$A=[a_{ij}]\in M_n(\mathbb{R})$ and $c=\max\{|a_{ij}|\}$

I need to show $|\det A|\le c^nn^{n/2}$

Let $\lambda_1,\dots,\lambda_n$ be the eigenvalues of $A$

I can say that $|\det A|^{2/n}=|\lambda_1\cdots\lambda_n|^{2/n}\le{|\lambda_1|^2+\dots+|\lambda_n|^2\over n} $

could anyone tell me how to proceed next?

  • $\begingroup$ See Hadamard's inequality. $\endgroup$ – Julien Nov 3 '13 at 15:08
  • $\begingroup$ What If I just want to prove myself from the given data? Thank you though :) $\endgroup$ – Marso Nov 3 '13 at 15:10

The absolute value of a determinant expresses the volume of the parallelepiped spanned by its columns. In this case each column has length no more than $c \sqrt{n}$ and therefore the volume is no more than $(c\sqrt{n})^n$.


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