I have encountered a problem of determining the condition for matrix $A+A^T$ to be positive definite, where

$$A= \operatorname{diag} \{a_1, a_2, \ldots, a_n\} + \mathbb{1}\cdot [b_1, b_2, \ldots, b_n]$$ where $\mathbb{1}$ is a column vector with element $1$, and $b_i>0$.

Is it true that $A$ is positive definite if and only if $\operatorname{diag} \{a_1, a_2, \ldots, a_n\}$ is positive definite, i.e., $\min(a_i)>0$? Or it has something to do with the $b_i$

Thanks in advance!

  • $\begingroup$ Try $n=2$, $a_1=a_2=1$, $b_1=1$, $b_2=100$. $\endgroup$ – Jonas Meyer Nov 1 '17 at 21:43
  • $\begingroup$ What is special for this example? $\endgroup$ – Lynn Nov 2 '17 at 23:08
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    $\begingroup$ What is $A+A^T$ for this example? It appears to me you can learn a lot about your question by experimenting with $n=2$ examples, my previous comment being an example. $\endgroup$ – Jonas Meyer Nov 2 '17 at 23:41
  • $\begingroup$ Ah, I see. I've always thought that if the eigenvalues of $A$ are all positive then $A+A^T$ should always be positive definite. Apparently, I was wrong. Thanks for this and I wondered if there was any sufficient (or even and necessary) condition for $A+A^T$ to be positive definite. $\endgroup$ – Lynn Nov 4 '17 at 0:37

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