determinant inequalities

There are a number of interesting inequalities bounding the determinant of a $n\times n$ complex matrix $A$, where $\rho$ is its spectral radius:

1) $\left|det(A)\right|\le {\rho }^n(A)$
2) $\left|det(A)\right|\le {\prod }_{i=1}^n\left({\sum }_{j=1}^n\left|a_{ij}\right|\right)={\prod }_{i=1}^n{\parallel a_i\parallel }_1$
3) $\left|det(A)\right|\le {\prod }_{j=1}^n\left({\sum }_{i=1}^n\left|a_{ij}\right|\right)={\prod }_{j=1}^n{\parallel a_j\parallel }_1$
4) $\left|det(A)\right|\le {\prod }_{i=1}^n{\left({\sum }_{j=1}^n{\left|a_{ij}\right|}^2\right)}^{\frac{1}{2}}={\prod }_{i=1}^n{\parallel a_i\parallel }_2$
5) $\left|det(A)\right|\le {\prod }_{j=1}^n{\left({\sum }_{i=1}^n{\left|a_{ij}\right|}^2\right)}^{\frac{1}{2}}={\prod }_{j=1}^n{\parallel a_j\parallel }_2$
6) if $A$ is Hermitian positive semidefinite, $det(A)\le {\prod }_{i=1}^n{a}_{ii}$, with equality if and only if $A$ is diagonal.

Inequalities 4)-6) are known as ”Hadamard’s inequalities”.

(Note that inequalities 2)-5) may suggest the idea that such inequalities could hold: $\left|det(A)\right|\le {\prod }_{i=1}^n{\parallel a_i\parallel }_p$ or $\left|det(A)\right|\le {\prod }_{j=1}^n{\parallel a_j\parallel }_p$ for any $p\in 𝐍$; however, this is not true, as one can easily see with and $p=3$. Actually, inequalities 2)-5) give the best possible estimate of this kind.)

Proofs:

1) $\left|det(A)\right|=\left|{\prod }_{i=1}^n{\lambda }_i\right|={\prod }_{i=1}^n\left|{\lambda }_i\right|\le {\prod }_{i=1}^n\rho (A)={\rho }^n(A).$

2) If $A$ is singular, the thesis is trivial. Let then $det(A)\ne 0$. Let’s define $B=DA$, $D=diag(d_{11},d_{22},\mathrm{\cdots },d_{nn})$,$d_{ii}={\left({\sum }_{j=1}^n\left|a_{ij}\right|\right)}^{-1}$. (Note that $d_{ii}$ exist for any $i$, because $det(A)\ne 0$ implies no all-zero row exists.) So ${\parallel B\parallel }_{\mathrm{\infty }}={\max }_i\left({\sum }_{j=1}^n\left|b_{ij}\right|\right)=1$ and, since $\rho (B)\le {\parallel B\parallel }_{\mathrm{\infty }}$, we have:

$\left|det(B)\right|=\left|det(D)\right|\left|det(A)\right|={\left({\prod }_{i=1}^n{\sum }_{j=1}^n\left|a_{ij}\right|\right)}^{-1}\left|det(A)\right|\le {\rho }^n(B)\le {\parallel B\parallel }_{\mathrm{\infty }}^n=1$,

from which:

$\left|det(A)\right|\le {\prod }_{i=1}^n\left({\sum }_{j=1}^n\left|a_{ij}\right|\right).$

3) Same as 2), but applied to $A^T$.

4)-6) See related proofs attached to ”Hadamard’s inequalities”.