Semidefinite relaxation of Quadratic equation

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I have read in various papers that we can write a Quadratic equation with symmetric matrix as a linear programming problem. For example

$$f(x)= x^T*Q*x + c$$

where Q=[2 0;0 3]; Now we can write $$Q=H^TH$$

So we can write

$$A=[I\quad Hx; x^TH^T \quad-c];$$ $$A= [1\quad 0\quad 1.4142x1; 0\quad 1\quad 1.7321x2; 1.4142x1\quad 1.7321x2\quad 1];$$

where I is identity matrix and the matrix A is linear in variables of vector x, rather than quadratic. Now determinant of A is exactly equal to the quadratic equation f(x).

But I have some problem in understanding the matrix A.

1) It has three equation. If I want to solve it for x1 and x2, the three equations doesn't hold.

2) How can I calculate the basis for the matrix A in x1 and x2?