Unique solution of quadratic expression

Question

I have a quadratic expression: $x^HAx - B$ in which I have to solve for $x$. $A$ is a 4 x 4 positive definite hermitian matrix and $x$ is a 4 x 1 vector. I have solved it by considering Singular Value Decomposition of A as well as Cholesky decomposition of A but each time I have obtained a different solution for this expression. Is it possible to obtain $x$ such that $x^HAx = B$ (or close) for this problem? I would appreciate suggestions.

Answer 1

If I understand correctly, you want to find $x$ such that $x^HAx = B$, where $B \geq 0$ is a real number. Let $v_1, \ldots, v_4$ be an orthonormal basis of eigenvectors of $A$ with $Av_i = \lambda_i v_i$. Write $x = a_1 v_1 + \ldots + a_4 v_4$. Then you have

$$ x^HAx = \left< Ax, x \right> = \sum_{i=1}^4 \lambda_i |a_i|^2 = B. $$

This gives you all possible solutions. If you work over the real numbers, you can see that the set of all solutions is an ellipsoid in $\mathbb{R}^n$ and in particular, the solutions are definitely not unique.