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

#Calculus

Easy Math Editor

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Comments

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.

Abhishek Sinha - 5 years, 3 months ago

nice short proof!

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

i'll mention that in the note

Hamza A - 5 years, 3 months ago

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

Abhishek Sinha - 5 years, 3 months ago

@Abhishek Sinha – you're right

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

i'll edit the note

Hamza A - 5 years, 3 months ago

Math	Appears as
Remember to wrap math in `\(` ... `\)` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$