0

Let $A \in \Bbb C^{n \times n}$ be a non-Hermitian matrix whose entries are denoted by $a_{i,j}$.

  1. What is the best upper bound that we have for $\det (A)$ in term of $\mbox{trace}(A)$?

  2. Does the following inequality hold for any complex matrix $A$?

$$\det(A) \leq \bigg(\frac{\mbox{trace}}{n}\bigg)^n$$

Rodrigo de Azevedo
  • 1
Abdo
  • 167
  • 1
    Related https://math.stackexchange.com/questions/202248/upper-bound-on-determinant-of-matrix-in-terms-of-trace – Peanut Aug 24 '20 at 17:43
  • 1
    You need to give more information about what you are looking for, the best upper bound is $\det A$, of course. There is Hadamard's inequality. – copper.hat Aug 24 '20 at 17:45
  • @Dude in your link the upper bound holds for A is Hermitian or symmetric .but I didi' t assume that? – Abdo Aug 24 '20 at 17:52
  • if $\det(A) \leq \bigg(\frac{Trace(A)}{n}\bigg)^n$ was true for any complex matrix then $\text{GM}\leq \text{AM}$ would hold for any selection of $x_i \in \mathbb C$ which of course is wrong. $\text{GM}\leq \text{AM}$ requires $x_i\geq 0$ for a reason. – user8675309 Aug 24 '20 at 20:06

1 Answers1

0

The best upper bound is $+\infty$. Consider the companion matrix $$ C=\pmatrix{0&&&&d\\ 1&\ddots&&&0\\ &\ddots&\ddots&&\vdots\\ &&\ddots&0&0\\ &&&1&t}. $$ Then $\operatorname{tr}(C)=t$ and $\det(C)=(-1)^{n-1}d$. Since $d$ and $t$ do not depend on each other, $|\det(C)|$ can assume any nonnegative real value, given $\operatorname{tr}(C)$.

user1551
  • 139,064
  • I ask whether this inequality holds for any complex matrix $$\det(A) \leq \bigg(\frac{Trace(A)}{n}\bigg)^n$$ – Abdo Aug 24 '20 at 18:12
  • @Abdo Of course it doesn't hold. Just pick a $d$ that is large enough in the counterexample above. If you want an inequality of the $AM\ge GM$ type, you need $A$ to have a nonnegative spectrum. – user1551 Aug 24 '20 at 18:37
  • do you know other upper bound for the complex matrix? – Abdo Aug 24 '20 at 19:00
  • 1
    @Abdo As copper.hat has pointed out in a comment under your question, we have Hadamard's inequality $|\det(A)|\le\prod_{j=1}^n|a_{\ast j}|_2$, which is a simple consequence of QR factorisation. – user1551 Aug 24 '20 at 19:51
  • in Wikipedia article said that Hadamard's inequality holds if and only if the columns vectors of $A$ are orthogonal is that true? – Abdo Aug 25 '20 at 00:37
  • 1
    @Abdo Yes and this also follows from QR factorisation. – user1551 Aug 25 '20 at 02:47
  • if we assume that A is a complex square matrix which also diagonalizable and $detA \in \mathbb{R}$, in this case, does the inequality of the trace(the first one) holds ? – Abdo Aug 26 '20 at 14:34
  • @Abdo No. Consider $A=\pmatrix{0&-a\ a&0}$ with $a>0$ for instance. It is diagonalisable over $\mathbb C$ because its two eigenvalues $\pm i$ are different. Its trace is zero, but $\det(A)=a^2$ is unbounded. – user1551 Aug 26 '20 at 14:49
  • actually I am looking for and an upper for the matrix of a linear transformation $m_{\alpha}(x)=\alpha x$ where the basis $S={ \zeta, \zeta^{2}, ....., \zeta^{p-1}}$ consist of the p-the root of unity, and $p$ is prime so obviously, the matrix representation $[m_{\alpha}]$ is compilex but it is hard to prove that it is Hermitian or orthogonal , so do you know any upper which help here ? – Abdo Aug 26 '20 at 15:38