proof of Hadamard’s inequality
Let’s first prove the second inequality. If $A$ is singular, the thesis is trivially verified, since for a Hermitian positive semidefinite matrix the right-hand side is always nonnegative, all the diagonal^{} entries being nonnegative. Let’s thus assume $det(A)\ne 0$, which means, $A$ being Hermitian positive semidefinite^{}, $det(A)>0$. Then no diagonal entry of $A$ can be $0$ (otherwise, since for a Hermitian positive semidefinite matrix, $0\le {\lambda}_{min}\le {a}_{ii}\le {\lambda}_{max}$, ${\lambda}_{min}$ and ${\lambda}_{max}$ being respectively the minimal and the maximal eigenvalue^{}, this would imply ${\lambda}_{min}=0$, that is $A$ is singular); for this reason we can define $D=diag({d}_{11},{d}_{22},\mathrm{\dots},{d}_{nn})$, with ${d}_{ii}={a}_{ii}^{-\frac{1}{2}}\in \mathbb{R}$, since all ${a}_{ii}\in {\mathbb{R}}^{+}$. Let’s furthermore define $B=DAD$. It’s easy to check that $B$ too is Hermitian positive semidefinite, so its eigenvalues ${\lambda}_{B}$ are all non-negative (actually, since $A={A}^{H}$ and since $D$ is real and diagonal, ${B}^{H}={(DAD)}^{H}={D}^{H}{(DA)}^{H}={D}^{H}{A}^{H}{D}^{H}=DAD=B$; on the other hand, for any $\mathbf{x}\ne \mathrm{\U0001d7ce}$, ${\mathbf{x}}^{H}B\mathbf{x}={\mathbf{x}}^{H}DAD\mathbf{x}=({\mathbf{x}}^{H}D)A(D\mathbf{x})={({D}^{H}\mathbf{x})}^{H}A(D\mathbf{x})={(D\mathbf{x})}^{H}A(D\mathbf{x})={\mathbf{y}}^{H}A\mathbf{y}\ge 0$). Moreover, we have obviously ${b}_{ii}={d}_{ii}{a}_{ii}{d}_{ii}=1$ so that $tr(B)=n$ and, recalling the geometric-arithmetic mean inequality (http://planetmath.org/ArithmeticGeometricMeansInequality), which holds in this case because the eigenvalues of $B$ are all non-negative,
$det(B)={\prod}_{i=1}^{n}{\lambda}_{B}^{(i)}\le {\left(\frac{1}{n}{\sum}_{i=1}^{n}{\lambda}_{B}^{(i)}\right)}^{n}={(\frac{1}{n}tr(B))}^{n}=1$,
and since $det(B)=det{(D)}^{2}det(A)={\left({\prod}_{i=1}^{n}{a}_{ii}\right)}^{-1}det(A)$, we have the thesis. Since $det(A)={\prod}_{i=1}^{n}{a}_{ii}$ if and only if $det(B)=1$ and since in the geometric-arithmetic inequality equality holds if and only if all terms are equal, we must have ${\lambda}_{B}^{(i)}={\lambda}_{B}$, so that ${\prod}_{i=1}^{n}{\lambda}_{B}^{(i)}={\lambda}_{B}^{n}=1$, whence ${\lambda}_{B}=1$ (${\lambda}_{B}$ having to be non-negative), and since $B$ is Hermitian and hence is diagonalizable, we obtain $B=I$, and so $A={D}^{-1}B{D}^{-1}={D}^{-2}=diag({a}_{11},{a}_{22},\mathrm{\dots},{a}_{nn})$. So we can conclude that equality holds if and only if $A$ is diagonal.
Let’s now derive the more general first inequality. Let $A$ be a complex-valued $n\times n$ matrix. If $A$ is singular, the thesis is trivially verified. Let’s thus assume $det(A)\ne 0$; then $B=A{A}^{H}$ is a Hermitian positive semidefinite matrix (actually, ${B}^{H}={(A{A}^{H})}^{H}={({A}^{H})}^{H}{A}^{H}=A{A}^{H}=B$ and, for any $\mathbf{x}\ne \mathrm{\U0001d7ce}$,${\mathbf{x}}^{H}B\mathbf{x}={\mathbf{x}}^{H}A{A}^{H}\mathbf{x}=({\mathbf{x}}^{H}A)({A}^{H}\mathbf{x})={({A}^{H}\mathbf{x})}^{H}({A}^{H}\mathbf{x})={\mathbf{y}}^{H}\mathbf{y}={\parallel \mathbf{y}\parallel}_{2}^{2}\ge 0$). Therefore, the second inequality can be applied to $B$, yielding:
${\left|det(A)\right|}^{2}=det(A){det}^{\ast}(A)=det(A)det({A}^{H})=det(A{A}^{H})=$
$=det(B)\le {\prod}_{i=1}^{n}{b}_{ii}={\prod}_{i=1}^{n}{\sum}_{j=1}^{n}{a}_{ij}{({a}^{H})}_{ji}={\prod}_{i=1}^{n}{\sum}_{j=1}^{n}{a}_{ij}{a}_{ij}^{\ast}={\prod}_{i=1}^{n}{\sum}_{j=1}^{n}{\left|{a}_{ij}\right|}^{2}$.
As we proved above, for $det(B)$ to be equal to ${\prod}_{i=1}^{n}{b}_{ii}$, $B$ must be diagonal, which means that ${\sum}_{k=1}^{n}{a}_{ik}{a}_{jk}^{\ast}={|{a}_{ii}|}^{2}{\delta}_{ij}$. So we can conclude that equality holds if and only if the rows of $A$ are orthogonal^{}. $\mathrm{\square}$
References
- 1 R. A. Horn, C. R. Johnson, Matrix Analysis, Cambridge University Press, 1985
Title | proof of Hadamard’s inequality |
---|---|
Canonical name | ProofOfHadamardsInequality |
Date of creation | 2013-03-22 15:37:02 |
Last modified on | 2013-03-22 15:37:02 |
Owner | Andrea Ambrosio (7332) |
Last modified by | Andrea Ambrosio (7332) |
Numerical id | 17 |
Author | Andrea Ambrosio (7332) |
Entry type | Proof |
Classification | msc 15A45 |