# I Affine Subspace

1. Jun 14, 2017

### Erik109

Hey,

I am struggling with developing an intuition behind 'Affine Subspaces'. So far I have read the theories concerning Affine Subspaces delivered by the course book and visited several websites, however none have made it 100% clear. I feel like I have some sort of intuition for it, but I fail to apply the intuition when it comes to solving problems.

At the moment, I am often required to show why a given set is an Affine Subspace of a certain space. Because I assume it is quite hard for you to convey the intuition behind it without writing a lot, I will try to convey the way I think by approaching a sample problem. I hope you can enlighten me on the errors/misconceptions I make or perhaps add something so I can learn more.

The sample problem:

Let X1 = {f : R → R : f(0) = 1 and f is continuous}. Prove that X1 is an affine subspace of C(R, R) (the space of all continuous function with domain R and mapping to R). Hint: You have to do something with the set X0 = {γ : R → R : f(0) = 0 and f is continuous}.

So, what I think they want me to do at first is two things:
- prove that the set X0 is a subspace of C(R, R) (Closed by addition/multiplication)
- define an arbitrary element β which is an element of C(R, R) but not an element of X0

Once I have done both, I feel like I have to define an arbitrary element of X1, for example, the function ƒ, and show that this function contains both an element of X0 (let's say γ) and the arbitrary element β. In other words:

ƒ(x) = γ(x) + β

If this holds then the set X1 is an Affine Subspace of C(R, R) because it contains the sum of a subspace of C(R, R) and an element of C(R, R)?

I feel like I don't fully understand how to construct this function ƒ(x) and which arbitrary β to define. I have scanned through the answer numerous times and I kinda get their reasoning why, but I can't develop my own reasoning. There must be a logical, stepwise approach to the problem.

Erik

2. Jun 14, 2017

### WWGD

As I understand it, an affine ( and a fine , haha) subspace is just a standard subspace followed by a translation. The subspace just does not go through the origin and there are issues with closure under operations, as in f(0)+g(0)=2 here. EDIT: Like you said, $X0$ is a subspace ( contains the origin) but $X1$ does not.

3. Jun 14, 2017

### Staff: Mentor

I'm not sure what the arbitrariness of $\beta$ has to do here. In general, an affine subspace is simply a linear subspace that got shifted from the origin $\vec{0}$ by a certain, fixed vector $\vec{f}_0\,$.

So you are right, the set $\{f \in C(\mathbb{R},\mathbb{R})\,\vert \,f(0)=0\}$ is the candidate for this subspace and you have to show that it is actually a subspace, i.e. closed under addition and scalar multiplication. (There are other multiplications on $C(\mathbb{R},\mathbb{R})$ so it's better to name them correctly for not to get confused later on.) You also have to make sure that $\vec{0}$ is in this set. This guarantees that this set isn't empty.

In the next step, you have to find this vector $\vec{f}_0\,$, which isn't an arbitrary $\beta$. But you are right as there is more than one possibility in this case. One $\beta = \vec{f}_0\,$ however, will do. What could be a candidate? Chose an easy one, this makes life easier if you want to do anything with this affine space.

4. Jun 14, 2017

### mathwonk

try translating by the constant function 1 or by the function cosine, or by any function at all with value 1 at 0.

5. Jun 14, 2017

### Erik109

Thanks a lot for the fast replies! They are all truly insightful and I surely understand the concept now.
@fresh_42 Special thanks for the extensive explanation.

The only thing is that I struggle a bit with proving a given affine subspace of $C(ℝ, ℝ)$ is indeed an affine subspace. I feel like I have to define 'arbitrary' elements to show that $X_1$ is an affine subspace in general. The answer I was provided started off with:
Fix an arbitrary $f ∈ X_1$. Let $γ: ℝ → ℝ$ be defined by $γ(x) = f(x) + h(x)$, where $h : ℝ → ℝ$ (the vector shifting from the origin, you mentioned) is given by $h ≡ −1$ Note that $γ$ is a continuous function with the property $γ(0) = 0$, i.e. $γ ∈ X_0$.

So, if I understand it correctly, the way they prove it is by showing that an arbitrary element from $X_1$ can be shifted back to $X_0$? So kind of 'deconstructing' an affine subspace to a linear subspace?

PS: Got the hang of the LaTeX structure on the forums, hope this helps with the layout.

6. Jun 14, 2017

### WWGD

Yes, essentially. You can " de-translate" the affine space into a vector space by finding the right amount to translate by.

7. Jun 14, 2017

### Staff: Mentor

Every affine subspace $A$ can be written $A=V +\vec{h}$ with a vector (linear) subspace $V$ and the shifting vector $\vec{h}$.
You can observe two things here:
• $\vec{h}$ is not unique. Every other vector $\vec{h}+\vec{v}$ with $\vec{v}\in V$ does the job as well.
• The difference between two elements $\vec{a}_1 = \vec{v}_1 + \vec{h} \, , \,\vec{a}_2 = \vec{v}_2 + \vec{h}$ in $A$ is $\vec{a}_1 -\vec{a}_2 = (\vec{v}_1 + \vec{h})- (\vec{v}_2 + \vec{h})= \vec{v}_1 - \vec{v}_2$ and always an element of $V$.
Now in your case, $V= \{f_V \in C(\mathbb{R},\mathbb{R})\,\vert \,f_V(0)=0\}$ and $A= \{f_A \in C(\mathbb{R},\mathbb{R})\,\vert \,f_A(0)=1\}$, so you need a function $h=\vec{h} \in C(\mathbb{R},\mathbb{R})$ that moves (shifts) all functions in such a way, that $1=\vec{f}_A(0) = \vec{f}_V(0) +\vec{h}(0) = 0 + h(0)$. @mathwonk mentioned a few candidates for $h$ in post #4. As said above, $\vec{h}$ isn't unique, only up to elements of $V$.

Affine subspaces are not closed under addition nor under scalar multiplication. However, all differences of two vectors are, as they are vectors in the corresponding linear subspace. This way you still have properties like flatness and can calculate similar to what you can do in vector spaces, but you don't have a zero and must be more careful with addition and scalar multiplication. E.g. all tangent spaces $T_p$ are linear spaces, but if you actually want to use them as tangents, you have to consider $\vec{p} + T_p$ which is affine. This is very important as usually it's spoken about the vector space $T_p\,$, e.g. a line $\mathbb{R} \cdot \vec{l}$. And as a line it is a vector space only if it contains the origin $\vec{0}$. The actual tangent are the points $\vec{p} + t\cdot \vec{l}$. I mention this, as it might be confusing if people talk about linearity of tangent spaces, e.g. when talking about a gradient $\nabla$, although it is strictly speaking the affine space shifted by $\vec{p}$.