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i've to prove that the following set is convex:

Let $A \in \mathbb{C}^{n \times n}$

$$S=\left\{ \frac{\underline{x}^HA\underline{x}}{\underline{x}^H\underline{x}} , \underline{x} \in \mathbb C ^n\right\}$$

i've already tried to use the definition without success and then i tried to express every $x \in S$ as a convex combination of some elements, without success too.

Can you help me?

Michael Hardy
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Alessandro
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  • A thought: it may help to write $A = P + Q$, where $P = P^H$ and $Q = -Q^H$. – Ben Grossmann Oct 08 '14 at 16:31
  • You say $\underline{x}\in\mathbb C^n$, but later you say $x\in S$. Could you clarify? ${}\qquad{}$ – Michael Hardy Oct 08 '14 at 16:47
  • When underlined i mean an element of $\mathbb C$ when not i mean an element of $S$, the vectors are used in the definition of $S$ but, as you can simply prove, $S$ is a scalar set so when later i say $x$ i mean something of the form $\frac{\underline{y}^HA\underline{y}}{\underline{y}^H\underline{y}}$. Is clear now? – Alessandro Oct 08 '14 at 18:07
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    http://math.stackexchange.com/questions/171729/why-is-the-numerical-range-of-an-operator-convex – daw Oct 08 '14 at 19:36
  • $S={ x^H A x|\ |x|=1}$ right ? – HK Lee Oct 10 '14 at 13:54

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