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?


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.

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

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