Do there exist complex square matrices that cannot be classified as either positive semi-definite, or negative semi-definite, or indefinite?

Here I am using the definitions:

  • Positive semi-definite: $u^* M u \ge 0\quad \forall u$.

  • Negative semi-definite: $u^* M u \le 0\quad \forall u$.

  • Indefinite: there exist $u,v$ such that $u^*Mu < 0 < v^* M v$.

Obviously these three cases cover all Hermitian matrices. But for non-Hermitian matrices it is not clear to me. For example, could there be matrices for which $u^* M u$ is always strictly complex and not real? Or perhaps sometimes not real, sometimes real and positive, but never real and negative.

  • 1
    $\begingroup$ The terminology of positive (or negative) definite or indefinite only apply to Hermitian matrix. $\endgroup$ – pisco Aug 22 '17 at 8:55
  • $\begingroup$ @pisco125 while that's true, it doesn't really answer OP's question. $\endgroup$ – 5xum Aug 22 '17 at 8:55
  • $\begingroup$ @pisco125 No, they apply more generally, and the definition written in the question is the standard generalization. See, for example, mathworld.wolfram.com/IndefiniteMatrix.html $\endgroup$ – Nick Alger Aug 22 '17 at 8:57

Obviously there are matrices where $u^*Mu$ are always non-real apart from $u=0$, for example $M=iM'$ where $M'$ is positive definite Hermitian.

In general one can write $M=M_1+iM_2$ where $M_1=\frac12(M+M^H)$ and $M_2$ are Hermitian, and then $$u^*Mu=u^*M_1u+i u^*M_2u$$ with the $u^*M_j u$ real. So to get $M$ sometimes non-real, sometimes positive, but never negative, take $M_1$ to be positive definite, and $M_2$ to be indefinite Hermitian, etc.

  • $\begingroup$ I'm accepting this answer because not only does it provide a counterexample, it also sheds light onto the general class of counterexamples. $\endgroup$ – Nick Alger Aug 22 '17 at 9:09

Take the $1 \times 1$ matrix $M=(i)$. Then for all $u \in \mathbb{C}$ we have $${}^t\bar{u}Mu=\bar{u}iu = |u|^2 i$$ which is always purely imaginary.

Of course, you have the same phenomenon in every dimension $n$, by taking the matrix $$M_n = i I_n.$$

  • $\begingroup$ That's probably as simple as it gets... $\endgroup$ – 5xum Aug 22 '17 at 8:56
  • $\begingroup$ Thanks, such a simple counter-example! $\endgroup$ – Nick Alger Aug 22 '17 at 9:08

Any matrix $i\cdot I_n\in\mathbb C^{n\times n}$ will do the trick ($I_n$ is the identity matrix).

You have $$v^*(iI_n)v=iv^*v=i|v|^2\notin \mathbb R$$

for $v\neq 0$.

  • $\begingroup$ @FrancescoPolizzi Thanks. Of course. $\endgroup$ – 5xum Aug 22 '17 at 9:04
  • $\begingroup$ Thanks for the answer. Dunno how I missed this one! $\endgroup$ – Nick Alger Aug 22 '17 at 9:10

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