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It is given that: $\left |\det(A) \right |\leq n^{\frac{-n}{2}}\left \| A \right \|^{n}$ where $A$ is an $n$ by $n$ matrix, and $\left \| A \right \|$ is the Hilbert Schmidt norm (i.e: $\left \| A \right \|=\left ( \sum_{i,j=1}^{n}a_{ij}^{2} \right )^{\frac{1}{2}}$).

Now, I want to prove the following inequality based on the above one:

$$\left |\det(A) \right |\leq \prod_{j=1}^{n}\left ( \sum_{i=1}^n a_{ij}^2\right )^{\frac 12}.$$

Any help?

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up vote 3 down vote accepted

Seems to me the second inequality is stronger than the first.

Write $x_j=\sum_ia_{ij}^2$. Then by the inequality of the arithmetic and geometric means, $$(1/n)\sum_j x_j\ge\prod_jx_j^{1/n}$$ Raise to the $n/2$: $$n^{-n/2}\left(\sum_jx_j\right)^{n/2}\ge\prod_jx_j^{1/2}$$ But the left side is $n^{-n/2}\|A\|^n$, and the right side is $\prod_j\left(\sum_ia_{ij}^2\right)^{1/2}$, so I seem to have proved the exact opposite of what you would need to get the 2nd inequality from the 1st.

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but you didn't prove that the right hand side is bigger than det(A) – M.Krov Mar 10 '12 at 5:37
What Gerry say is that if what you ask for is true, then it will not follow from the first inequality. – AD. Mar 10 '12 at 6:17

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