14. Affine subspaces of a linear space

Affine subspaces of a linear space

Till now we have dealt with linear spaces, in particular in all our lines, planes and other subspaces the zero vector was always included. But obviously it makes sense to study also such spaces but translated from zero by a given vector. Such subspaces (are no longer linear) are called affine subspaces of a linear space. The term hyperplane is used to describe affine subspaces of dimension one less then the whole linear space we work in.

So if V is a linear subspace and v is a vector, then M=v+V (meaning subspace V translated by v), is an affine subspace. The space V will be called the tangent space (or the direction) to M and denoted by T(M) (or \vec{M}).

An affine subspace of a linear space can be described in the following ways:

  • giving a vector and its direction, e.g.: (1,0,-1)+lin((1,2,0),(0,1,1)).
  • giving points it goes through (we need one point more then the dimension of considered subspace), e.g.: (1,0,-1),(2,2,-1),(1,1,0) (sometimes denoted by af((1,0,-1),(2,2,-1),(1,1,0))).
  • giving a parametrization, e.g..: \{(1+t,2t+s,-1+t)\colon t,s\in\mathbb{R}\}
  • giving a system of linear equations, e.g.:

        \[2x-y+z=0\]

As usually we would like to be able to change the form of description.

A vector with a the tangent space and a set of points

Given a translation vector and the tangent space, we can easily get the points, which are included in the considered affine subspace. E.g. if H=(1,0,0,-1)+lin((1,-1,0,1),(2,-1,1,0)), it can be also defined as the only affine subspace including points: (1,0,0,-1),(1,0,0,-1)+(1,-1,0,1)=(2,-1,0,0), (1,0,0,-1)+(2,-1,1,0)=(3,-1,1,-1).

Reversely, if H goes through: (1,0,0,-1),(2,-1,0,0),(3,-1,1,-1), choose one of those points as a translation vector, and we get that H=(1,0,0,-1)+lin((2,-1,0,0)-(1,0,0,-1),(3,-1,1,-1)-(1,0,0,-1))=(1,0,0,-1)+lin((1,-1,0,1),(2,-1,1,0)).

A translation vector with the direction and parametrization

Given a translation vector and vectors spanning the tangent space we can easily get a parametrization. E.g. if H=(1,0,0,-1)+lin((1,-1,0,1),(2,-1,1,0)), then every vector of H is of form (1,0,0,-1)+t(1,-1,0,1)+s(2,-1,1,0), so H=\{(1+t+2s,-t-s,s,-1+t)\colon s,t\in\mathbb{R}\}.

Reversely, given H=\{(1+t+2s,-t-s,s,-1+t)\colon s,t\in\mathbb{R}\} we know that all the vectors of H are of form (1,0,0,-1)+t(1,-1,0,1)+s(2,-1,1,0), so H=(1,0,0,-1)+lin((1,-1,0,1),(2,-1,1,0)).

From a translation vector and the direction to a system of linear equations

The system of linear equations describing a given affine subspace differ from the system of equation describing its tangent space only by the column of free coefficients (because the difference between two points from H always is in T(H), so the difference of two solutions of the system of linear equations we are looking for is a solution of the system of linear equations describing the tangent space) — and in the second space we have an uniform system which we already know how to find. Therefore start with finding the system describing T(H). For example, let H=(1,0,0,-1)+lin((1,-1,0,1),(2,-1,1,0)). First find the system describing T(H)=lin((1,-1,0,1),(2,-1,1,0)) — we already know how to find such a system and so T(H) is given by:

    \[\begin{cases}-x-y+z=0\\x+2y+w=0\end{cases}\]

The given translation vector has to be a solution of the system we are looking for, so we have to choose the free coefficients in such a way that it will be true. So we can simply substitute the translation vector into the left sides of the equations: -1-0+0=-1, 1+0+-1=0. Therefore, the system of equations we are looking for is the following:

    \[\begin{cases}-x-y+z=-1\\x+2y+w=0\end{cases}\]

From a system of equations to a parametrization

Notice, that simply given a system of equations for H, its general solution is a parametrization of H!

Projections and reflections with respect to affine subspaces

How to calculate a projection of a vector onto an affine subspace and its image under reflection with respect to such a subspace? Simply bring it to the know case of linear subspaces, calculate, and go back to the initial setting. In other words translate everything to make the considered affine subspace go through zero, calculate the projection and translate everything backwards.

E.g. let us calculate the projection of (2,2,1) onto (2,1,0)+lin(-1,-1,0). First we calculate the projection of (2,2,1)-(2,1,0)=(0,1,1) onto lin(-1,-1,0):

    \[\frac{\left<(0,1,1),(-1,-1,0)\right>}{\left<(-1,-1,0),(-1,-1,0)\right>}(-1,-1,0)=\frac{-1}{2}(-1,-1,0)=\left(\frac{1}{2},\frac{1}{2},0\right).\]

And translate this projection backwards, so the final result is: \left(\frac{1}{2},\frac{1}{2},0\right)+(2,1,0)=\left(\frac{5}{2},\frac{3}{2},0\right).