3

I have two $n \times n$ matrices $A$ and $B$, where $B$ is symmetric and p.s.d and $A$ is symmetric , rank $2$ and its two dominant eigenvalues have different signs. Considering the following inequality:

$$\{B : \mbox{tr} (AB)<0\}$$

Meaning that we have $A$, but looking for feasible set of matrices $B$. Apparently, the $\inf \{tr(AB)\}$ is bounded for each specific matrix $A$, but the volume of the solution set $\{B\}$ is unbounded, when $A$ is not p.s.d.

However, as: $$\lambda_{\min} (A) \cdot \mbox{trace} (B) \leq \mbox{trace} (AB) \leq \lambda_{\max} (A) \cdot \mbox{trace} (B) $$ we are looking for solutions to: $$\lambda_{\min} (A) \cdot \mbox{trace} (B) \leq \mbox{trace} (AB) \leq 0$$ Also in order to have a feasible solution set, it is required to: $$\lambda_{\min} (A) \cdot \mbox{trace} (B) \lt 0$$ So if $\lambda_{\min} (A)$ becomes smaller in its absolute value, the two last inequalities become tighter, meaning that $\|\lambda(B)\|_1$ will shrinks when $\left | \lambda_{\min} (A) \right|$ gets smaller.

Q: According to the above, what can i interpret about the form/shape of the solution set's spectrahedron, if we make $\left| \lambda_{\min} (A) \right|$ bigger or smaller? Or any other relevant interpretation?

Bob
  • 700

1 Answers1

2

We have the intersection of the positive semidefinite cone with an open half-space. If we relax the openness of the half-space and make it closed, we can determine if this intersection is empty via semidefinite programming (SDP) using an arbitrary objective function, say, the zero function

$$\begin{array}{ll} \text{minimize} & \langle \mathrm O_n ,\mathrm X \rangle\\ \text{subject to} & \langle \mathrm A ,\mathrm X \rangle \leq 0\\ & \mathrm X \succeq \mathrm O_n\end{array}$$

If the semidefinite program is infeasible, then the intersection is empty.

The feasible set of the semidefinite program is a spectrahedron. If computing the volume of polytopes is not easy, I expect that computing the volume of spectrahedra is even harder.

Rodrigo de Azevedo
  • 23,223
  • If you have MATLAB, I would recommend CVX to numerically solve SDP's. – Rodrigo de Azevedo Oct 13 '16 at 08:50
  • can we also have the solution set in the answer? not just the feasibility flag! – Bob Oct 13 '16 at 11:27
  • @Bob The solution set is the spectrahedron defined by the constraints of the SDP. If the matrices are $2 \times 2$, then the inner product of the two matrices gives us the inequality constraint $$a_{11} x_{11} + 2 a_{12} x_{12} + a_{22} x_{22} \leq 0$$ and $\mathrm X \succeq \mathrm O_n$ gives us three additional inequality constraints $$x_{11} \geq 0 \qquad \qquad x_{22} \geq 0 \qquad \qquad x_{11} x_{22} - x_{12}^2 \geq 0 $$ – Rodrigo de Azevedo Oct 13 '16 at 12:01