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), 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

questionsare:Which methods can I use to check if the defined set is convex?

If so, is it possible to represent it as an LMI?