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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.

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Hint
$A$ 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 $A$.

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  • $\begingroup$ 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 $x^TAx=-3$. $\endgroup$ – user1551 May 21 '15 at 14:59
  • $\begingroup$ @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 $A=A^T$. $\endgroup$ – AlexR May 21 '15 at 15:02
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You can also use Gaussian Elimination Method. Show that in each step of elimination, elements of main diagonal stay positive. So, you will have an upper triangular matrix in which all the elements of main diagonal are positive. Now use the fact that in an upper triangular matrix, determinant is equal to the product of elements of main diagonal.

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