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 $\det(A)\neq 0$, 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, $0\leq\lambda_{min}\leq a_{ii}\leq\lambda_{max}$, $\lambda_{min}$ 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 $D=diag(d_{11},d_{22},\ldots,d_{nn})$, 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 $\mathbf{x}\neq\mathbf{0}$, $\mathbf{x}^{H}B\mathbf{x}=\mathbf{x}^{H}DAD\mathbf{x}=(\mathbf{x}^{H}D)A(D% \mathbf{x})=(D^{H}\mathbf{x})^{H}A(D\mathbf{x})=(D\mathbf{x})^{H}A(D\mathbf{x}% )=\mathbf{y}^{H}A\mathbf{y}\geq 0$). 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,

$\det(B)=\prod_{i=1}^{n}\lambda_{B}^{(i)}\leq\left(\frac{1}{n}\sum_{i=1}^{n}% \lambda_{B}^{(i)}\right)^{n}=\left(\frac{1}{n}tr(B\right))^{n}=1$,

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 $A=D^{-1}BD^{-1}=D^{-2}=diag(a_{11},a_{22},\ldots,a_{nn})$. 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 $\det(A)\neq 0$; then $B=AA^{H}$ is a Hermitian positive semidefinite matrix (actually, $B^{H}=(AA^{H})^{H}=(A^{H})^{H}A^{H}=AA^{H}=B$ and, for any $\mathbf{x}\neq\mathbf{0}$,$\mathbf{x}^{H}B\mathbf{x}=\mathbf{x}^{H}AA^{H}\mathbf{x}=(\mathbf{x}^{H}A)(A^{% H}\mathbf{x})=(A^{H}\mathbf{x})^{H}(A^{H}\mathbf{x})=\mathbf{y}^{H}\mathbf{y}=% \|\mathbf{y}\|_{2}^{2}\geq 0$). Therefore, the second inequality can be applied to $B$, yielding:

$\left|\det(A)\right|^{2}=\det(A)\det^{\ast}(A)=\det(A)\det(A^{H})=\det(AA^{H})=$
$=\det(B)\leq\prod_{i=1}^{n}b_{ii}=\prod_{i=1}^{n}\sum_{j=1}^{n}a_{ij}(a^{H})_{% ji}=\prod_{i=1}^{n}\sum_{j=1}^{n}a_{ij}a_{ij}^{\ast}=\prod_{i=1}^{n}\sum_{j=1}% ^{n}\left|a_{ij}\right|^{2}$.

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. $\square$

## References

• 1 R. A. Horn, C. R. Johnson, Matrix Analysis, Cambridge University Press, 1985
Title proof of Hadamard’s inequality ProofOfHadamardsInequality 2013-03-22 15:37:02 2013-03-22 15:37:02 Andrea Ambrosio (7332) Andrea Ambrosio (7332) 17 Andrea Ambrosio (7332) Proof msc 15A45