1
$\begingroup$

Let $(A,B) \in S_n(\mathbb R)^2$ $\forall X \ne 0, (AX|X) \ge 0 $ Show that $ sp(AB) \subset \mathbb R $

My work

I took an eigen value of $AB$ to show that it was equal to its conjugate

To do that I considered the product $ \bar{X}^T ABX$ with X the vector associated to the value $\lambda $

It is on the one hand equal to $\lambda \bar{X}^T X$ and on the other hand to $(BA\bar{X})^T X$. However I am blocked here. I don’t know where to use the inner product from here.

$\endgroup$
2
  • $\begingroup$ You might check the assumptions: there is no $B$ in the inequality. Also: eigenvalue has nothing to do with the number eight ;) $\endgroup$
    – daw
    Jan 25 at 12:50
  • $\begingroup$ @daw sorry, I’m new to doing maths in english :) $\endgroup$
    – Julien
    Jan 25 at 12:52
0
$\begingroup$

$A$ is positive semidefinite so it has a positive semidefinite square root $A^{1/2}$. We get

$$\sigma(AB) =\sigma(A^{1/2}A^{1/2}B) = \sigma(A^{1/2}BA^{1/2}) \subseteq \Bbb{R}$$

since $A^{1/2}BA^{1/2}$ is symmetric.

$\endgroup$

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.