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Let $A$ , $B$ be two real symmetric matrices and $A$ is positive definite. Then show that $AB$ has real eigenvalues.

Symmetric matrices have real eigenvalues and product of two symmetric matrices need not be symmetric. How can the positive definiteness of $A$ be used here to show that the eigenvalues of $AB$ are real?

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1 Answer 1

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Let $A = S^t S.$ Then $AB = S^t S B.$ The eigenvalues of $AB$ are thus the same as those of $S^tBS,$ which is a symmetric matrix.

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  • $\begingroup$ I'm probably being dense here, but how did you go from $S^t S B$ to $S^t B S$? $\endgroup$
    – user169852
    Apr 27, 2021 at 21:25
  • $\begingroup$ @Bungo by typing too fast, it's $SBS^t.$ $\endgroup$
    – Igor Rivin
    Apr 27, 2021 at 21:26
  • $\begingroup$ @IgorRivin What kind of a matrix is S and can every positive definite matrix be written in that form ? $\endgroup$
    – user766787
    Apr 27, 2021 at 21:34
  • $\begingroup$ @user766787 Nonsingular, and yes (exercise). $\endgroup$
    – Igor Rivin
    Apr 27, 2021 at 21:50
  • $\begingroup$ @IgorRivin Okay, I will try to prove that. Thank you so much. $\endgroup$
    – user766787
    Apr 27, 2021 at 22:05

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