Bounding the determinant of a matrix with bounded coefficients

Question

Suppose that $A$ is a real matrix of dimension $n \times n$ and that its coefficients are bounded by $c\ge0$ ($\vert a_{ij} \vert \le c$ for all $1\le i,j \le n$).

How to prove that $$\vert \det A \vert \le c^n n^{n/2}$$

Answer 1

If we consider $A/c$, we see that it is equivalent to show that for $c = 1$, we have $\det(A) \leq n^{n/2}$. Moreover, since $|\det(A)| = \sqrt{\det(A^TA)}$, it suffices to note that $\det(A^TA) \leq n^n$.

With this in mind, suppose that $A$ is such that $|a_{ij}| \leq 1$. We note that $M = A^TA$ satisfies $$ |M_{ij}| = \left|\sum_{k=1}^n a_{ik}a_{kj}\right| \leq \sum_{k=1}^n |a_{ik}| \,|a_{kj}| \leq \sum_{k=1}^n 1 = n $$ Moreover, $M$ is symmetric positive definite. Let $\lambda_i$ denote the eigenvalues of $M$. We have $$ \operatorname{tr}(M) = \sum_i \lambda_i = \sum_{i} M_{ii} \leq n^2 $$ Now, consider the problem of maximizing $\prod_{i=1}^n \lambda_i$ under the constraint that we have $\lambda_i \geq 0$ and $\sum \lambda_i \leq n^2$. We find that this product is maximized when $$ \lambda_1 = \cdots = \lambda_n = n^2/n = n $$ for a maximum product of $n^n$. Or, if you prefer, the AM-GM inequality tells us that $$ \prod_{i=1}^n \lambda_i = \left(\left[\prod_{i=1}^n \lambda_i\right]^{1/n}\right)^n \leq \left(\left[\sum_{i=1}^n \lambda_i/n\right]\right)^n \leq n^n $$

Thus, it follows that $$ \det(A^TA) = \det(M) = \prod_{i=1}^n \lambda_i \leq n^n $$ As was desired.

Answer 2

Since

is ture for every element in the matrix, let us consider a matrix where all of the coefficients are the value of c.

Now factor out the c throughout the entire matrix.

Using a formula we obtain

Which reduces to

Using another formula

Reduces to

Since all elements in the matrix where of value c then all of the elements left are valued as 1. The result will be that the determinant of a matrix times a matrix transpose will be zero.