For any real square matrix $M$ let $\lambda^+(M)$ be the number of positive eigenvalues of $M$ counting multiplicity. Let $A$ be an $n\times n$ real symmetric matrix and $Q$ be an $n\times n$ real invertible matrix. Then which are correct?

  1. Rank $(A)$ = Rank $Q^TAQ$

  2. Rank $(A)$ = Rank $Q^{-1}AQ$

  3. $\lambda^+(A)$ = $\lambda^+(Q^TAQ)$

  4. $\lambda^+(A)$ = $\lambda^+(Q^{-1}AQ)$

I can see 2 is correct but what about the rest?


(1) is true and is in fact a consequence of the more general rule ${\sf rank}(PAQ)={\sf rank}(A)$ whenever $P$ and $Q$ are invertible.

(4) is also true because the eigenvalues of $Q^{-1}AQ$ are exactly the eigenvalues of $A$ (indeed, $v$ is an eigenvector of $Q^{-1}AQ$ iff $Qv$ is an eigenvector of $A$, and the eigenvalue is the same).

(3) is a consequence of Sylvester’s law of inertia as pointed out in user44197's answer.

  • $\begingroup$ So all four are correct? $\endgroup$ – user113578 Dec 27 '13 at 6:27
  • 1
    $\begingroup$ @user113578 Yes. $\endgroup$ – Ewan Delanoy Dec 27 '13 at 6:28
  • $\begingroup$ (1)${\sf rank}(PAQ)={\sf rank}(A)$ how does the statement true $\endgroup$ – Unknown x Dec 15 '17 at 17:44
  • $\begingroup$ can you point out reference. please help me. $\endgroup$ – Unknown x Dec 15 '17 at 17:45
  • $\begingroup$ @ManeeshNarayanan wikipedia $\endgroup$ – Ewan Delanoy Dec 15 '17 at 17:55

2, 4) is similarity transformation, so $\cdots$

3) is the law of inertia. If you have not come across it, you should read it up

1) also follows from law of inertia

Wikipedia Article on Sylvester's Law of Inertia


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.