1
$\begingroup$

This question already has an answer here:

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 :)

$\endgroup$

marked as duplicate by Martin R, Jyrki Lahtonen, mechanodroid, Paul Frost, Cesareo Dec 2 '18 at 13:42

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

  • $\begingroup$ 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? $\endgroup$ – C. Brendel Dec 2 '18 at 9:31
  • 1
    $\begingroup$ Note that there is (at least) one answer which does not use Gershgorin's circle theorem, but does an explicit estimate on $x^T A x$. $\endgroup$ – Martin R Dec 2 '18 at 9:42
  • $\begingroup$ Oh yeah right! Thanks :) That's quite a coefficient match! Either way, the proof I accepted seems waaay easier :P $\endgroup$ – C. Brendel Dec 2 '18 at 10:53
1
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ Perfect :D Thank you! Quite elegant! :) $\endgroup$ – C. Brendel Dec 2 '18 at 10:51

Not the answer you're looking for? Browse other questions tagged or ask your own question.