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}?

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- Apr 10th 2009, 08:01 PMantmaneigenvalues
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, 12:02 AMmr fantastic
$\displaystyle \det (A - \lambda I) = \det (A - \lambda I)^T = \det (A^T - \lambda I) $.

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

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

**Proof:**

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

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