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If $S$ is an affine independent subset of $R^n$ and $x \notin$ the affine span of $S$ then $S \cup \{x\}$ is affine independent.

I start by letting $S=\{a_0, \dots, a_m\}$ and therefore for by the definition of linear independence: $$(\forall \{t_i\})[\sum_{i=1}^m t_i(a_i-a_0)=0 \Rightarrow t_i = 0, \forall i]$$ So then I need to show that $S \cup \{x\} = \{a_0, \dots , a_n, x\}$ is affine independent, or $$(\forall \{j_i\})[\sum_{i=1}^m j_i(a_i-a_0) + j_x(x-a_0)=0 \Rightarrow j_i = 0, \forall i]$$

But I'm a little confused as to how I'd continue from here after letting $\{j_i\}$ be arbitrary and assuming that $\sum_{i=1}^m j_i(a_i-a_0) + j_x(x-a_0)=0$.

Anyone have any ideas?

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Let $x$ be outside the span of $S$. Suppose that $\alpha_1 v_1 + \dotsb + a_n v_n - \beta x = 0$ with $S = \{ v_1, \dotsc, v_n \}$ and $\alpha_1 + \dotsb + \alpha_n - \beta = 0$.

Suppose that $\beta \neq 0$, then $x = \frac{\alpha_1}{\beta} v_1 + \dotsb + \frac{\alpha_n}{\beta}$, and since we supposed $\alpha_1 + \dotsb + \alpha_n - \beta = 0$, we deduce that $\frac{\alpha_1}{\beta} + \dotsb + \frac{\alpha_n}{\beta} = 1$. This contradicts the hypothesis that $x$ is not in the span of $S$.

Therefore $\beta = 0$, we have $\alpha_1 v_1 + \dotsb + \alpha_n v_n = 0$ and $\alpha_1 + \dotsb + \alpha_n = 0$. The affine independence of $S$ implies that $\alpha_i = 0$ for all $i$. From this we deduce that $S \cup \{ x\}$ is affinely independent.

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Starting from

$ \sum_{i=1}^m j_i(a_i-a_0) + j_x(x-a_0)=0 $, you could write $ \sum_{i=1}^m j_i(a_i-a_0) = - j_x(x-a_0) $

Now, the left-side term is a combination of elements in $S$ and therefore is an element of $span(S)$. But, $x$ being not in $span(S)$, the right-hand term is not in $span(S)$ unless $j_x = 0$. Given that the two terms are equal, they should both belong to the same vector spaces; in particular, they should belong to $span(S)$.

Hence, the right hand term is $0$, and you can conclude that all $j_i$ are $0$ from here.

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  • $\begingroup$ Why is it true that the right-hand term is not in $\text{span}(S)$ unless $j_x = 0$? Because for an element to be in $\text{span}(S)$ it must be an affine combination of the form $\sum_{i=0}^{m} t_is_i$ such that $\sum_{i=0}^{m}t_i=1$? $\endgroup$ – Oliver G May 1 '17 at 18:45
  • $\begingroup$ And why is the left side in $S$ too? What about the equation implies the sum of each $j_i=1$? $\endgroup$ – Oliver G May 1 '17 at 20:28
  • $\begingroup$ the right-hand term is not in $span(S)$ unless $j_x=0$ because we know that $x$ is not in $span(S)$, and that $a_0$ is, so their difference $x-a_0$ cannot be in $span(S)$. The left side is in $span(S)$ because it is a linear combination of elements in $span(S)$. $\endgroup$ – Salem May 2 '17 at 8:34
  • $\begingroup$ But for an element to be in the affine span of $S$, it has to be the affine combination of elements from $S$ such that the sum of the coefficients equals $1$. How does the fact that they're a linear combination imply that they're also an affine combination? $\endgroup$ – Oliver G May 2 '17 at 11:34
  • $\begingroup$ Sorry, I didn't catch the "affine" in your question $\endgroup$ – Salem May 2 '17 at 12:12

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