Convexity of a specific semialgebraic set - MathOverflow most recent 30 from http://mathoverflow.net2013-05-24T00:01:47Zhttp://mathoverflow.net/feeds/question/99758http://www.creativecommons.org/licenses/by-nc/2.5/rdfhttp://mathoverflow.net/questions/99758/convexity-of-a-specific-semialgebraic-setConvexity of a specific semialgebraic setCDSousa2012-06-15T23:44:31Z2012-06-19T17:20:27Z
<p>I have an engineering problem which maybe resolved with semi-definite programming optimization.
I have a set which I would like to know if is convex:</p>
<p>Being $m \in \mathbb{R}^+$ a positive real scalar, $l \in \mathbb{R}^3$ a size 3 real vector, and $L \succ 0$ a size 3x3 real positive-definite matrix,
does the expression</p>
<p>$L - m\ S\left(\frac{l}{m}\right)^T S\left(\frac{l}{m}\right) \succ 0$</p>
<p>where $S(\dot{})$ means the skew-symmetric matrix operator, and $\succ 0$ means positive-definite,
defines a convex set?</p>
<hr>
<p>I.e.:</p>
<p>Being</p>
<p>$$m \in \mathbb{R}$$</p>
<p>$$l \equiv \left[l_x\ l_y\ l_z\right]^T$$</p>
<p>
$$
L \equiv \left[\begin{matrix}
L_{xx} & L_{xy} & L_{xz} \\
L_{xy} & L_{yy} & L_{yz} \\
L_{xz} & L_{yz} & L_{zz} \\
\end{matrix}\right]
$$
</p>
<p>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.</p>
<p>The constraints</p>
<p>
$$
\left\{
\begin{matrix}
m &> 0 \\
L &\succ 0\\
L - m\ S\left(\frac{l}{m}\right)^T S\left(\frac{l}{m}\right) &\succ 0
\end{matrix}
\right.
$$
</p>
<p>which, since $m>0$, are equivalent to</p>
<p>
$$
\left\{
\begin{matrix}
m &> 0 \\
L &\succ 0\\
m L - S\left(l\right)^T S\left(l\right) &\succ 0
\end{matrix}
\right.
$$
</p>
<p>define a semialgebraic set on the $\mathbb{R}^{10}$ variables space.</p>
<p>Here, $\succ 0$ means that the left argument is a positive-definite matrix, and,</p>
<p>
$$
S(x) = \left[\begin{smallmatrix}
0 & -x_3 & x_2 \\
x_3 & 0 & -x_1 \\
-x_2 & x_1 & 0
\end{smallmatrix}\right]\quad\text{with}\quad
x = \left[x_1\ x_2\ x_3\right]^T
$$
</p>
<hr>
<p>I did some, manipulation and rewrote the last constraint as a polynomial inequalities system:</p>
<p>being</p>
<p>
$$
mI = m L - S\left(l\right)^T S\left(l\right) =
\left[\begin{smallmatrix}
L_{1xx} m_{1} - l_{1y}^{2} - l_{1z}^{2} & L_{1xy} m_{1} + l_{1x} l_{1y} & L_{1xz} m_{1} + l_{1x} l_{1z} \\
L_{1xy} m_{1} + l_{1x} l_{1y} & L_{1yy} m_{1} - l_{1x}^{2} - l_{1z}^{2} & L_{1yz} m_{1} + l_{1y} l_{1z} \\
L_{1xz} m_{1} + l_{1x} l_{1z} & L_{1yz} m_{1} + l_{1y} l_{1z} & L_{1zz} m_{1} - l_{1x}^{2} - l_{1y}^{2}
\end{smallmatrix}\right]
$$
</p>
<p>then, through Sylvester's criterion,</p>
<p>
$$
mI \succ 0 \Leftrightarrow
\left\{
\begin{matrix}
\det\left(mI_{1,1}\right) = L_{1xx} m_{1} - l_{1y}^{2} - l_{1z}^{2} &> 0\\
\det\left(mI_{1:2,1:2}\right) &> 0\\
\det\left(mI\right) &>0
\end{matrix}
\right.
$$
</p>
<p>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.)</p>
<p>I also tried to write the set as a <a href="http://www.math.uni-konstanz.de/~schweigh/presentations/dcssblmi.pdf" rel="nofollow">linear matrix inequality (LMI)</a>, but I couldn't (my knowledge in this area is really short). </p>
<hr>
<p><strong>Update</strong>:</p>
<p>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.</p>
<hr>
<blockquote>
<p>Now, the <strong>questions</strong> are:</p>
<p>Which methods can I use to check if the defined set is convex?</p>
<p>If so, is it possible to represent it as an LMI?</p>
</blockquote>
http://mathoverflow.net/questions/99758/convexity-of-a-specific-semialgebraic-set/100018#100018Answer by BS for Convexity of a specific semialgebraic setBS2012-06-19T15:55:33Z2012-06-19T17:20:27Z<p>Your set is indeed a convex cone.</p>
<p>Since it is a cone, it suffices to show that the $m=1$ section is convex. </p>
<p>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" $(u,L)$ : $L\succ P(u)$.</p>
<p>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.</p>
<p>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 <em>defined by a Linear Matrix Inequality</em>.</p>
<p>Indeed this is equivalent to </p>
<p>$$\left(\begin{matrix}
S(y) & A(x)^* \\
A(x) & I \end{matrix}\right) \succ 0$$</p>
<p>as easily seen by row and column operations. In fact substituting $I$ by $\lambda I$, you can "re-homogenize" the problem.</p>