Let $\Theta \subset \mathbb{R}^k$, $\theta_0 \in \Theta$. Show that $\operatorname{aff}(\Theta) = \operatorname{aff}(\Theta - \theta_0) + \theta_0$

Question

Here is my working. So I am trying to show $\operatorname{aff}(\Theta) = \theta_0 + \operatorname{aff}(\Theta - \theta_0)$ for any $\theta_0 \in \Theta \subset \mathbb{R}^k$.

We know $\operatorname{aff}(\Theta) = \big\{\sum_{i=1}^n\alpha_i\theta_i : \sum_{i=1}^n\alpha_i = 1, \alpha_i \in \mathbb{R}\big\}$. So

$\sum_{i=1}^n\alpha_i\theta_i = \sum_{i=1}^{n-1}\alpha_i\theta_i + \alpha_n\theta_n = \sum_{i=1}^{n-1}\alpha_i\theta_i + (1 - \sum_{I=1}^{n-1})\theta_n = \sum_{i=1}^{n-1}\alpha_i(\theta_i - \theta_n) + \theta_n$

Now I am tempted to say that the right-hand side is $\operatorname{aff}(\Theta - \theta_n) + \theta_n$ but how can I say that when I cannot claim that $\sum_{i=1}^{n-1}\alpha_i = 1$?

Additionally, can someone also help me show that for any $\Theta \subseteq \mathbb{R}^k$ satisfying $0_k \in \mathbb{R}^k$ we have that $\operatorname{aff}(\Theta) = \operatorname{span}(\Theta)$

Answer 1

Actually you are near here :

Showing $ aff(\theta) = \theta_0 + aff(\theta-\theta_0)$ by simple double inclusion :

1. Let $x \in aff(\theta), \exists a_i, x=\sum_{i=1}^na_i\theta_i$

So $$ x=\sum_{i=1}^na_i(\theta_i-\theta_n)+\theta_n$$

2. Let $y \in \theta_n + aff(\theta-\theta_n) $

So $$ \exists b_i, y=\theta_n+\sum_{i=1}^nb_i(\theta_i-\theta_n) $$

with by definition $\sum_{i=1}^n b_i=1$.

Here the result.