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While I was reading a Convex Optimization book, I found an example, which is attached below. What is confused me is that how the authors derived/concluded that the hyperbolic cone is the image inverse of f(x).

enter image description here

Learner
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1 Answers1

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Let $C = \{(x, t) | z^Tz \le t^2 | t \ge 0\}$ and $K := \{x|x^TPx \le (c^Tx)^2, c^Tx \ge 0\}$. It's straightforward to see that $C$ has the representation $C = \{(z, t) | \|z\|_2 \le t, t \ge 0\} = \text{epi}(\|.\|_2) \cap (\mathbb R^n \times \mathbb R_+)$, and is thus convex.

It should be clear that $f$ is affine, and it's a rather straightforward exercise to show that the pre-image of a convex set under an affine function is again convex. So we just need to show that $K = f^{-1}C$. Now, noting that $ x^TPx = (P^{1/2}x)^TP^{1/2}x = ||P^{1/2}x||_2^2$ (since $P \in \mathbb S^n_+$), we get

$$ K = \{x|\|P^{1/2}x\|_2 \le c^Tx, c^Tx \ge 0\} = \{x | f(x) \in C\} = f^{-1}C,$$

and we're done.

dohmatob
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  • One more question: how ( transpose(x)Px = transpose(sqrt(P)x)sqrt(P)x ) ? – Learner Jul 07 '16 at 06:08
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    $x^TPx = x^TP^{1/2}P^{1/2}x = x^T(P^{1/2})^TP^{1/2}x = (P^{1/2}x)^TP^{1/2}x$, where we've used the fact that $P^{1/2}$ is be symmetric (in fact, it is positive semi-definite too!). – dohmatob Jul 07 '16 at 06:10