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Suppose $A \in \mathbb{R}^{n\times n}$ is symmetric and diagonally dominant with positive diagonal entries. I have to prove that $A$ is positive definite but without using theorems, just algebraically.

I´ve started with: $$x^T A x = \sum_{i=1}^n a_{ii} x_i^2 + \sum_{i=j} a_{ij} x_i x_j > \sum_{i=1}^n \sum_{i\neq j} |a_{ij}| x_i^2 + \sum_{i\neq j} a_{ij} x_i x_j$$ but I could much further. I was thiking about: $$ \sum_{1=1}^n \sum_{j>i} |a_{ij}| (x_i^2 + x_j^2) + 2 a_{ij} x_i x_j$$ but I´m not sure how to continue.

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  • $\begingroup$ Doesn't positive definite enclose symmetric? Hence, can't you just write down a non-symmetric matrix which satisfies your assumptions? $\endgroup$ – Klaus Nov 14 '19 at 16:44
  • $\begingroup$ @Klaus In some circles, a real matrix is said to be positive definite if and only if $x^TAx > 0$ for all real vectors $x$, which is equivalent to the condition that the symmetric matrix $A + A^T$ is "symmetric and" positive definite. $\endgroup$ – Omnomnomnom Nov 14 '19 at 17:10
  • $\begingroup$ @Klaus it was my mistake, A is SYMMETRIC and diagonally dominant. $\endgroup$ – variableXYZ Nov 14 '19 at 17:45
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I guess you write $x^TAx=\sum_{i=1}^n a_{i,i}x_i^2+\sum_{i\neq j}a_{i,j}x_ix_j\ge\sum_{i=1}^n (\sum_{i\neq j}|a_{i,j}|)x_i^2-\sum_{i\neq j}|a_{i,j}||x_i|x_j| =\sum_{j>i }(|a_{i,j}|(x_i^2+x_j^2-2|x_i||x_j|))\ge 0$

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