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Given $A,B$ self adjoint matrices in $M_{n\times n}(\mathbb C)$ with positive real eigenvalues. Show that the eigenvalues of $AB$ are positive.

I think this can either be done by showing that $A$ and $B$ can be simultaneously diagonolized (not 100% sure) or maybe showing that $<ABv,v>$ is bigger than $0$ somehow?

Edit: I dont think this question is the same as the possoble duplicate sonce mine is not about real matrices specifically

Jason
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$AB$ is similar to $A^{-1/2}(AB)A^{1/2}=A^{1/2}BA^{1/2}$, which is positive definite.

user1551
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  • What does $A^{1/2}$ mean? Sqrt? There must be some properties about Positive definte matrices which I don't know about :/ – Jason Aug 10 '18 at 15:58
  • @Jason It's the unique positive definite square root of $A$. – user1551 Aug 10 '18 at 16:30
  • Hmm we have not learnt that. I'll have to look it up thanks! If by any chance you know of a more elementary method to show this I'd love to hear it – Jason Aug 10 '18 at 16:34
  • @Jason You may use a Cholesky decomposition instead. Let $A=LL^\ast$ be the Cholesky decomposition of $A$. Then $AB$ is similar to $L^{-1}(AB)L=L^\ast BL$, which again is positive definite. – user1551 Aug 10 '18 at 17:05
  • @Jason : If $A$ is selfadjoint it has an orthonormal basis of eigenvectors ${ e_1,e_2,\cdots,e_N }$. If $\lambda_1,\cdots,\lambda_N$ are the corresponding eigenvalues, and if these are positive, then $A^{1/2}x=\sum_{k=1}^N \lambda_k^{1/2}\langle x, e_k\rangle e_k$ is a positive square root of $A$, which is all you need to justify user1551's argument. – Disintegrating By Parts Aug 10 '18 at 17:45
  • @DisintegrationByParts Thanks! That makes sense, I understand this answer now. – Jason Aug 10 '18 at 20:10