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I am doing exercise 2.15 of "convex optimization". While the solution confused me a lot.

as indicated by the solution: $\sum_{i=1}^{n} p_i =1$ defines a hyperplane.

while it's different from the dinition of hyperplane in page 27, 2.2.1 of the book:

A hyperplane is a set of the form $ \{x | a^T x=b\}, $ where $a \in \mathbb{R}^n, a \neq 0$, and $b \in \mathbb{R}$.

by the definition, solution of x could be considered as a line orthogonal to y axis in the x-y 2-dimensional coordinate, and also could be considered as a hyperplane in the other spaces.

how $\sum_{i=1}^{n} p_i =1$ defines a hyperplane ? it's quite weird!

LucasYang
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    For future reference: $\mathbb{R}$, or equivalently, $\Bbb{R}$ both produce "$\mathbb R$" which is what you probably meant when you say $b\in R$. – 5xum Oct 06 '21 at 07:04

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Consistently with the notation in use, $\{p\in\Bbb R^n\,:\, \sum_{j=1}^n p_j=1\}$ is the hyperplane $\{p\in\Bbb R^n\,:\, \mathbf 1^\top p=1\}$, where $\mathbf 1$ is the vector such that $\mathbf 1_j=1$ for all $j$.

5xum
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  • so, the hyperplane determined by this constraint is n dimensional ? each solution of p could be considered as one dimension. – LucasYang Oct 06 '21 at 06:48
  • @YangGaoguang Given an $m\times n$ matrix $A$, the dimension of the affine subspace $H_{A,b}={x\in\Bbb R^n,:, Ax=b}$ is, when the set is non-empty, $\dim \ker A=n-\operatorname{rk}A$. In this case, the dimension is $n-1$. –  Oct 06 '21 at 10:01
  • thank you for your help, thank you very much. – LucasYang Oct 06 '21 at 13:51