Am looking for a lower bound for determinant, $\det(X^TY)$ where $X^T$ is $p \times n$ and $Y$ is $n \times p$. Is it $Tr(X^TY)^{-1}$? Regardless, what are other lower bounds for this? $X,Y$ are real-valued matrices.

  • $\begingroup$ What is $Z$ in this context? $\endgroup$ – tch Mar 27 at 17:30
  • $\begingroup$ sry, typo. changed it to $Y$ $\endgroup$ – hearse Mar 27 at 17:31
  • $\begingroup$ If $X = -cI$, $c>0$ and $Y=I$ when $c$ is negative and $n$ is odd then the $\det(X^T)$ will be $-c^n$ while $1/Tr(X^TY)$ will be $-1/(cn)$. For $c$ large enough the determinant would be arbitrarily smaller than the bound from the trace. $\endgroup$ – tch Mar 27 at 19:54
  • $\begingroup$ You might want to check out en.wikipedia.org/wiki/Cauchy%E2%80%93Binet_formula $\endgroup$ – daw Mar 27 at 20:36
  • $\begingroup$ If $n < p$, then the determinant is always $0$, For $n \ge p$, the lower bound of the absolute value of the determinant is $0$. The lower bound for the determinant itself is $-\infty$. It is independent of the value of the trace for $p > 1$. $\endgroup$ – Paul Sinclair Mar 28 at 0:38

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.