It is currently Fri Aug 23, 2019 2:02 pm

All times are UTC [ DST ]

Post new topic Reply to topic  [ 2 posts ] 
Author Message
 Post subject: Hadamard's inequality
PostPosted: Mon Dec 28, 2015 10:45 am 
User avatar

Joined: Sat Nov 07, 2015 6:12 pm
Posts: 841
Location: Larisa
Continuing the Hermite - Hadamard's inequality let us see the Hadamard's inequality for matrices.

Let $\mathbf{a_1, a_2, \dots, a_N}$ be column vectors in $\mathbb{R}^N$ and let $A=(a_1, a_2, \dots, a_N)$ be the corresponding $N \times N$ real matrix. Then the following inequality holds:

\left | \det A \right |\leq \prod_{n=1}^{N}\left \| a_n \right \|

where $\left \| \cdot \right \|$ is the Euclidean norm on vectors in $\mathbb{R}^N$. In continuity , give the geometrical interpretation of the inequality above.

Imagination is much more important than knowledge.

Offline Profile  
Reply with quote  

PostPosted: Wed Apr 06, 2016 7:37 pm 

Joined: Sat Nov 14, 2015 6:32 am
Posts: 159
Location: Melbourne, Australia
By the Gramm - Schmidt process we can establish the existence of an orthonormal basis $\mathbf{b_1, b_2, \dots, b_N}$ for $\mathbb{R}^N$ such that

{\rm span}_{\mathbb{R}} \left \{ \mathbf{a_1, a_2, \dots, a_N} \right \}={\rm span}_{\mathbb{R}}\left \{ \mathbf{a_1, a_2, \dots, b_N} \right \}

for each $n=1, 2, \dots, N$. Now, we may write $B=(\mathbf{b_1, b_2, \dots, b_N})$ for the corresponding $N \times N$ real and orthogonal matrix. By orthogonality each vector $\mathbb{\xi}$ in $\mathbb{R}^N$ has an expansion as:

$$\mathbf{\xi} = \sum_{n=1}^{N}\langle \mathbf{\xi}, \mathbf{b}_n\rangle \mathbf{b}_n \Rightarrow \left \| \mathbf{\xi} \right \|^2 = \sum_{n=1}^{N}\left | \langle \mathbf{\xi}, \mathbf{b}_n \rangle \right |^2 $$

On the other hand $(2)$ implies that each vector $\mathbf{a}_m$ has a shorter expansion of the form:

\mathbf{a}_m = \sum_{n=1}^{m}\langle \mathbf{a}_m, \mathbf{b}_n \rangle \mathbf{b}_n

Alternatively let $C=(c_{k\ell})$ be the $N \times N$ upper triangular matrix defined as:

$$c_{k\ell}= \langle \mathbf{a}_\ell, \mathbf{b}_k \rangle \;\; \text{if} \; 1 \leq k \leq \ell \;\; \textbf{and} \;\; c_{k \ell}=0 \; \; \text{if} \; \ell < k \leq N$$

Then $(3)$ is restated as $A=BC$ and using again the fact that $B$ has orthonormal columns and the fact that $C$ is upper triangular we get:

\left ( \det A \right )^2 &=\det \left ( A^T A \right ) \\
&= \det \left ( C^T B^T BC \right )\\
&= \det \left ( C^T C \right )\\
&= \left ( \det C \right )^2 \\
&= \prod_{n=1}^{N}\left | \langle \mathbf{a}_n , \mathbf{b}_n \rangle \right |^2 \\
&\leq \prod_{n=1}^{N} \left ( \sum_{m=1}^{n}\left | \langle \mathbf{a}_n, \mathbf{b}_n \rangle \right |^2 \right ) \\
&=\prod_{n=1}^{N} \left \| \mathbf{a}_n \right \|^2


  1. The above argument also shows that there exists equality if and only if

    $$\left | \langle \mathbf{a}_n, \mathbf{b}_n \rangle \right |^2 = \sum_{m=1}^{n}\left | \langle \mathbf{a}_m , \mathbf{b}_n \rangle \right |^2$$

    for each $n$. That is , if and only if, $\mathbf{a}_n= \langle \mathbf{a}_n, \mathbf{b}_n \rangle \mathbf{b}_n$. This can only be achieved if the vectors $\mathbf{a}_n$ are pairwise orthogonal.
  2. The geometrical interpretation of this inequality is the following: The volume of an $n$ dimensional parallelepiped produced by $n$ vectors can not exceed the product of their measures.

$\displaystyle \sum_{n=1}^{\infty}\frac{1}{n^s}= \prod_{p \; \text{prime}}\frac{1}{1-p^{-s}}$

Offline Profile  
Reply with quote  

Display posts from previous:  Sort by  
Post new topic Reply to topic  [ 2 posts ] 

All times are UTC [ DST ]

Mathimatikoi Online

Users browsing this forum: No registered users and 1 guest

You cannot post new topics in this forum
You cannot reply to topics in this forum
You cannot edit your posts in this forum
You cannot delete your posts in this forum
You cannot post attachments in this forum

Search for:
Jump to:  
Powered by phpBB® Forum Software © phpBB Group Color scheme created with Colorize It.
Theme created