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.

  • $\begingroup$ 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

$A$ 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.


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