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, 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 $\mathbf{x}\neq\mathbf{0}$ $\mathbf{x}^{H}B\mathbf{x}=\mathbf{x}^{H}AA^{H}\mathbf{x}=(\mathbf{x}^{H}A)(A^{H}\mathbf{x})=(A^{H}\mathbf{x})^{H}(A^{H}\mathbf{x})=\mathbf{y}^{H}\mathbf{y}=\Vert \mathbf{y}\Vert _{2}^{2}\geq 0$ . Therefore, the second inequality can be applied to $B$ yielding:

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

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 $

Bibliography

1

R. A. Horn, C. R. Johnson, Matrix Analysis, Cambridge University Press, 1985