Are this an affine subset of R^n?

**gotmejerry** · November 19th 2011, 08:54 AM

$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!

**HallsofIvy** · November 19th 2011, 11:07 AM

A subset S, of vector space V, is an "affine" set if and only if for $\vec{v}$ a member of S, the set $\{\vec{u}- \vec{v}|u\in X\}$ is a subspace of V.

**gotmejerry** · November 20th 2011, 03:21 AM

What is your X here?

**HallsofIvy** · November 20th 2011, 04:19 AM

Whatever the overall space is. In your problem, it would be $R^n$ . As long as you are working in $R^n$ 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 $\sum \alpha_i^2= 615$ simply says that this is the set of all vectors in $R^n$ with length $\sqrt{615}$ . Is that a plane (of whatever dimension) in $R^n$ ?

**gotmejerry** · November 20th 2011, 05:30 AM

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 $L \subseteq \mathbb{R}^n$ is affine if $\alpha x+\beta y \in L$ for all $x,y\in L$ and $\alpha + \beta=1$ .

And what I did:
$\sum_{i=1}^n \alpha {x_i}^2 + \sum_{i=1}^n (1-\alpha) {y_i}^2$
then follows
$\alpha \sum_{i=1}^n {x_i}^2 + (1-\alpha) \sum_{i=1}^n {y_i}^2$
and because $\sum_{i=1}^n {x_1}^2 =615$ and $\sum_{i=1}^n {y_1}^2=615$

$\alpha \sum_{i=1}^n {x_i}^2 + (1-\alpha) \sum_{i=1}^n {y_i}^2 =615$
So it is an affine set.

**albi** · November 21st 2011, 03:51 PM

You should have: $\sum_{i = 1} (\alpha x_i + (1-\alpha) y_i)^2$

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