# Determinant of a matrix with positive diagonal entries is greater than 1

Let $$A$$ be a $$n\times n$$ matrix with entries on its diagonal are positive and other entries are negative with sum of entries in every column is 1. Prove that $$\det(A) > 1.$$

I got no idea to begin with. Any suggestion or hint?

## 1 Answer

By your assumptions, $$A^T$$ is a matrix with positive entries on the diagonal and negative off diagonal entries such that each row sums to 1. Let $$B$$ denote any matrix satisfying these conditions, we'll prove that $$\det(B)>1$$. First, notice that $$B$$ has an eigenvector $$(1,\dots,1)$$ with eigenvalue 1. Now suppose $$\textbf{x}=(x_1,\dots,x_n)$$ is an eigenvector with complex entries, not all the same, and eigenvalue $$\lambda$$, we'll show that $$|\lambda|>1$$. Suppose $$x_i$$ has maximal modulus among the entries of $$\textbf{x}$$ and suppose WLOG that $$x_i>0$$ (otherwise just multiply by the appropriate phase). Now we know that $$|\lambda x_i|=|\sum_{j\neq i}b_{i,j}x_j+b_{i,i}x_i| \geq |b_{i,i}x_i|-\sum_{j\neq i}|b_{i,j}x_j| = b_{i,i}x_i+\sum_{j\neq i}b_{i,j}|x_j| \geq \sum_{j=1}^n b_{i,j}x_i=x_i$$ But one of the above inequalities must be a strong inequality (the first one if $$x_j$$ all have the same modulus and the second if not). This implies that $$|\lambda|>1$$, as desired.

• Spent some time to catch your first ineq, but still struggling. Would you mind to elaborate on it? – Hanno Apr 14 at 14:32
• @Hanno No problem, I've added an intermediate step. Sorry if it wasn't clear. – Amichai Lampert Apr 14 at 14:53
• Now I got it & you got it too, +1 – Hanno Apr 15 at 13:06
• @Hanno haha thanks – Amichai Lampert Apr 15 at 13:38