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Hadamard Inequality

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:

(1)   \begin{equation*} \left | \det A \right |\leq \prod_{n=1}^{N}\left \| a_n \right \| \end{equation*}

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

Solution

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

(2)   \begin{equation*} {\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 \} \end{equation*}

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 \xi in \mathbb{R}^N has an expansion as:

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

(3)   \begin{equation*} \mathbf{a}_m = \sum_{n=1}^{m}\langle \mathbf{a}_m, \mathbf{b}_n \rangle \mathbf{b}_n \end{equation*}

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

\displaystyle 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:

    \begin{align*} \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 \end{align*}

Notes:

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

    \displaystyle \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.

  • 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.

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