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Let $A \in \mathbb{R^{n\times n}}$ be a positive definite matrix and $U \in \mathbb{R^{n\times n}}$ a diagonal matrix with positive entries,

$A=A^\top \succ 0 $, $\:\:U=\textrm{diag}(u_1, \cdots, u_n) \quad(\forall i, \:u_i >0).$

I'm curious to see if the following is true.

$B:=U A + A U \succ 0$

So far, I found these

  1. $UA +AU$ is symmetric
  2. For any $x\in\mathbb{R}^n, \:\:x^\top UA x = x^\top AU x$
  3. $B_{ij}=(u_i + u_j)A_{ij}$

Any advice would be appreciated!!!!!!

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1 Answer 1

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Try $$A = \pmatrix{1 & 1\cr 1 & 2\cr},\ U = \pmatrix{1 & 0\cr 0 & t}$$ for small positive $t$.

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  • $\begingroup$ Yup, thanks, it becomes clear that $B$ is not positive definite! $\endgroup$ Jan 24, 2021 at 3:21

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