prove Hadamard's inequality for any matrix \(\left[ { a }_{ n }^{ i } \right] \)

\({ det }\left[ { a }_{ n }^{ i } \right] \le\displaystyle\prod _{ n=1 }^{ k }{ (\displaystyle\sum _{ i=1 }^{ k }{ \left[ { a }_{ n }^{ i } \right] ) } } \)

Hint:Differential calculus

**Note**:all the entries in the matrix are non-negative

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TopNewestFor the above inequality to be true, it is required that the all entries of the matrix are non-negative. Assuming that to be the case, we can upper-bound the determinant of the matrix \(A\) by the permanent of the matrix \(A\) to obtain \[\text{det}(A) \stackrel{(a)}{\leq} \text{per}(A) = \sum_{\sigma \in \pi} \prod_{n=1}^{k}a_{n \sigma({n})}\stackrel{(b)}{\leq} \prod_{n=1}^{k}(\sum_{i=1}^{k}a_{i,n}),\] where \(\pi\) is the set of all permutations on \(n\) elements and the inequalities (a) and (b) follows from the non-negativity of the elements. – Abhishek Sinha · 9 months, 2 weeks ago

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and yes i think that the entries should all be non-negative

i'll mention that in the note – Hummus A · 9 months, 2 weeks ago

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– Abhishek Sinha · 9 months, 2 weeks ago

Also, the LHS should be \(\text{det(A)}\) as I proved, instead of \(\text{det(A)}^2\). As a counter-example, take \[A=\begin{pmatrix} 1 & 1 \\ 0 & 2 \end{pmatrix} \]Log in to reply

i don't know why i thought it was the square of it

i'll edit the note – Hummus A · 9 months, 2 weeks ago

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