set of linearly independent vectors that can be used to represent any vector in a vector space. The basis vectors form the building blocks for all other vectors in the space.

3. Can you explain the concept of basis conversion?

Basis conversion is the process of changing the basis of a vector without changing its direction or magnitude. This allows us to express the same vector in terms of different basis vectors, which is useful in many applications, including computer graphics and linear algebra.

4. What is the importance of knowing how to take vectors from one basis to another?

Knowing how to take vectors from one basis to another is important because it allows for more efficient and accurate calculations in linear algebra and other fields. It also allows for a better understanding of vector spaces and their properties.

5. Are there any shortcuts for taking vectors from one basis to another?

Yes, there are some shortcuts for taking vectors from one basis to another, such as using the change of basis formula or using specific properties of the transformation matrix. However, it is important to understand the underlying concepts and methods in order to accurately and efficiently convert vectors between bases.

Similar threads

  • Linear and Abstract Algebra
Replies
3
Views
1K
  • Linear and Abstract Algebra
Replies
3
Views
1K
Replies
27
Views
1K
  • Linear and Abstract Algebra
Replies
4
Views
1K
  • Linear and Abstract Algebra
Replies
2
Views
1K
  • Linear and Abstract Algebra
Replies
5
Views
2K
  • Linear and Abstract Algebra
Replies
23
Views
1K
  • Special and General Relativity
Replies
1
Views
429
  • Linear and Abstract Algebra
Replies
17
Views
3K
  • Linear and Abstract Algebra
Replies
11
Views
3K
Back
Top