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Say I have a positive semi-definite matrix with least positive eigenvalue x. Are there always principal minors of this matrix with eigenvalue less than x?

(Here "semidefinite" can not be taken to include the case "definite" -- there should be a zero eigenvalue.)

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2 Answers 2

A counterexample (to the unedited question): $$ A = \begin{pmatrix}1+x&1\\\1&1+x\end{pmatrix}. $$ Eigenvalues of $A$ are $2+x$ and $x$, principal minors have one eigenvalue $1+x$.

Voting to close.

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The poster clearly left out the condition that the matrix should be semidefinite and not definite (or else the interlacing inequalities make the condition impossible). –  Allen Knutson Mar 5 '13 at 2:25

No, it is not true; Consider the matrix

$$A= \begin{pmatrix} \frac{9}{2} & \frac{9}{20} & \frac{21}{20} & -\frac{3}{2}\\ -\frac{79}{11} & -\frac{3}{110} & \frac{23}{110} & \frac{31}{11}\\ \frac{6}{11} & \frac{21}{55} & \frac{114}{55} & \frac{6}{11}\\ \frac{16}{11} & \frac{12}{55} & \frac{128}{55} & \frac{5}{11} \end{pmatrix} $$ which has eigenvalues 0,1,2,3 (so it is positive semi-definite, but not definite.) The four principal minors are $$\frac{27}{22},\frac{189}{110},\frac{153}{55},\frac{36}{11}$$ sorted in increasing order. This should give a definite negative answer to your question.

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On the other hand, if the matrix is assumed to be real symmetric (or Hermitian), the claim is true (because one can find a linear combination of the two eigenvectors corresponding to zero and the least positive eigenvalue respectively which are supported on only n-1 of the n coordinates). It may be that this is another hypothesis that was omitted by the OP. –  Terry Tao Mar 5 '13 at 16:40
Terry Tao: Right, that is a possibility, but I can see why the poster believed that the current version was true; I did some quick numerics (that is how I got the example), and the counterexamples are quite rare, (with appropriate definition on "rare"). –  Per Alexandersson Mar 5 '13 at 17:26

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