# Hadamard's inequality - alternative proof(the use of Lagrange multiplier method)

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

can be proved by the use of Lagrange multiplier methods. I saw and understand the proof by the use of methods from linear algebra. However, I do not see how to apply such tool from mathematical analysis to detive the aforementioned proof.

If anyone would be able to provide proof, which uses Lagrange multiplier method, I would be very thankful!

• Check your indices Commented Nov 16, 2016 at 4:04
• Could anyone give outline/strategy of the proof, please?
– user379168
Commented Nov 16, 2016 at 17:14
• A proof using this approach is given at the following link: afhalifax.ca/bete/DALEMBERTIMAGES/DESCARTES/… Commented Nov 21, 2016 at 0:12

I haven't seen a method that uses Lagrange multipliers, but since it's related to convexity, maybe this proof could be of some use. You should be able to find a more detailed (complete) version of this in Chapter 9 of Marshall, Olkin, Arnold 1979.

The proof is based on Majorization.

Definition 1: The vector ${\bf h} \in R^n$ is said to be majorized by $\boldsymbol{\lambda}\in R^n$ if $\sum_{i=1}^k{\bf h}_i \le \sum_{i=1}^k{\boldsymbol\lambda}_i$ for $k < n$ and $\sum_{i=1}^n{\bf h}_i = \sum_{i=1}^n{\boldsymbol\lambda}_i$, where $\bf h_i$ is the $i^{th}$ entry of ${\bf h}$ (similarly for ${\boldsymbol\lambda}$).

Example 1: it is well-known that the diagonal elements of an $n\times n$ Hermitian matrix ${\bf A}$ are majorized by its vector of eigenvalues, i.e. $eig({\bf A})$ majorizes $[{\bf A}]_{ii}$. Needless to say, $[{\bf A}]_{ii}$ is the $n\times 1$ vector of diagonal elements of ${\bf A}$.

Definition 2: A function $g: R^n \rightarrow R$ is Schur-concave if knowing that ${\boldsymbol\lambda}$ majorizes ${\bf h}$ implies that $g({\bf h})\ge g(\boldsymbol{\lambda})$.

Example 2: the function $g({\bf h}) = \prod_{i=1}^n {\bf h}_i$ is Schur-concave.

By now you may have guessed that the two examples were deliberately chosen. Setting ${\boldsymbol\lambda}=eig({\bf A})$ and ${\bf h}=[{\bf A}]_{ii}$ and combining Examples 1,2 proves Hadamard's inequality. We also use the fact that $det({\bf A})$ is the product of its eigenvalues.