Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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?

share|cite|improve this question
Please specify, what is $A$ exactly? (A linear transformation between inner product spaces?) Perhaps it will help to note that $(A'A)^2 = A'A$. – Jonas Meyer Dec 9 '12 at 23:36
up vote 6 down vote accepted

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.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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