I know that an affine subspace is a translation of a linear subspace. I also know that $\{\lambda_0 v_0 + \lambda_1 v_1 + ... + \lambda_n v_n : \sum_{k=0}^{n}\lambda_k = 1\}$ for vectors $v_i$ is an affine subspace.

A book I'm reading discusses affine subspaces as follows:

1) We take for granted that affine subspaces can be described by affine equations.

2) As the affine image of some vector space $R^k$.

3) As the kernel of some affine map

4) As the intersection of affine hyperplanes (hyperplanes are defined as affine subspaces of 1 less dimension than the ambient space)

I'm not sure I understand these 4 descriptions of affine spaces. For (1) By affine equations do they mean the set of affine combinations? Or maybe like how the equation of a plane is $ax + by + cz + d = 0$? And for a affine space of higher dimension we add more variables in the equation?

For (2) I am guessing we find a linear map from $R^k$ onto the subspace of the same dimension as our affine space, and then translate to make our affine space the image?

For (3) I suppose we similarly first find a linear map that vanishes on the subspace associated with our affine space, and then translate to 0?

For (4) I suppose in $R^3$ an affine line can be produced as the intersection of affine planes? I suppose they claim this is a general phenomenon?


An affine hyperplane $A$ in $\Bbb R^n$ can be given by an equation $u^Tx=c$ where $u$ is a normal vector for $A$ and $c\in\Bbb R$.
Consequently, an intersection of affine hyperplanes can be given by the system of affine equations $$u_1^Tx=c_1\\ \ \ \ \dots \\u_k^Tx=c_k$$ which can be rewritten in matrix form: $$Ux=c$$ This shows 1)$\iff$4), moreover considering the affine map $\phi:\Bbb R^k\to\Bbb R^n\ \ x\mapsto Ux-c$, this also shows equivalence with 3).

We can easily verify that any of the properties 1)-4) ensure that the set in question is closed under affine combinations.

Conversely, if a nonempty $A$ is closed under affine combinations, pick any $a_0\in A$, then $A-a_0$ is a linear subspace, then e.g. choosing a basis for $A-a_0$ we can get a linear map $\psi_0:\Bbb R^m\to\Bbb R^n$ such that $\mathrm{im}(\psi_0)=A-a_0$, hence the image of the affine map $\psi:=\psi_0+a_0$ is just $A$. This connects to 2).

Finally, assuming 2), $A=\mathrm{im}(\psi)$, then again, $B:=A-\psi(0)$ is a linear subspace, so (by e.g. extending a basis of $B$) we can define a linear map $\phi_0$ which vanishes exactly on $B$, and then the affine map $\phi:=\phi_0+\psi(0)$ will vanish exactly on $A$, proving 3).

  • $\begingroup$ Great. Very helpful. But shouldn't the map $\phi$ be $\phi = \phi_0 - \phi_0(\psi(0))$. $\endgroup$ – trynalearn Sep 23 '18 at 22:01
  • $\begingroup$ Yes, indeed.. We just adequately translate affine maps/subspaces to obtain linear ones. $\endgroup$ – Berci Sep 24 '18 at 7:44

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