$\displaystyle L=\{x= \[\alpha_1,\ldots,\alpha_n\]^\top\ | \sum_{i=1}^n \alpha_i^2 =615\} $

And in general how can I show wether a subset of a vector space is affine or not?

Thank you!

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- Nov 19th 2011, 08:54 AMgotmejerryAre this an affine subset of R^n?

$\displaystyle L=\{x= \[\alpha_1,\ldots,\alpha_n\]^\top\ | \sum_{i=1}^n \alpha_i^2 =615\} $

And in general how can I show wether a subset of a vector space is affine or not?

Thank you!

- Nov 19th 2011, 11:07 AMHallsofIvyRe: Are this an affine subset of R^n?
A subset S, of vector space V, is an "affine" set if and only if for $\displaystyle \vec{v}$ a member of S, the set $\displaystyle \{\vec{u}- \vec{v}|u\in X\}$ is a sub

**space**of V. - Nov 20th 2011, 03:21 AMgotmejerryRe: Are this an affine subset of R^n?
What is your X here?

- Nov 20th 2011, 04:19 AMHallsofIvyRe: Are this an affine subset of R^n?
Whatever the overall space is. In your problem, it would be $\displaystyle R^n$. As long as you are working in $\displaystyle R^n$ you can charactize an "affine set" (that is not a subspace), geometrically, as a line or plane or "hyper-plane" (of whatever dimension) that does NOT include the origin. The condition that $\displaystyle \sum \alpha_i^2= 615$ simply says that this is the set of all vectors in $\displaystyle R^n$ with length $\displaystyle \sqrt{615}$. Is that a plane (of whatever dimension) in $\displaystyle R^n$?

- Nov 20th 2011, 05:30 AMgotmejerryRe: Are this an affine subset of R^n?
Hmm, yes I tihnk that is a plane. By the way I found an other way of solution but Im not sure wether it is right or not. So $\displaystyle L \subseteq \mathbb{R}^n$ is affine if $\displaystyle \alpha x+\beta y \in L$ for all $\displaystyle x,y\in L$ and$\displaystyle \alpha + \beta=1$.

And what I did:

$\displaystyle \sum_{i=1}^n \alpha {x_i}^2 + \sum_{i=1}^n (1-\alpha) {y_i}^2 $

then follows

$\displaystyle \alpha \sum_{i=1}^n {x_i}^2 + (1-\alpha) \sum_{i=1}^n {y_i}^2 $

and because $\displaystyle \sum_{i=1}^n {x_1}^2 =615$ and $\displaystyle \sum_{i=1}^n {y_1}^2=615$

$\displaystyle \alpha \sum_{i=1}^n {x_i}^2 + (1-\alpha) \sum_{i=1}^n {y_i}^2 =615 $

So it is an affine set. - Nov 21st 2011, 03:51 PMalbiRe: Are this an affine subset of R^n?
You should have: $\displaystyle \sum_{i = 1} (\alpha x_i + (1-\alpha) y_i)^2$