Let $A$ be an $n \times n$ matrix whose diagonal entries are strictly positive and off-diagonal entries are negative. The sum of the entries on each column is $1$. Prove that $\det(A) > 1$.
I think to find all of eigenvalue to prove the determinant is greater than $1$ but I only get $1$ as eigenvalue.
Best Answer
Let $a_{ij}$ be the entries of the matrix. Then $a_{ii}=1+\sum_j|a_{ij}|$ since the other terms in the same column are negative.
Now consider $Ax=\lambda x$, $$\lambda x_j=\sum_ia_{ij}x_i=a_{jj}x_j+\sum_{i\ne j}a_{ij}x_i$$ so if we pick the largest coefficient of $x$, say $\alpha=x_k$, then $$|\lambda-a_{kk}||\alpha|\le\sum_{i\ne k}|a_{ik}||x_i|\le\sum_{i\ne k}|a_{ik}| |\alpha|=(a_{kk}-1)|\alpha|$$ $$\therefore\quad |\lambda-a_{kk}|\le a_{kk}-1$$ in particular $1\le\lambda$ (or $\mathrm{Re}(\lambda)\ge1$ if complex). So the determinant, as the product of the eigenvalues, is larger than $1$. (Note: $\lambda\bar{\lambda}=|\lambda|^2\ge1$.)
The general result along these lines is Gershgorin's theorem.