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Let $A$ be an $n\times n$ matrix, not necessarily symmetric, and suppose all principal minors of $A$ are positive. How to show that any real eigenvalue of $A$ is must be positive?

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  • $\begingroup$ Are you sure it's true ? $\endgroup$
    – Jean Marie
    Aug 8, 2016 at 22:19
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    $\begingroup$ Yes. In fact, these matrices are called $P-$ matrices. $\endgroup$
    – Babai
    Aug 8, 2016 at 22:22
  • $\begingroup$ @JeanMarie This matrix doesn't have any real eigenvalue. $\endgroup$
    – Babai
    Aug 8, 2016 at 22:24
  • $\begingroup$ All right. Pardon me. I misread your question. I delete what I have said in order not to hide my error but not to confuse the reader in a century. $\endgroup$
    – Jean Marie
    Aug 8, 2016 at 22:25
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    $\begingroup$ For reference: p-matrix $\endgroup$ Aug 8, 2016 at 22:45

1 Answer 1

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You should be able to find this result on some reference books on advanced linear algebra. I haven't any relevant book at hand, but this result seems obvious: as $A$ is a P-matrix, its adjugate matrix has a positive diagonal; therefore, by Jacobi's formula and by mathematical induction on the size of $A$, we have $\frac d{dt}\det(A+tI)=\operatorname{tr}(\operatorname{adj}(A+tI))>0$ when $t\ge0$. Since $\det(A)$ is also positive, the polynomial $\det(A+tI)$ cannot have a positive real root and hence the characteristic polynomial $\det(tI-A)$ cannot have any negative real root.

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    $\begingroup$ "obvious" is a relative term, but in any case I'm impressed $\endgroup$ Aug 9, 2016 at 23:01
  • $\begingroup$ Your obvious arguments seems quite complicated for me. For example, the first argument itsellf, the adjugate of a P-matrix has a positive diagonal. Why is it true? Can you please elaborate your answer in detail? $\endgroup$
    – Babai
    Aug 17, 2016 at 20:00
  • $\begingroup$ @Babai Let $A_1$ be the trailing principal submatrix of $A$ of size $n-1$. The first diagonal element of $\operatorname{adj}(A+tI)$ is then $\det(A_1+tI)$. By induction hypothesis, $\det A_1>0$ and all real eigenvalues (if any) of $A_1$ are nonnegative. Therefore $\det(A_1+tI)>0$ for every $t\ge0$. Simiilarly for every other diagonal element of $A+tI$. $\endgroup$
    – user1551
    Aug 17, 2016 at 22:44
  • $\begingroup$ @user1551 I understood your explanation. Very nice answer. Thank you. $\endgroup$
    – Babai
    Aug 19, 2016 at 21:05

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