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Can anything be said about a (not necessarily symmetric) matrix $A$, all of whose principal minors (upper-left squares) have positive determinant? Do these matrices have a name?

I would like to know if this implies another condition usually found in positive-definite matrices, such as $x^T A x > 0$ for all $x \ne 0$, or that all eigenvalues are positive / have positive real part, etc.

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  • $\begingroup$ I am not entirely sure if this what you mean, but I think it might be what you're looking for: en.m.wikipedia.org/wiki/Sylvester's_criterion $\endgroup$
    – dreamer
    Aug 9, 2014 at 7:19
  • $\begingroup$ @dreamer Sylvester's criterion deal with symmetric (or Hermitian) matrices. $\endgroup$ Aug 9, 2014 at 11:36
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    $\begingroup$ A square matrix with every principal minor > 0 is called a P-matrix. The real eigen values of such a matrix are positive. $\endgroup$
    – La Rias
    Mar 2, 2016 at 9:50

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This condition does not imply $x^TAx>0$ for all $x\ne 0$.

Define $A$ by $$ A = \begin{pmatrix} 1 & 2 \\ 0 & 1 \end{pmatrix}. $$ Then all principal minors of $A$ have determinant $1$. But for $x\ne 0$ $$ x^T Ax = \frac12 x^T(A+A^T)x = x^T \begin{pmatrix} 1 & 1 \\ 1 & 1 \end{pmatrix}x. $$ For $x=\begin{pmatrix}1\\-1\end{pmatrix}$ this expression is zero.

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  • $\begingroup$ The O.P has asked whether one can any such property. I think the O.P is well aware of the fact that $x^TAx>0$ is not true. $\endgroup$
    – Babai
    Aug 8, 2016 at 22:09
  • $\begingroup$ @Babai Please read again the question, in particular the part ``I would like to know if this implies another condition usually found in positive-definite matrices, such as xTAx>0 for all x≠0''. $\endgroup$
    – daw
    Aug 9, 2016 at 6:08
  • $\begingroup$ If you read it fully the sentence doesn't stop with $x^TAx>0$ for all $x\neq 0$, it says more "......or eigen values are positive or positive real part." $\endgroup$
    – Babai
    Aug 9, 2016 at 6:23

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