The following problems is from the book Berkeley Problems in the Mathematics. The solution is not given for this problem.
Problem:Let $A$ be a real $n \times n$ real symmetric matrix with diagonal entries non-negative and all other entries non-positive. Also any column sum is strictly positive.
Show $det(A)>0$
I am only able to see this for $2 \times 2 $ case. Since $a_{11} > -a_{21}$ and $a_{22} > -a_{12}$ from the positive column sum conditions.
Now we can multiply the inequalities since both sides are positive, to get $a_{11}a_{22} > a_{21}a_{12}$.
But this idea doesn't generalize when we go to higher dimensions since we get many other terms than 'product over permutations'.
I tried writing this as: $A^{t}(1, 1, 1)^t > 0 , e_iAe_j \leq 0$ and $e_iAe_i \geq 0$ but this also doesn't seem to lead anywhere. This matrix need not be positive definite as the given conditions don't extend to submatrices.
Am I missing some property of Linear Transformations with positive determinant?
In the $2 \times 2$ case, I see the following geometrical interpretation: $A$ has positive determinant IFF the angle between first row and second row (going anticlockwise starting from the first row) is less than $180^{\circ}$ but I couldn't generalize this to higher dimensions either.
I am looking for hints. Thank you.
Best Answer
First we show that $$\langle Ax,x\rangle >0,\qquad x\neq 0\qquad\ \ (*)$$ $$\displaylines{\langle Ax,x\rangle=\sum_{i,j=1}^n a_{ij}x_ix_j=\sum_{i=1}^n a_{ii}x_i^2 +\sum_{i\neq j}a_{ij}x_ix_j \\ \ge \sum_{i=1}^n a_{ii}x_i^2-\sum_{i\neq j}|a_{ij}|\,|x_i|\,|x_j|}$$ Next $$\displaylines{\sum_{i\neq j}|a_{ij}|\,|x_i|\,|x_j|\le {1\over 2} \left (\sum_{i\neq j}|a_{ij}|\,x_i^2+\sum_{i\neq j}|a_{ij}|\,x_j^2\right )\\ =\sum_{i\neq j}|a_{ij}|\,x_i^2\le \sum_{i=1}^nx_i^2\sum_{j=1, j\neq i}^n|a_{ij}|< \sum_{i=1}^n a_{ii}x_i^2}$$ This implies $(*),$ which in turn gives that all the eigenvalues of $A$ are positive. Therefore $\det A>0.$
Remark The proof can be slightly modified to capture the case of complex space $\mathbb{C}^n$ and a self-adjoint matrix $A$ such that $$\sum_{j=1,j\neq i}^n|a_{ij}|<a_{ii}$$