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

In summary, the conversation discusses the concept of eigenvectors and eigenvalues in a 2 by 2 matrix. It is determined that for symmetric or hermitian matrices, eigenvectors from distinct eigenspaces are orthogonal. A proof is provided using the transpose property of symmetric matrices.
  • #1
Hassan2
426
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.
 
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  • #2
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.
 
  • #3
Thanks a lot. I think its clear to me now. Bellow is my proof. I hope it's correct.

[itex]Ax_{1}=\lambda_{1} x_{1}[/itex]
[itex]Ax_{2}=\lambda_{2} x2[/itex]

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

[itex]x^{T}_{2}Ax_{1}=\lambda_{1} x^{T}_{2} x_{1}[/itex] (1)

[itex]x^{T}_{1}Ax_{2}=\lambda_{2} x^{T}_{1} x_{2}[/itex]

transposoing the latter, we have

[itex]x^{T}_{2}A^{T}x_{1}=\lambda_{2} x^{T}_{2} x_{1}[/itex] (2)

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

Do you have an intuitive reason as why for a symmetric matrix, two eigenvectors from distinct eigenspaces are orthogonal?
 

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

What are orthogonal eigenvectors?

Orthogonal eigenvectors are a set of vectors that are perpendicular to each other and satisfy certain mathematical properties. They play an important role in linear algebra and are often used to solve systems of linear equations.

How are orthogonal eigenvectors related to eigenvalues?

Orthogonal eigenvectors are associated with eigenvalues, which are scalar values that represent how the vector changes when multiplied by a specific matrix. Each eigenvalue has a corresponding eigenvector, and these eigenvectors are orthogonal to each other when the matrix is symmetric.

What is the significance of orthogonal eigenvectors?

Orthogonal eigenvectors are important because they can simplify the analysis of systems of linear equations and help identify important properties of a matrix. They also have applications in various fields such as physics, engineering, and computer science.

How do you find orthogonal eigenvectors?

To find orthogonal eigenvectors, you first need to determine the eigenvalues of the matrix. Then, for each eigenvalue, you can use a specific algorithm, such as Gram-Schmidt or Householder, to find an orthogonal set of eigenvectors. These algorithms involve performing calculations and transformations on the matrix.

Can a matrix have more than one set of orthogonal eigenvectors?

Yes, a matrix can have multiple sets of orthogonal eigenvectors. This occurs when the matrix has repeated eigenvalues, meaning that two or more eigenvectors correspond to the same eigenvalue. In this case, any linear combination of the eigenvectors will also be an orthogonal eigenvector.

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