1
$\begingroup$

In some presentation the speaker said that pushing AA'A-A to 0 makes encourages singular values of A to be either 0 or 1, can anyone tell me where this follows from?

$\endgroup$
1
  • 1
    $\begingroup$ Please specify, what is $A$ exactly? (A linear transformation between inner product spaces?) Perhaps it will help to note that $(A'A)^2 = A'A$. $\endgroup$ Dec 9, 2012 at 23:36

1 Answer 1

6
$\begingroup$

Your statement implies that $(AA')^2 = AA'$, so $AA'$ is a projection. Projections only have eigenvalues of 1 or 0. Since the singular values of $A$ are the eigenvalues of $AA'$ the original statement follows.

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .