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

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In a 2x2 matrix with distinct eigenvalues, the eigenvectors corresponding to the same eigenvalue cannot have different directions, as they are linearly dependent within a one-dimensional eigenspace. Consequently, eigenvectors from the same eigenvalue cannot be orthogonal. However, for symmetric or Hermitian matrices, it is possible to have orthogonal eigenvectors corresponding to distinct eigenvalues. The discussion highlights the distinction between general matrices and symmetric matrices regarding the orthogonality of eigenvectors. An intuitive understanding of this phenomenon in symmetric matrices is sought, emphasizing the unique properties of such matrices.
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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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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...

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