1
$\begingroup$

Suppose that $\mathbf{A}\in\mathrm{R}^{m\times m}$ is a square but not necessarily symmetric matrix whose eigenvalues and eigenvectors are $\lambda_i$ and $\mathbf{x}_i,$ $i = 1,2,\cdots,m$.

  1. Is there any relation between the eigenvalues of $\mathbf{A}$ and those of $\mathbf{A'A}$?

  2. Is there any relation between the eigenvectors of $\mathbf{A}$ and those of $\mathbf{A'A}$?

$\endgroup$
0
$\begingroup$

If $A$ is normal (that is, if $A'A = AA'$), then the relation is "what you'd expect":

  • $\lambda$ is an eigenvalue of $A\iff |\lambda|^2$ is an eigenvalue of $A'A$
  • $A$ and $A'A$ have an identical set of orthonormal eigenvectors.

Also, $A$ is normal if and only if trace$(A'A) = \sum_{i=1}^k |\lambda_i(A)|^2$.

More generally, these are not true. However, we can say that $$ \sqrt{\lambda_1(A'A)} \leq |\lambda_i(A)| \leq \sqrt{\lambda_n(A'A)} $$ where $\lambda_1(X) \leq \cdots \leq \lambda_n(X)$ are the eigenvalues of $X$.

There is no general relation between the eigenvectors.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.