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Let $A=(a_{ij}) \in \mathbb{R}^{n \times n}$ be a symmetric matrix. For all $i=1, \dots ,n$ we have $a_{ii} > \sum_{i \ne j} \vert{a_{ij}}\vert$. I now have to show that $A$ is positive definite. I tried to look at the general case $x^\top Ax$, using the estimate above. Unfortunately without any luck... In the end I got something like $x^\top Ax > \sum_{m=1}^n (x_m ^2 + x_m \min\{x_k : k \ne m\}) \sum_{m \ne k}a_{mk}$.

I am unfortunately not allowed to use Gershgorins Circle Theorem as we did not discuss it up to this date :)

Any ideas to prove the statement?

~Cedric :)

Cedric Brendel
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  • Indeed, but unfortunately I don't have access to Gershgorins Circle Theorem... Do you know of a way to prove the statement without the theorem? – Cedric Brendel Dec 02 '18 at 09:31
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    Note that there is (at least) one answer which does not use Gershgorin's circle theorem, but does an explicit estimate on $x^T A x$. – Martin R Dec 02 '18 at 09:42
  • Oh yeah right! Thanks :) That's quite a coefficient match! Either way, the proof I accepted seems waaay easier :P – Cedric Brendel Dec 02 '18 at 10:53

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It's actually a simple estimation. We can see that $$\begin{eqnarray}x^T Ax &=&\sum_{1\leq i\leq n} a_{ii}x_i^2 + \sum_{1\leq i,j\leq n,i\neq j} a_{ij}x_ix_j\\&\geq &\sum_{1\leq i\leq n} a_{ii}x_i^2 - \sum_{1\leq i,j\leq n,i\neq j} |a_{ij}||x_i||x_j|\\ &\geq &\sum_{1\leq i\leq n} a_{ii}x_i^2 - \frac{1}{2}\sum_{1\leq i,j\leq n,i\neq j} |a_{ij}|(|x_i|^2+|x_j|^2)\\&=&\sum_{1\leq i\leq n} a_{ii}x_i^2 - \sum_{1\leq i,j\leq n,i\neq j} |a_{ij}||x_i|^2 \\&=&\sum_{1\leq i\leq n} (a_{ii}- \sum_{1\leq j\leq n,j\neq i} |a_{ij}|)|x_i|^2 >0, \end{eqnarray}$$ for all $x = (x_1,x_2,\ldots,x_n)\in \mathbb{R}^n\setminus\{0\}.$ The fact that $|a_{ij} | = |a_{ji}|$ is used in the second equality.

Myunghyun Song
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