linear algebra

[mrcustard]

Let M be an affine subset of V.
(i) Prove that M+a is affine for every a in V and that, if 0 is in M, then M is a subspace.
(ii) Deduce that there exists a subspace U of V and a in V such that: M=U+a. [Hint: what can be said about such an a, assuming that it exists?] Show further that the subspace U is uniquely determined by M and describe the extent to which a is determined by M.

I can do (i), but have no idea how to do (ii)

note- By 'in V', I mean an element of V.

Any help would be greatly appreciated.

 

[Deveno]

well of course there is.

suppose first that 0 is in M. then we can take U = M, and a to be the 0-vector of V.

but, if not, then since an affine set is non-emtpy, we may choose some a in M. since, by supposition, 0 is not in M, a does not equal 0.

let U = M - a = M + (-a).

by part i) U is an affine set, and since 0 is in U (because 0 = a - a), U is a subspace of V. and then:

U + a = (M - a) + a = M + (-a + a) = M + 0 = M.

(technically, you don't even need to decide whether or not 0 is in M, but it makes the reasoning clearer:

the thing is, if M contains 0, and a is in M, then M - a = M. to see this, suppose u is in M - a,

so that u = m - a for some m in M. since 0 is in M, M is a subspace, so u = m - a is in M.

on the other hand if m is in M, and a is in M, then so is m+a = m', in which case we have:

m = m' - a, for m' in M, that is: m is in M - a.

so we could have just argued that U could be M - a in any case, but it is not obvious that in the case

where 0 is in M, that U is actually just M)