Are the eigenvectors of A and A^T related?

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SUMMARY

The eigenvalues of a matrix A and its transpose A^T are identical, as established by their characteristic polynomials being the same. However, the eigenvectors associated with these eigenvalues can differ significantly. For instance, given the matrix A = [[2, 1], [3, 0]], the eigenvalues are 3 and -1 with corresponding eigenvectors (1, 1) and (1, -3). In contrast, for A^T = [[2, 3], [1, 0]], the eigenvalues remain 3 and -1, but the corresponding eigenvectors are (3, 1) and (1, -1), demonstrating that knowledge of the eigenvector v of A does not provide information about the eigenvector of A^T.

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I have an (unknown) matrix A and with real non-negative values. I know its largest eigenvalue [tex]\lambda[/tex] and the associated eigenvector, v. (I know nothing about the other eigenvectors). Does this give me any information about the eigenvector of AT associated with [tex]\lambda[/tex] or is it completely independent of v?
 
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The eigenvalues of A and A^T are the same. Consider the characteristic polynomial as the expansion of the determinant of A. The characteristic polynomials of A and A^T are identical so they have the same set of roots.

However you can't say anything about the left and right eigenvectors. For example let A =

2 1
3 0

Eigenvalues are 3 and -1
Correponding vectors are (1 1) and (1 -3)

A^T =
2 3
1 0

Eigenvalues are again 3 and -1
Corresponding eigenvectors are (3 1) and (1 -1)
 

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