hermit matrices and real spectrum

eugene · Yang-Mills Theory Offline Joined: 16 May 2004 Posts: 970

Let $A,B$ -be hermit(symmetric) matrices, one of which,say $A$ , is positively defined. Prove that $Spec(AB)$ -real

**Posted:** Fri Apr 15, 2005 1:55 am

Leva1980 · Poincare Conjecture Offline Joined: 08 Jan 2005 Posts: 175

Suppose that $\lambda$ is an eigenvalue of of $AB$ . Then there exists
$v \in \mathbb{C}^{n}$ such that $ABv = \lambda v$ . Then
$(ABv,Bv) = \lambda (v,Bv)$ . Now, because $A,B$ are symmetric we have that
$(v,Bv) , (ABv,Bv) \in \mathbb{R}$ . Now, if $(v,Bv) = 0$ then
$(ABv,Bv) = 0$ and because $A$ is positive we get that $Bv = 0$ , therefore
$\lambda v = ABv = 0$ , so $\lambda = 0$ , and hence real.
Now, if $(v,Bv) \neq 0$ , then $\lambda = \frac{(ABv,Bv)}{(v,Bv)} \in \mathbb{R}$ .
So in any case $\lambda$ is real.

**Posted:** Fri Apr 15, 2005 9:05 am

_________________
Gromov is the best!

oxyz · P versus NP Offline Joined: 08 Mar 2009 Posts: 22

$A$ is positively defined show that the exist matrix $C$ is hermit(symmetric) matrices and positively defined such that : $A=C^2$ since
$SP_c(AB)=SP_c(C^2B)$ = $SP_c(C(CB))=SP_c(CBC) \subset R$ by $CBC$ is hermit(symmetric) matrices (use $SP_c(MN)=SP_c(NM)$ )

**Posted:** Sun Mar 08, 2009 7:21 am