Matrix A & A^t: Same Eigenvalues, Different Eigenvectors?

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The discussion confirms that a matrix A and its transpose A^t share the same eigenvalues due to the property that det(A) = det(A^t) and the characteristic polynomial being det(A - xI). However, they can possess different eigenvectors, as demonstrated by the eigenvalue equation A\mathbf{v} = \lambda\mathbf{v} and its transpose \mathbf{v}^TA^T = \lambda\mathbf{v}^T, which indicates that the eigenvector of A becomes a left eigenvector for A^t. A generic nonsymmetric matrix serves as an example where A and A^t have different eigenvectors.

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I am trying to show that a matrix A and A^t have the same eigenvalues. Well I am sure its because det(a) = det(A^t), the characteristic polynomial is det(A-xI). But this question I am trying to solve also asks for an example of a 2x2 matrix A where A and A^t have different eigenvectors. I am kinda lost of this one. Wouldn't they have the same eigenvectors because they have the same eigenvalues?
 
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Originally posted by gimpy
I am trying to show that a matrix A and A^t have the same eigenvalues. Well I am sure its because det(a) = det(A^t), the characteristic polynomial is det(A-xI). But this question I am trying to solve also asks for an example of a 2x2 matrix A where A and A^t have different eigenvectors. I am kinda lost of this one. Wouldn't they have the same eigenvectors because they have the same eigenvalues?

no. consider the eivenvalue equation:

A\mathbf{v}=\lambda\mathbf{v}
then take the transpose of that equation:

\mathbf{v}^TA^T=\lambda\mathbf{v}^T

so the eigenvector of A becomes something you might call a "left eigenvector" of A^T, but there is no reason to think that it should also be a regular eigenvector.

the eigenvalues must be the same, however, for the reason you stated.

i imagine, that to find an example of a matrix whose transpose has different eigenvectors, you should just grab any old generic nonsymmetric matrix, it will probably do.
 
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thanks lethe,

I feel kinda stupid that i overlooked that equation. Its really fundamental stuff[b(]
 
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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