Prove that $\det A$ does not exceed $1$

Question

Let $A =(a_{ij}) ∈ M_n(\Bbb R)$ be a matrix with nonnegative entries such that the sum of the entries in each row does not exceed $1$. Prove that $|\det A|$ does not exceed $1$ too.

This is one of my exercise,I tried to induction,but i'm stuck. Help me please! Thanks

Answer 1

You should be able to use expansion by minors in an inductive proof.

Answer 2

I think that this is a direct consequence of Hadamard's inequality (determinant is just a volume). Google it!

Answer 3

You don't need induction. Prove/use the following two results: 1) the determinant is the product of the eigenvalues, and 2) any eigenvalue of a nonnegative matrix has magnitude at most the largest row sum (just write out the eigenvalue equation to see why).

Answer 4

You can adapt part (b) of this proof about stochastic matrics: https://yutsumura.com/eigenvalues-of-a-stochastic-matrix-is-always-less-than-or-equal-to-1/

Copying most their proof here per site policy on self contained answers:

For eigenvector $\lambda$, $AV = \lambda V$ for some nonzero vector $V$, and looking at the $i'th$ row of that matrix equation gets:

$$\sum_{j} A_{i,j}V_{j} = \lambda V_{i}$$

Let $V_k$ be the element of that $V$ with the largest magnitude, that is

$$|V_k| = \max( |V_1|, |V_2| \dots)$$

Note $|V_k > 0|$ by the fact $V$ is an eigenvector as stated above.

Now use the triangle inequality $|\sum_k m_k n_k| \le \sum |m_k| |n_k|$ and the above facts to prove that $|\lambda| \cdot |V_k| = |\lambda \cdot V_k| \le |V_k|$. Then the fact that the determinant is the product of the eigen values to finish your proof.

Answer 5

$\lvert \det(A) \rvert \stackrel{\text{row exp.}}= \lvert \sum_{j=1}^n (-1)^{j+1} \cdot a_{1j}\cdot \det(A_{1,j}) \rvert \le \sum_{j=1}^n \lvert a_{1j} \rvert \cdot \lvert \det(A_{1,j}) \rvert \stackrel{\text{ind. hyp.}}\le \sum_{j=1}^n \lvert a_{1j} \rvert \stackrel{\text{assumption}}\le 1$