lower bound for determinant of $(X^TY)$

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.

• What is $Z$ in this context? – tch Mar 27 at 17:30
• sry, typo. changed it to $Y$ – hearse Mar 27 at 17:31
• 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. – tch Mar 27 at 19:54
• You might want to check out en.wikipedia.org/wiki/Cauchy%E2%80%93Binet_formula – daw Mar 27 at 20:36
• 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 $p > 1$. – Paul Sinclair Mar 28 at 0:38