How to show that det(A)≤1?

Question

Let $A = (a_{ij})_n$ where $a_{ij} \ge 0$ for $i,j=1,2,\ldots,n$ and $\sum_{j=1}^n a_{ij} \le 1$ for $i = 1,2,\ldots,n$.
Show that $|\det(A)| \le 1$.

Should I use the definition of matrix:

$$\det(A)=\sum \textrm{sgn}(\sigma)a_{1,\sigma(1)}a_{2,\sigma(2)}...a_{n,\sigma(n)}$$

I don't understand what is $a_{i,\sigma(i)}$ ? Where $i=1,2, \ldots,n$. Or is there another way to solve it?

Answer 1

Correct. Use Leibniz formula for determinants. See here for an example.

As for the a's (from the link above, formatting mine):

Here the sum is computed over all permutations σ of the set {1, 2, ..., n}. A permutation is a function that reorders this set of integers. The value in the ith position after the reordering σ is denoted $σ_i$. For example, for n = 3, the original sequence 1, 2, 3 might be reordered to σ = [2, 3, 1], with $σ_1 = 2, σ_2 = 3$, and $σ_3 = 1$.

Answer 2

Here is the sneaky approach:

First note, we can decide whether we want the column-sums or the row-sums to be $\leq 1$, since we can of course transpose our matrix. I want the column-sums to be $\leq 1$.

The $1$-norm $\| \cdot \|_1$ on $\mathbb C^n$ induces via

$$\| A \| := \sup_{x \neq 0} \frac{\|Ax\|_1}{\|x\|_1}$$

the column-sum norm on the $n \times n$-matrices. The assumption is precisely $\|A\| \leq 1$ then.

For any $x \in \mathbb C^n$ we have $\|Ax\|_1 \leq \|A\|\|x\|_1 \leq \|x\|_1$. This holds in particular for any eigenvector of $A$, which shows that any (possibly complex) eigenvalue $\lambda$ of $A$ satisfies $|\lambda| \leq 1$. This shows the assertion, since the determinant is obtained as the product of all eigenvalues.

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We also have a geometric reason: The edges of the unit cubes are mapped into the unit cube, hence any convex combination of them is mapped into the unit cube. This shows $A[0,1]^n \subset [0,1]^n$. Since the determinant is geometrically defined to be the volume of the image of the unit cube, we immediately get the desired result.

Answer 3

You can prove that by induction on the dimension $n$ of the matrix. It is clearly true for $n = 1$.

Now assume that $n > 1$ and the statement is true for all matrices of dimension $(n-1, n-1)$. Using the Laplace formula, you have $$ \det(A) = \sum_{j=1}^n a_{1j} \det(A_j) $$ where $A_1, ..., A_n$ are submatrices of the dimension $(n-1, n-1)$ which have the same properties as $A$: All entries are positive and the row sums are less or equal to one, so $|\det(A_j)| \le 1$ for $j=1,..., n$.

Then $$ |\det(A)| \le \sum_{j=1}^n a_{1j} |\det(A_j)| \le \sum_{j=1}^n a_{1j} \le 1 $$

Answer 4

Just take modulus the determinant expression you wrote. Note each term on the right under sum can be bounded by using G.M. $\le$ A.M.