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

Let $B^{-}$ be a generalized inverse of a symmetric matrix $B$ and assume $B^{-}$ is also symmetric. Show that if $P = BB^{−}$ , then rank of $B$ is the same as trace of $P$

share|cite|improve this question
Welcome to math.SE: since you are new, I wanted to let you know a few things about the site. In order to get the best possible answers, it is helpful if you say in what context you encountered the problem, and what your thoughts on it are; this will prevent people from telling you things you already know, and help them give their answers at the right level. If this is homework, please add the homework tag; people will still help, so don't worry. Also, many find the use of imperative ("Prove", "Solve", "Show"...) to be rude when asking for help; please consider rewriting your post. – amWhy Dec 3 '12 at 2:51

Hint: Let $\mathrm{rank}(B)=r$. We can write $B=Q(D_r\oplus 0_{n-r})Q^T$ where $Q$ is a real orthogonal matrix and $D_r$ is an $r\times r$ nonzero diagonal matrix. The Moore-Penrose generalized inverse is then $Q(D_r^{-1}\oplus 0_{n-r})Q^T$.

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.