# Is $K =\{ S: \exists \text{ positive diagonal} D, D^TSD \;\text{diagonally dominant}\}$ convex?

I am doing some convex cone optimization and wonder whether the following set $K_1$ is convex or not.

Assume the following matrices are all in $\mathbb{R}^{n\times n}$ and symmetric. Let the set of all diagonally dominant matrices with non-negative diagonal to be $K_0$, i.e. $$K_0 = \{ A: A= [a_{ij}], \forall i,a_{ii} \geq \sum_{j\not = i}|a_{ij}|\}.$$

I wonder $$K =\{ S: \exists \;\text{positive diagonal} \;D, D^TSD \in K_0 \},$$

is convex or not(positive diagonal, $D_{ii}>0,\forall i, D_{ij}=0,\forall i\not= j)$

So far, what I have tried is brutal computation and this question seems to be that whether $$\exists d_i, d_s (\alpha a_{ss}+(1-\alpha) b_{ss}) \geq \sum_{j\not=s } ( d_s (\alpha a_{sj}+(1-\alpha) b_{sj} )$$

for any matrices $A=[a_{ij}],B=[b_{ij}] \in K$ and $\alpha \in [0,1]$. I have no clue in choosing the $d_i$s.

Any comments or ideas?

• $K$ is trivially convex because $0 \in K_0$ and $0^T S 0 = 0 \in K_0$ for all $S$. Presumably you have some other constraints in mind? – copper.hat Jun 28 '15 at 6:23
• Right, I should put $D$ to be positive diagonal. Thanks, @copper.hat! – Brian Ding Jun 28 '15 at 18:16
• See sandia.gov/~egboman/papers/factorwidth2.pdf theorem 8 and 9. It should not be hard with this theorem. – Brian Ding Jan 19 '16 at 3:32