# Show that $\det(A) > 0$

Let $(a_{ij})$ be a real $n \times n$ matrix satisfying,

1. $a_{ii} > 0 \space (1 \leq i \leq n) ,$
2. $a_{ij} \leq 0 \space (i \ne j, 1 \leq i,j \leq n) ,$
3. $\sum_{i=1}^ {i=n} \space a_{ij} > 0 (1 \leq j \leq n).$

Then $\det (A) > 0$

How to prove this? I have no idea.

$A$ has positive eigenvalues because it is diagonally dominant (why?) and the diagonal entries are positive. This suffices to show that $\det A = \prod_{i=1}^n \lambda_i > 0$ where $\lambda_i$ are the eigenvalues of $A$.
• Uhh, it's a typo. My example $A$ should be $\pmatrix{6&0\\-5&1}$. The definition in your book is equivalent to that $\frac12(A+A^T)$ has a positive spectrum. With $x=(4,9)^T$, we have $x^TAx=-3$. – user1551 May 21 '15 at 14:59
• @user1551 Thanks for your patience. I've fixed the error now. I forgot that $A\succ 0$ is only equivalent to $\sigma(A)\subset\mathbb R_+$ if $A=A^T$. – AlexR May 21 '15 at 15:02