Sign up ×

I had two doubts

  • Let $A$ be a hermitian matrix, what is the set of all unit norm vectors such that $x^{H}Ax\geq 0$.
  • Let $A_1$,$A_2$ be 2 hermitian matrices, what is the set of all unit norm vectors such that $x^{H}A_1 x\geq 0$ and $x^{H} A_2 x\geq 0$.
share|cite|improve this question
For the first question, if you diagonalize the matrix you can at least write up a simple equation on the form $\sum\lambda_j\lvert x_j\rvert^2\ge0$, where the $\lambda_j$ are the eigenvalues. Keep them in order to get a better handle on what that means. The second question is harder, unless the two matrices commute of course. – Harald Hanche-Olsen Nov 7 '12 at 16:58

1 Answer 1

up vote 1 down vote accepted

I am not sure what exactly do you expect, but basically every answer to your question is just a reformulation of the required point sets. Suppose we are talking about vectors in $\mathbb{C}^n$. Let $U$ be a unitary matrix such that $UAU^H = \mathrm{diag}(d)$ for some $d\in\mathbb{R}^n$ ($d$ is a real vector because every eigenvalue of a Hermitian matrix is real). Define a polytope $P$ and a halfspace $H$ by \begin{align} P &= \left\{x=(x_1,\ldots,x_n)^T\in\mathbb{R}^n: x_i\ge 0\ \forall i,\ \sum_kx_k= 1\right\},\\ H &= \left\{x\in\mathbb{R}^n: x^Td\ge 0\right\}. \end{align} Note that $u=(u_1,\ldots,u_n)\in\mathbb{C}^n$ is a unit vector if and only if $$ f(u)\equiv u\circ \overline{u}\equiv (|u_1|^2,\ldots,|u_n|^2)\in P. $$ Then $$ \left\{x\in\mathbb{C}^n: \|x\|=1,\ x^HAx\ge 0\right\} = U^H\left(f^{-1}(H\cap P)\right). $$ This is because \begin{align} x^HAx\ge 0 &\Leftrightarrow (Ux)^H\mathrm{diag}(d)(Ux) \ge 0\\ &\Leftrightarrow y^Td \ge 0,\ \textrm{ where } y=f(Ux)\\ &\Leftrightarrow y\in H\cap P\\ &\Leftrightarrow Ux\in f^{-1}(H\cap P)\\ &\Leftrightarrow x\in U^H\left(f^{-1}(H\cap P)\right). \end{align} So, the set $\{x\in\mathbb{C}^n: \|x\|=1,\ x^HA_1x, x^HA_2x\ge0\}$ would be something like $U_1^H\left(f^{-1}(H_1\cap P)\right) \cap U_2^H\left(f^{-1}(H_2\cap P)\right)$, but I wonder if this formulation could be useful.

share|cite|improve this answer
its a really good answer from a optimization point, you essentially reformulated it as intersections of convex sets I believe., but I was expecting something in the line of subspaces. But now I realize it will not be a subspace. – dineshdileep Nov 8 '12 at 3:23
@dineshdileep: While $H\cap P$ is convex, $f^{-1}(H\cap P)$ is not. So I guess this formulation doesn't help :-( – user1551 Nov 8 '12 at 3:58
your formulation is indeed useful!! – dineshdileep Nov 21 '12 at 17:50

Your Answer


