Rank-one optimal solution conjecture for dissipative bilinear control semidefinite programs

Consider the source semidefinite program

$$\max_{N\succeq0}\langle A_4,N\rangle, \qquad \langle A_i,N\rangle=-p_i(0)\quad(1\le i\le3),$$

where $\langle C,N\rangle=\operatorname{tr}(CN)$ and, for a real matrix $A=(a_{ij})$, the symmetric matrices $A_i$ are defined by

$$(A_i)_{ii}=2a_{ii},\qquad (A_i)_{ij}=(A_i)_{ji}=a_{ij}\quad(j\ne i),$$

with all remaining entries zero.

Take

$$A= \begin{pmatrix} -1&-4/3&-7/6&5\\ 4/3&-1&-4&20/17\\ 7/6&4&-1&1\\ -5&-20/17&-1&-1 \end{pmatrix}.$$

Every off-diagonal entry is nonzero, so $A$ is irreducible. Moreover,

$$A+A^{\mathsf T}=-2I_4\prec0.$$

Define

$$U= \begin{pmatrix} -1/5&-1\\ -3/5&-1/2\\ 1&0\\ 0&1 \end{pmatrix}, \qquad R= \begin{pmatrix} 1&-1\\ -1&4 \end{pmatrix}, \qquad M=URU^{\mathsf T}.$$

Since $\det R=3$ and $R_{11}=1$, we have $R\succ0$, so $M\succeq0$ has rank two. Explicitly,

$$M= \begin{pmatrix} 91/25&71/50&4/5&-19/5\\ 71/50&19/25&-1/10&-7/5\\ 4/5&-1/10&1&-1\\ -19/5&-7/5&-1&4 \end{pmatrix}.$$

Its four constraint values are

$$\bigl( \langle A_1,M\rangle, \langle A_2,M\rangle, \langle A_3,M\rangle, \langle A_4,M\rangle \bigr) = \left( -\frac{764}{15}, -\frac{58}{255}, -\frac{44}{15}, \frac{600}{17} \right).$$

Thus $M$ is feasible for strictly positive initial data

$$(p_1(0),p_2(0),p_3(0)) = \left( \frac{764}{15}, \frac{58}{255}, \frac{44}{15} \right).$$

To certify global optimality, let

$$B= \begin{pmatrix} 5&3\\ 0&4\\ 1&3\\ 5&5 \end{pmatrix}, \qquad S=\frac1{25}BB^{\mathsf T},$$

and

$$(w_1,w_2,w_3) = \left(\frac{17}{25},\frac8{25},\frac15\right).$$

Direct rational calculation gives

$$S=-w_1A_1-w_2A_2-w_3A_3-A_4\succeq0, \qquad B^{\mathsf T}U=0.$$

In particular, $SM=0$. Every feasible $N\succeq0$ satisfies

$$0\le\langle S,N\rangle = \sum_{i=1}^3w_ip_i(0)-\langle A_4,N\rangle = \frac{600}{17}-\langle A_4,N\rangle.$$

Equality holds at $N=M$, so the exact optimum is $600/17$.

Finally, if $N$ is any optimal feasible matrix, then

$$0=\langle S,N\rangle = \frac1{25}\operatorname{tr}(B^{\mathsf T}NB).$$

Since $B^{\mathsf T}NB\succeq0$, it vanishes, and hence $NB=0$. Therefore

$$\operatorname{im}N\subseteq\ker B^{\mathsf T} =\operatorname{im}U,$$

so $N=UXU^{\mathsf T}$ for a symmetric $2\times2$ matrix $X$. The three feasibility equations have coefficient matrix

$$Q= \begin{pmatrix} 1/15&-7/3&-40/3\\ 22/5&166/51&-35/102\\ -109/15&-13/3&0 \end{pmatrix}, \qquad \det Q=-\frac{1140}{17}\ne0.$$

Thus they uniquely force $X=R$. Consequently $M$ is the unique optimum, and

$$\operatorname{rank}M=2.$$

There is no rank-one optimum. The conjecture is therefore false even for an irreducible strictly dissipative system with strictly positive constrained initial data.