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Let $c∈\Bbb R$ and $U⊂\Bbb R^n$,$U \ne∅$ ($U$ is a nonempty subset). Further let $〈·,·〉:\Bbb R^n×\Bbb R^n→\Bbb R$ be the standard inner product. Define

$$U := \{ v ∈ \Bbb R^n : ∀ u ∈ U : 〈 v , u 〉 = c \} .$$

Show that $Uc$ is an affine subspace.

I think I should use the Inner standard product.

Edit, solution :

we have the Characterisation of affine subspaces

Let S ⊂ R^n, such that a,b ∈ S ,λ∈R ==> λa + (1−λ)b∈ S .Then S is already an affine subspace.

Solution :

we assume u1,u2 ∈ U and λ∈R so we have ⟨v,λu2 + (1−λ)u1⟩=c. ==> the property of Inner Product ⟨v,λu2⟩+⟨v,(1−λ)u1⟩ then we pull out lambda out λ⟨v,u2⟩+(1−λ)⟨v,u1⟩ ==> ⟨v,u2⟩ = c and ⟨v,u1⟩ = c ==> λc + (1−λ) c = λc + c - λc = c

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  • $\begingroup$ Hint: A subset is an affine subspace if and only if it differs from a linear subspace by a constant. $\endgroup$ – user3482749 Nov 18 '18 at 13:59
  • $\begingroup$ Think of a hyperplane H that gores through the origin with unit normal vector $u$. Thus all vectors $v \in$ H have the property that $<v,u>$ = 0. If we consider H$^{\,\prime}$ = $\{H + u\}$, the translate H by $u$, then all h $\in$ H$^{\prime}$ have the property $<h,u> = <v+u,u> = <v,u>+<u,u>$ = 1. So if <h,u>~=~c, then you have translated the hyperplane by adding $cu$. $\endgroup$ – Joel Pereira Feb 17 at 22:41

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