$\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!
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 $\displaystyle R^n$?
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.