My question is: Is a symmetric, invertible and positive semi-definite matrix automatically positive definite?

Any solution or hints on how this can be answered would be appreciated.

  • $\begingroup$ Hint: if any of the principal minors is $0$, then there exists a nonzero vector $x $ with $x^T Ax =0$ (why?). $\endgroup$ – darij grinberg May 17 '18 at 22:38

Note that for a positive semi-definite matrix ($\lambda_i\ge 0$) and invertible ($\lambda_i\neq 0$) we have that $\lambda_i> 0$ then the matrix is positive definite.

  • $\begingroup$ Do you have any reference where i can connect the notions that Positive (semi-)definite is equivalent to $\lambda_i (\geq) >0$? $\endgroup$ – Martin May 17 '18 at 19:21
  • $\begingroup$ @Martin Symply recall that a symmetric matrix is diagonalizable, then $x^TAx$ with respect to a basis of orthogonal eigenvectors becomes $$y^TM^TAMy=y^TDy=\sum \lambda_iy_i^2$$ $\endgroup$ – gimusi May 17 '18 at 19:28

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