# Prove it!

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

Note by Hummus A
2 years, 4 months ago

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For 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.

- 2 years, 4 months ago

nice short proof!

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

i'll mention that in the note

- 2 years, 4 months ago

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

- 2 years, 4 months ago

you're right

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

i'll edit the note

- 2 years, 4 months ago