Can Eigenvectors of the Same Eigenvalue Be Orthogonal in a 2x2 Matrix?

Click For Summary
SUMMARY

In the discussion regarding the orthogonality of eigenvectors in a 2x2 matrix, it is established that eigenvectors corresponding to the same eigenvalue cannot be orthogonal if the matrix has distinct eigenvalues. The eigenspaces in this case are of dimension 1, making all eigenvectors from the same eigenvalue linearly dependent. However, for symmetric or Hermitian matrices, it is confirmed that eigenvectors from distinct eigenspaces can indeed be orthogonal. The proof provided utilizes the properties of symmetric matrices to demonstrate this relationship.

PREREQUISITES
  • Understanding of eigenvalues and eigenvectors in linear algebra
  • Knowledge of symmetric and Hermitian matrices
  • Familiarity with eigenspaces and their dimensions
  • Basic proficiency in matrix multiplication and transposition
NEXT STEPS
  • Study the properties of symmetric matrices and their eigenvectors
  • Learn about the spectral theorem for real symmetric matrices
  • Explore the concept of orthogonality in vector spaces
  • Investigate the implications of eigenvalue multiplicity on eigenspace dimensions
USEFUL FOR

Mathematicians, students of linear algebra, and anyone interested in the properties of matrices and eigenvector relationships.

Hassan2
Messages
422
Reaction score
5
This seems a simple question but I can't find the solution by myself. Please help.

Say we have a 2 by 2 matrix with different eigenvalues. Corresponding to each eigenvalue, there are a number of eigenvectors.

Q1. Could the eigenvectors corresponding to the same eigenvalue have different directions?
Q2. Could the eigvenvectors corresponding to the same eigenvalue be orthogonal?

Q3. How can we prove that there is a pair of orthogonal eigenvalues for the 2 by 2 matrix, each for one eigenvalue?

Your help would be appreciated.
 
Physics news on Phys.org
Hassan2 said:
This seems a simple question but I can't find the solution by myself. Please help.

Say we have a 2 by 2 matrix with different eigenvalues. Corresponding to each eigenvalue, there are a number of eigenvectors.

So there are 2 distinct eigenvalues. This means to me that there are 2 distinct eigenspaces each of dimension 1.

Q1. Could the eigenvectors corresponding to the same eigenvalue have different directions?
Q2. Could the eigvenvectors corresponding to the same eigenvalue be orthogonal?

No. The eigenspaces have dimension 1 in this case, so every two eigenvectors from the same eigenvalue are linearly dependent.

Q3. How can we prove that there is a pair of orthogonal eigenvalues for the 2 by 2 matrix, each for one eigenvalue?

We can't because it is not true. It is true for symmetric/hermitian matrices however.
 
Thanks a lot. I think its clear to me now. Bellow is my proof. I hope it's correct.

Ax_{1}=\lambda_{1} x_{1}
Ax_{2}=\lambda_{2} x2

multiplying the first equation with x^{T}_{2} and the second one with x^{T}_{1} , we have

x^{T}_{2}Ax_{1}=\lambda_{1} x^{T}_{2} x_{1} (1)

x^{T}_{1}Ax_{2}=\lambda_{2} x^{T}_{1} x_{2}

transposoing the latter, we have

x^{T}_{2}A^{T}x_{1}=\lambda_{2} x^{T}_{2} x_{1} (2)

if A is symmetric then (1) and (2) yields x^{T}_{2} x_{1} =0 or \lambda_{1} = \lambda_{2}

Do you have an intuitive reason as why for a symmetric matrix, two eigenvectors from distinct eigenspaces are orthogonal?
 
I am studying the mathematical formalism behind non-commutative geometry approach to quantum gravity. I was reading about Hopf algebras and their Drinfeld twist with a specific example of the Moyal-Weyl twist defined as F=exp(-iλ/2θ^(μν)∂_μ⊗∂_ν) where λ is a constant parametar and θ antisymmetric constant tensor. {∂_μ} is the basis of the tangent vector space over the underlying spacetime Now, from my understanding the enveloping algebra which appears in the definition of the Hopf algebra...

Similar threads

  • · Replies 13 ·
Replies
13
Views
3K
  • · Replies 2 ·
Replies
2
Views
3K
  • · Replies 33 ·
2
Replies
33
Views
2K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 3 ·
Replies
3
Views
3K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 2 ·
Replies
2
Views
9K
  • · Replies 10 ·
Replies
10
Views
3K
  • · Replies 2 ·
Replies
2
Views
3K
Replies
5
Views
4K