I am not too sure how to prove that a hyperplane in $\mathbb{R}^{n}$ is convex? So far I know the definition of what convex is, but how do we add that hyperplane in $\mathbb{R}^{n}$ is convex?
Thanks in advance!
Hyperplane is a convex
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linear-algebra
geometry
convex-analysis
linear-programming
1 Answers
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Roughly speaking, you need to show that any two points of the hyperplane of $\mathbb{R}^n$ can be joined by a line segment.
First, lets define what this hyperplane $H$ is: $$ H=\left\{ \pmatrix{x_1\\ \vdots\\ x_n} \in \mathbb{R}^n \;|\; a_1x_1+\cdots+a_n x_n = c \right\} $$ where $a_1,\cdots,a_n \neq 0$ and $c\in \mathbb{R}$.
So you need to show that $$ \forall X,Y \in H, \; \lambda X+(1-\lambda)Y \in H, \; 0\le \lambda \le 1 $$
Lets do it:
$$ \lambda X+(1-\lambda)Y =\pmatrix{\color{red}{\lambda x_1+(1-\lambda)y_1}\\ \vdots \\ \color{red}{\lambda x_n+(1-\lambda)y_n}} $$ Since $X,Y \in H$, we can write $$ \begin{cases} a_1x_1+\cdots+a_n x_n = c \\ a_1y_1+\cdots+a_n y_n = c \end{cases} $$ and so $\forall \lambda \in [0,1]$: $$ \begin{cases} \lambda a_1x_1+\cdots+\lambda a_n x_n = \lambda c \\ (1-\lambda)a_1y_1+\cdots+(1-\lambda)a_n y_n = (1-\lambda)c \end{cases} $$ Summing these two equations yields \begin{align} &\lambda a_1x_1+(1-\lambda)a_1y_1+\cdots+\lambda a_n x_n+(1-\lambda)a_n y_n = \lambda c+(1-\lambda)c \\ \Rightarrow \quad &a_1\left(\color{red}{\lambda x_1+(1-\lambda)y_1}\right)+\cdots+a_n\left(\color{red}{\lambda x_n+(1-\lambda)y_n}\right)= c \\ \Rightarrow \quad &\lambda X+(1-\lambda)Y \in H \end{align}