# 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) $\left|\det(A)\right|\leq\rho^{n}(A)$
2) $\left|\det(A)\right|\leq\prod_{i=1}^{n}\left(\sum_{j=1}^{n}\left|a_{ij}\right|% \right)=\prod_{i=1}^{n}\|a_{i}\|_{1}$
3) $\left|\det(A)\right|\leq\prod_{j=1}^{n}\left(\sum_{i=1}^{n}\left|a_{ij}\right|% \right)=\prod_{j=1}^{n}\|a_{j}\|_{1}$
4) $\left|\det(A)\right|\leq\prod_{i=1}^{n}\left(\sum_{j=1}^{n}\left|a_{ij}\right|% ^{2}\right)^{\frac{1}{2}}=\prod_{i=1}^{n}\|a_{i}\|_{2}$
5) $\left|\det(A)\right|\leq\prod_{j=1}^{n}\left(\sum_{i=1}^{n}\left|a_{ij}\right|% ^{2}\right)^{\frac{1}{2}}=\prod_{j=1}^{n}\|a_{j}\|_{2}$
6) if $A$ is Hermitian positive semidefinite, $\det(A)\leq\prod_{i=1}^{n}a_{ii}$, 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: $\left|\det(A)\right|\leq\prod_{i=1}^{n}\|a_{i}\|_{p}$ or $\left|\det(A)\right|\leq\prod_{j=1}^{n}\|a_{j}\|_{p}$ for any $p\in\mathbf{N}$; however, this is not true, as one can easily see with $A=\begin{bmatrix}1&1\\ -1&1\end{bmatrix}$ and $p=3$. Actually, inequalities 2)-5) give the best possible estimate of this kind.)

Proofs:

1) $\left|\det(A)\right|=\left|\prod_{i=1}^{n}\lambda_{i}\right|=\prod_{i=1}^{n}% \left|\lambda_{i}\right|\leq\prod_{i=1}^{n}\rho(A)=\rho^{n}(A).$

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

$\left|\det(B)\right|=\left|\det(D)\right|\left|\det(A)\right|=\left(\prod_{i=1% }^{n}\sum_{j=1}^{n}\left|a_{ij}\right|\right)^{-1}\left|\det(A)\right|\leq\rho% ^{n}(B)\leq\|B\|_{\infty}^{n}=1$,

from which:

$\left|\det(A)\right|\leq\prod_{i=1}^{n}\left(\sum_{j=1}^{n}\left|a_{ij}\right|% \right).$

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

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

Title determinant inequalities DeterminantInequalities 2013-03-22 15:34:46 2013-03-22 15:34:46 Andrea Ambrosio (7332) Andrea Ambrosio (7332) 12 Andrea Ambrosio (7332) Result msc 15A15