A problem on positive semi-definite quadratic forms/matrices

Question

Suppose $a+b+c=0$ and (without loss of generality) $a\leq b\leq 0\leq c$, $a^2+b^2+c^2=1$, is the following quadratic form positive semi-definite? Thank you very much.

\begin{equation*} \begin{split} I(x,y,z)=&(2a^2+5b^2+5c^2)a^2x^2+(5a^2+2b^2+5c^2)b^2y^2\\ +&(5a^2+5b^2+2c^2)c^2z^2\\ -&ab(a^2+b^2+c^2+6ab)xy-ac(a^2+b^2+c^2+6ac)xz\\ -&bc(a^2+b^2+c^2+6bc)yz\\ =&(2+3b^2+3c^2)a^2x^2+(2+3a^2+3c^2)b^2y^2+(2+3a^2+3b^2)c^2z^2\\ -&ab(1+6ab)xy-ac(1+6ac)xz-bc(1+6bc)yz. \end{split} \end{equation*}

We can write the matrix for $I$ as a product \begin{equation*} \left(\begin{array}{ccc} a&0&0\\ 0&b&0\\ 0&0&c \end{array}\right) \left(\begin{array}{ccc} 2+3b^2+3c^2&3ab+\frac{1}{2}&3ac+\frac{1}{2}\\ 3ab+\frac{1}{2}&2+3a^2+3c^2&3bc+\frac{1}{2}\\ 3ac+\frac{1}{2}&3bc+\frac{1}{2}&2+3a^2+3b^2 \end{array}\right) \left(\begin{array}{ccc} a&0&0\\ 0&b&0\\ 0&0&c \end{array}\right) \end{equation*}

Therefore we need to show the middle matrix is positive semi-definite, is it correct?

Answer 1

A sum-of-squares decomposition of the polynomial is possible (the constraints are not needed).

Here is an implementation in the MATLAB Toolbox YALMIP (developed by me). YALMIP (together with a semidefinite programming solver) easily derives a decomposition $I(a,b,c,x,y,z) = v(a,b,c,x,y,z)^TQv(a,b,c,x,y,z)$ where $Q$ is positive definite (the fact that $Q$ is rather far away from being singular can be used to show that the numerical approach actually yields a certificate, despite any possible numerical issue).

sdpvar a b c x y z
I = (2*a^2+5*b^2+5*c^2)*a^2*x^2+(5*a^2+2*b^2+5*c^2)*b^2*y^2+(5*a^2+5*b^2+2*c^2)*c^2*z^2-a*b*(a^2+b^2+c^2+6*a*b)*x*y-a*c*(a^2+b^2+c^2+6*a*c)*x*z-b*c*(a^2+b^2+c^2+6*b*c)*y*z

[~,v,Q]=solvesos(sos(I))

However, we can do better and generate a nicer certificate. One way to do that is to search for an integer matrix $Q$. This will not work here (the problem is infeasible). We can however search for simple rational decompositions. Hence, trying the most simple rationals, we enforce all elements of $Q$ to be an integer divided by $2$. Alternatively, we search for a sum-of-squares decomposition of $2I$. This leads to an integer semidefinite program, but luckily YALMIP has a built-in solver for that. We do however have to formulate the sum-of-squares problem manually, but it is pretty easy when we already have run the module once and generated a sufficient set of monomials $v$.

Q = intvar(9,9);
Match = coefficients(I*2-v{1}'*Q*v{1},[a b c x y z])==0;
optimize([Match, Q >=0])

The integer SDP is solved in no-time, and the solution is given by $2I = v^TQv$ where

 value(Q)
 4     0    -1    -1     0    -1     0     0    -1
 0    10    -5    -1     0     0     0    -1     0
-1    -5    10     0     0    -1     0     0     0
-1    -1     0     4     0     0     0    -1    -1
 0     0     0     0    10    -5    -1     0    -1
-1     0    -1     0    -5    10     0     0     0
 0     0     0     0    -1     0    10    -5    -1
 0    -1     0    -1     0     0    -5    10     0
-1     0     0    -1    -1     0    -1     0     4

sdisplay(v{1})

ans = 

'c^2*z'
'b*c*z'
'b*c*y'
'b^2*y'
'a*c*z'
'a*c*x'
'a*b*y'
'a*b*x'
'a^2*x'