I have an engineering problem that may be solved using semidefinite programming. I would like to know whether a given set is convex.


Let $m \in \mathbb{R}^+$ be a positive real scalar, $l \in \mathbb{R}^3$ be a real vector, and $L \succ 0$ be a size $3 \times 3$ real positive definite matrix. Does the following inequality


$$L - m\ S\left(\frac{l}{m}\right)^T S\left(\frac{l}{m}\right) \succ 0$$


where $S(\cdot)$ denotes the skew-symmetric matrix operator and $\succ 0$ denotes positive definiteness, define a convex set?




---


I.e.:


Being


$$m \in \mathbb{R}$$


$$l \equiv \left[l\_x\ l\_y\ l\_z\right]^T$$



 $$
L \equiv \left[\begin{matrix}
 L\_{xx} & L\_{xy} & L\_{xz} \\
 L\_{xy} & L\_{yy} & L\_{yz} \\
 L\_{xz} & L\_{yz} & L\_{zz} \\
 \end{matrix}\right]
$$



The variables $m,l\_x,l\_y,l\_z,L\_{xx},L\_{xy},L\_{xz},L\_{yy},L\_{yz},L\_{zz}$ define a $\mathbb{R}^{10}$ space.


The constraints



 $$
\left\{
\begin{matrix}
 m &> 0 \\
 L &\succ 0\\
 L - m\ S\left(\frac{l}{m}\right)^T S\left(\frac{l}{m}\right) &\succ 0
\end{matrix}
\right.
$$



which, since $m>0$, are equivalent to



 $$
\left\{
\begin{matrix}
 m &> 0 \\
 L &\succ 0\\
 m L - S\left(l\right)^T S\left(l\right) &\succ 0
\end{matrix}
\right.
$$



define a semialgebraic set on the $\mathbb{R}^{10}$ variables space.


Here, $\succ 0$ means that the left argument is a positive-definite matrix, and,



 $$
S(x) = \left[\begin{smallmatrix}
 0 & -x\_3 & x\_2 \\
 x\_3 & 0 & -x\_1 \\
 -x\_2 & x\_1 & 0
\end{smallmatrix}\right]\quad\text{with}\quad
x = \left[x\_1\ x\_2\ x\_3\right]^T
$$





---


I did some, manipulation and rewrote the last constraint as a polynomial inequalities system:


being



 $$
mI = m L - S\left(l\right)^T S\left(l\right) =
\left[\begin{smallmatrix}
 L\_{1xx} m\_{1} - l\_{1y}^{2} - l\_{1z}^{2} & L\_{1xy} m\_{1} + l\_{1x} l\_{1y} & L\_{1xz} m\_{1} + l\_{1x} l\_{1z} \\
 L\_{1xy} m\_{1} + l\_{1x} l\_{1y} & L\_{1yy} m\_{1} - l\_{1x}^{2} - l\_{1z}^{2} & L\_{1yz} m\_{1} + l\_{1y} l\_{1z} \\
 L\_{1xz} m\_{1} + l\_{1x} l\_{1z} & L\_{1yz} m\_{1} + l\_{1y} l\_{1z} & L\_{1zz} m\_{1} - l\_{1x}^{2} - l\_{1y}^{2}
\end{smallmatrix}\right]
$$



then, through Sylvester's criterion,


 
 $$
mI \succ 0 \Leftrightarrow
\left\{
\begin{matrix}
\det\left(mI\_{1,1}\right) = L\_{1xx} m\_{1} - l\_{1y}^{2} - l\_{1z}^{2} &> 0\\
\det\left(mI\_{1:2,1:2}\right) &> 0\\
 \det\left(mI\right) &>0
\end{matrix}
\right.
$$



It would be sufficient that the polynomials were concave to guarantee set convexity, however they are not concave.
Although not being concave, it does not imply that set is not convex; for example, the first polynomial $L\_{1xx} m\_{1} - l\_{1y}^{2} - l\_{1z}^{2}$ is not concave itself but defines a convex set if constraint $m>0$ is taken into account.
(This representation also gave me some suspicions that maybe the $L \succ 0$ constraint is implicit on the other.)


I also tried to write the set as a [linear matrix inequality (LMI)](http://www.math.uni-konstanz.de/~schweigh/presentations/dcssblmi.pdf), but I couldn't (my knowledge in this area is really short). 




---


**Update**:


I was able to check that this set is close under positive scalar multiplication, since
$$ (\gamma\ m) (\gamma\ L) - S\left(\gamma\ l\right)^T S\left(\gamma\ l\right) = \gamma^2 \ \left( m L - S\left(l\right)^T S\left(l\right)\right) \succ 0 \quad \text{for} \quad \gamma > 0$$
then it is a cone. If one can prove the set is close under addition then it will be proven to be a convex cone.




---



> 
> Now, the **questions** are:
> 
> 
> Which methods can I use to check if the defined set is convex?
> 
> 
> If so, is it possible to represent it as an LMI?
> 
> 
> 



Convexity of a specific semialgebraic set
Your set is indeed a convex cone.


Since it is a cone, it suffices to show that the $m=1$ section is convex. 


But this is equivalent to show that the (symmetric matrix valued) "function" $u\mapsto P(u)=S(u)^\*S(u)$ is "convex", i.e. $P((u+v)/2)\prec (P(u)+P(v))/2$, because the set is basically the "epigraph" $(u,L)$ : $L\succ P(u)$.


Now it is easily checked that the quadratic function $P$ satisties the parallelogram identity $P((u+v)/2)+P((u-v)/2)=(P(u)+P(v))/2$, which does the job since $P$ is positive.


EDIT : in fact any set of $(x,y)$ defined by an inequality $S(y)-A(x)^\*A(x) \succ 0$, with $S(y)$ $n\times n$ symmetric and linear in $y$, and $A(x)$ $n\times d$ and linear in $x$, is convex and moreover *defined by a Linear Matrix Inequality*.


Indeed this is equivalent to 


$$\left(\begin{matrix}
 S(y) & A(x)^\* \\\
 A(x) & I \end{matrix}\right) \succ 0$$


as easily seen by row and column operations. In fact substituting $I$ by $\lambda I$, you can "re-homogenize" the problem.


