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$A$ is indefinite iff $A$ fits none of the above criteria. Equivalently, $A$ has both positive and negative eigenvalues. Also equivalently, $x^TAx$ is positive for at least one $x$ and negative for at least another $x$.

Note that the leading principal minors refer to the determinants of the northwest-corner submatrices, and are merely a subset of all the principal minors.

Now, suppose that a symmetric $n\times n$ matrix $M$ is neither positive definite nor negative definite. $$$$ From the facts highlighted above, and possibly using linear algebra, then is statement (2) true? If not, is at least statement (1) true? I have seen both assertions separately in different texts (e.g. http://people.ds.cam.ac.uk/iar1/teaching/Hessians-DefinitenessTutorial.pdf and http://www.econ.ucsb.edu/~tedb/Courses/GraduateTheoryUCSB/BlumeSimonCh16.PDF), but am unable to prove either:

(1) If $M$'s leading principal minors are all nonzero, then $M$ is indefinite.

(2) If $M$ has some nonzero leading principal minor, then $M$ is indefinite.

EDIT: (1) is true can actually be simplified:

(1) If det $M$ is nonzero, then $M$ is indefinite.

We don't need to check all the leading principal minors because once det M is nonzero, we can immediately deduce that M has no zero eigenvalues, and since it is also given that M is neither positive definite nor negative definite, then M can only be indefinite.


Edited per Ryan's clarification below.

Statement 1: Yes, this is fine. If $M$ is neither positive nor negative definite, and has no zero eigenvalues, then it must have at least one positive and one negative eigenvalue. Notice that this is a sufficient but not necessary condition on $M$ being indefinite. $\left[\begin{array}{ccc}0 & 0 &0\\0 & 1 & 0\\0 & 0 & -1\end{array}\right]$ is indefinite, for instance.

Statement 2: No, this is false. Consider for instance $\left[\begin{array}{cc}1 & 0\\0 & 0\end{array}\right]$ which is positive-semidefinite.

It is impossible to characterize indefinite matrices from the leading minors alone. For example, if the first row and column of a symmetric matrix $M$ is zero, the matrix might be positive-semidefinite, negative-semidefinite, or indefinite, yet all of the leading minors will be zero.

A complete, correct statement requires looking at all principal minors, for example: a symmetric matrix $M$ is indefinite (has positive and negative eigenvalues) if and only if:

  1. $\Delta_k < 0$ for some even $k$; or
  2. $\Delta_{k_1} > 0$ and $\Delta_{k_2} < 0$ for two different odd $k_1$ and $k_2$.

Knowing that $M$ is not strictly positive- or negative-definite does not really help. You can check that if $M$ satisfies neither of these conditions, then it must satisfy one of the rows of the purple box.

EDIT: Proof of the "only if" direction. Let $M$ be indefinite. Suppose, for contradiction, that neither of the above two hold. Then either all of the odd-dimensional minors are nonnegative, or all are nonpositive.

In the former case, $M$ satisfies the third row of the purple box above, and $M$ is positive-semidefinite, a contradiction.

In the latter case, $M$ satisfies the fourth row of the purple box above, and $M$ is negative-semidefinite, a contradiction.

EDIT 3: Proof of the "if" direction. Suppose one of the even-dimensional minors is negative, and suppose, for contradiction, that $M$ is positive-semidefinite or negative-semidefinite. Then by row three or four of the purple box (as appropriate), that minor is in fact positive, a contradiction. Therefore $M$ is neither positive- nor negative-semidefinite, and so is indefinite.

Suppose instead one of the odd-dimensional minors is positive, and another is negative, and suppose $M$ is positive-semidefinite. Then both of those minors are positive, a contradiction. Now suppose $M$ is negative-semidefinite. Then both of those minors are negative, a contradiction. The only remaining possibility is that $M$ is indefinite.

  • $\begingroup$ Ok, I've made some edits. $\endgroup$ – user7530 Nov 4 '13 at 15:16
  • $\begingroup$ If there were a zero eigenvalue, then $\det M$, which is the product of the eigenvalues, would be zero, and $\det M$ is a principal minor. $\endgroup$ – user7530 Nov 4 '13 at 15:42
  • $\begingroup$ That matrix isn't symmetric? $\endgroup$ – user7530 Nov 4 '13 at 19:14
  • $\begingroup$ Hey, doesn't this matrix (1 0 0 , 0 0 0 , 0 0 -1) contradict your characterisation of an indefinite symmetric matrix? The leading principal minors are 1,0,0, none of which are negative (thus violating the conditions you specified), yet the matrix is indefinite because its eigenvalues are 1,0,-1, i.e. both positive and negative. $\endgroup$ – Ryan Nov 4 '13 at 19:30
  • $\begingroup$ @Ryan right you are, those should be principal minors, not just leading principal minors ($\Delta$ instead of $D$ I guess) $\endgroup$ – user7530 Nov 4 '13 at 20:40

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