# I have difficulty understanding functions forming vector space.

I have knowledge of basic linear algebra, so I can understand the finite vector space as linear combinations of vectors of $R^n$.

However, when it comes function as vector and functions form a vector space, I just cannot go over.

Below are two pieces of definitions I hardly got the essences. Could you help to explain how should I understand it. I am preparing the fundamentals so that I can understand wavelet better.

My wrong idea is:

Try set the domain of a function fixed, say $R$, then the image of the function is a infinite-dimension vector.

• Your question is rather vague. Exactly what part of the excerpt don't you understand? Is it the definitions of the addition and scalar multiplication? Vector spaces are rather abstract objects, so naturally there are examples that don't fit the "usual" idea as $\mathbb{F}^n$ with $\mathbb{F}$ a field and $n\in\mathbb{N}$. Jan 26, 2015 at 2:57
• Sorry for that my question is vague. Since I really have many doubts, and I really cannot express clearly. Thank you for your comments. Jan 26, 2015 at 3:29
• Which of the conditions in the definition are you having trouble verifying? Jan 26, 2015 at 3:34
• @WillO If I see a function as a vector in the vector space, I cannot see the dimension of the vector space. I just know I am composing new functions by addition and scalar-multiplication. Jan 26, 2015 at 3:43
• That doesn't answer my question. Jan 26, 2015 at 3:45

This is simply applying a vector space structure to functions from a given set to $\mathbb{C}$. For instance, take $A = \mathbb{R}$. The set of all complex-valued functions on $\mathbb{R}$ form a vector space where scalar multiplication and vector addition are defined as given there. This is simply showing how we can make vector spaces out of functions.

If you have seen the notation $T \in \mathcal{L}(V,W)$, this might be a little easier to grasp. That is read as "The function $T$ is in the set of all linear transformation from $V$ to $W$", where $\mathcal{L}(V,W)$ is the set of all such transformations. This set has a natural vector space structure on it, namely the one given in the snippet you posted:

For $\alpha \in F$ (the underlying field of both $V$ and $W$) define $\alpha T$ as $v \mapsto \alpha T(v)$ and $S+T$ as $v \mapsto S(v) + T(v)$. This gives us a vector space structure, and the objects (vectors) in this vector space are linear transformations (i.e., functions).

• First, thank you for your kind reply. With respect to my question, I can understand that functions equipped with addition and scalar multiplication result in a vector space with linearity( as defined $\alpha T$ as $v \rightarrow \alpha T(v)$ and $S + T$ as $v \rightarrow S(v) + T(v)$ ). But in definition shown in the first image, it says these functions form a vector space. Where are the vectors? dimension? Can I do element-wise addition for the functions? Jan 26, 2015 at 3:26
• The vectors are the functions themselves. We begin teaching linear algebra by having you think of vectors as points in Euclidean space or as arrows in a a plane, etc, but really vectors are just objects with this nice special structure. We could consider a finite vector space of various fruits and as long as we define things in the right way, it doesn't matter that we use fruits instead of arrows, what really matters about a vector space is the structure of the operations. That is the essence of algebra as a whole. Jan 26, 2015 at 3:29
• As for the dimension, we need to be careful in setting things up to answer that question. For instance, what is the dimension of the vector space $\mathbb{C}$? There is no well-defined answer until we say what our underlying field is, that is, where our scalars are coming from. Over $\mathbb{C}$, $\mathbb{C}$ is a 1-dimensional vector space. Over $\mathbb{R}$, it is 2-dimensional. Jan 26, 2015 at 3:30
• Ok, let me be more specific: (in second image) $\phi \in L^2(R)$ . What is the dimension of a vector in this vector space spanning by $\phi$ and their translations $\phi(.-k), k \in Z$ (I am not familiar with $L^2(R)$) It is difficult for me to understand this. Jan 26, 2015 at 3:39
• To be honest, I'm not 100% sure what's going on in the second image. I think $L^2(\mathbb{R})$ is something to do with measure theory and Hilbert spaces, neither of which I'm terribly familiar with. I can only speak to the general linear algebra, with which I'm much more familiar. Sorry! Jan 26, 2015 at 3:41

All that it means to be a vector space is that you can add and subtract vectors, and scale them by elements of your field. This is clearly true about functions! What makes this vector space different than, say, $\mathbb{R}^n$ is that it doesn't have a standard basis to relate a given vector to.

Actually if $A$ is finite there is a reasonable basis. Let $A = \{a_1,\ldots,a_n\}$, then this vector space is the same as $\mathbb{R}^n$, your more familiar example. You can think of each element $a\in A$ as a basis vector and the the value $f(a)$ is the component in that "direction".

More precisely, if you take the functions $f_1,\ldots,f_n$ where $f_i(a_j) = \delta_{ij}$ then these form a basis for the space, and any function $f$ decomposes as $f = \sum_{i=1}^n f(a_i)f_i$.

[Truthfully the two vector spaces just described are dual. If you have seen dual spaces before then this function space is naturally isomorphic to the dual space to the free vector space on $A$.]

The big difference when $A$ is infinite is that these vectors $\{f_i\}$ do not form a basis. Instead, they span a proper subspace consisting of functions which are non-zero on only finitely many values $a \in A$. If we take a general function and try to write $f = \sum_{i\in I}f(a_i)f_i$ we will not get a finite sum, and thus we have left the theory of vector spaces.

It is possible to work with infinite sums of vectors, but you need to add a topology to do so. This is done especially in Hilbert spaces.

The Hilbert space $L^2[0,1]$ possesses a particularly nice (orthonormal) basis, which is $\{e^{2\pi i nx}\}_{n\in\mathbb{Z}}$. By computing a function's Fourier coefficients you can relate an arbitrary vector to this basis as an (infinite) linear combination.

However, in a general function space such as yours there is no nice basis to compare vectors to, and all you need to understand about this example is that you can add, subtract and scale functions, thus they form a vector space.

Try not to think the way you are thinking now, you've probably focused too much on the input space and output space of a function. Try to think of a function as an object, just like a vector. And some answers from this question might help:

Why is a function space considered to be a "vector" space when its elements are not vectors?

Please let me know if I'm wrong, and I will remove this answer. Here is a sketch for the associativity proof.

$V$ we define as $\{f: f: A \to \mathbb{C}\}$.

I claim $V$ is a vector space, based on the definition here.

Let $u, v, w \in V$. Then given an arbitrary $x$ in the corresponding field, \begin{align} ((u+v)+w))(x) &= (u+v)(x)+w(x) \\ &=u(x)+v(x)+w(x) \\ &= u(x)+(v+w)(x) \\ &=(u+(v+w))(x)\text{.}\end{align} Hence $((u+v)+w) = (u+(v+w))$.