For any real square matrix which are correct?

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.

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