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 Post subject: Hadamard's inequality
PostPosted: Mon Dec 28, 2015 10:45 am 
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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.

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PostPosted: Wed Apr 06, 2016 7:37 pm 

Joined: Sat Nov 14, 2015 6:32 am
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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.

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

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