Thread: proof by induction of affine subsets

1. proof by induction of affine subsets

Hi, I've got a project at uni and part of it is this proof by induction which i am terrible at. I can never work out where to sub in n+1 or n+2 n-1 or whatever it may be so any help much appreciated!!

An affine subset of $V$ is a non-empty subset $M$ of $V$ with the property that $\lambda x+(1-\lambda )y \in M$ whenever $x, y \in M$ and $\lambda \in \Re$

i) Let $M$ be an affine subset of $V$. Prove by induction on $n$ that, if $x_{1}, x_{2}, ... x_{n} \in M$ and $\lambda_{1}, \lambda_{2}, ... \lambda_{n} \in \Re$ with $\sum_{i=1}^{n} \lambda_{i} = 1$, then

$x=\sum_{i=1}^{n} \lambda_{i}x_{i}$ (1)

belongs to $M$.

ii) A sum of the form (1) is called an affine combination of $x_{1}, x_{2}, ... x_{n}$. Prove that, given a non-empty subset $S$ of $V$, the set consisting of all afine combinations of elements of $S$ is an affine subset of $V$ and is the smallest affine subset of $V$ containing $S$. This set is called the affine span of $S$.

2. Originally Posted by leshields
Hi, I've got a project at uni and part of it is this proof by induction which i am terrible at. I can never work out where to sub in n+1 or n+2 n-1 or whatever it may be so any help much appreciated!!

An affine subset of $V$ is a non-empty subset $M$ of $V$ with the property that $\lambda x+(1-\lambda )y \in M$ whenever $x, y \in M$ and $\lambda \in \Re$

i) Let $M$ be an affine subset of $V$. Prove by induction on $n$ that, if $x_{1}, x_{2}, ... x_{n} \in M$ and $\lambda_{1}, \lambda_{2}, ... \lambda_{n} \in \Re$ with $\sum_{i=1}^{n} \lambda_{i} = 1$, then

$x=\sum_{i=1}^{n} \lambda_{i}x_{i}$ (1)

belongs to $M$.

ii) A sum of the form (1) is called an affine combination of $x_{1}, x_{2}, ... x_{n}$. Prove that, given a non-empty subset $S$ of $V$, the set consisting of all afine combinations of elements of $S$ is an affine subset of $V$ and is the smallest affine subset of $V$ containing $S$. This set is called the affine span of $S$.
Two hints for the first part:

1. The base case "works" quite easily.

2. Assuming the result holds for $n-1$, as $\sum_{i=1}^{n} \lambda_{i} = 1$ you want to set $\lambda = \lambda_n$ for your induction step. You then have to show that the rest of the sum is of the form $(1 - \lambda)y$ with $y \in M$. This, I believe, requires a bit of a trick, but think about it for a bit first.