# eigenvalues

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• Apr 10th 2009, 09:01 PM
antman
eigenvalues
Show that A and A^{T} have the same eigenvalues. what if anything can we say about the associated eigenvectors of A and A^{T}?
• Apr 11th 2009, 01:02 AM
mr fantastic
Quote:

Originally Posted by antman
Show that A and A^{T} have the same eigenvalues. what if anything can we say about the associated eigenvectors of A and A^{T}?

$\det (A - \lambda I) = \det (A - \lambda I)^T = \det (A^T - \lambda I)$.

The eigenvectors of $A^T$ are generally different to those of $A$. However, there is a relationship between the two:

If $v_i$ is an eigenvector of $A$ corresponding to the eigenvalue $\lambda_i$ and $w_j$ is an eigenvector of $A^T$ corresponding to the eigenvalue $\lambda_j$ then $v_i^T w_j = 0 ~ (\lambda_i \neq \lambda_j)$.

Proof:

$A v_i = \lambda_i v_i \Rightarrow v_i^T A = \lambda_i v_i^T \Rightarrow v_i^T A w_j = \lambda_i v_i^T w_j$ .... (1)

$A^T w_j = \lambda_j w_j \Rightarrow v_i^T A^T w_j = \lambda_j v_i^T w_j$ .... (2)

(1) - (2): $0 = \lambda_i v_i^T w_j - \lambda_j v_i^T w_j$ and the result is easily seen.