# When and why can functions “take on” the role of vectors in defining vector spaces?

In what I call "advanced" linear algebra, we examine the properties of vectors in a vector space like an inner product space by checking that they satisfy e.g. the Cauchy-Schwarz inequality, the Triangle Inequality, and similar tests. A few pages later in a typical textbook, we are doing the same tests on functions, rather than vectors (that is, substituting functions for vectors) in these inequalities.

What allows us to do this? My limited understanding is that this can be done if the functions in question can be mapped onto vectors? Is this true, and if so why?

• If a set of functions satisfies the axioms of a vector space, then it is a vector space. That sentence remains true if I replace the word "functions" with any other type of object. Indeed, the ability to treat functions and sequences and other objects as "vectors" is the main reason why the abstract notion of a vector space was introduced. – Bungo Aug 21 '14 at 23:13
• You might want to look up the concept of an $R$-module, where $R$ is a ring. A vector space over a field $K$ is then just a $K$-module. – Shawn O'Hare Aug 21 '14 at 23:35
• @Bungo: Then what makes a set of functions satisfy the axioms of a vector space. Linearity? Or "mappability"? – Tom Au Aug 22 '14 at 0:28

There are many different vector spaces, such as $\mathbb{Q}^n$, $\mathbb{R}^n$, $\mathbb{C}^n$, $C^k(M, \mathbb{R})$, $L^p(X, \mu)$ just to name a few infinite families. As I described above, you could write out a 'concrete' proof of any vector space result for any one of these vector spaces to convince yourself the result holds for that vector space, but why would you want to? In fact, 'abstract' algebraic notions like group, ring, field, vector space, etc. were conceived because this is unnecessary. The idea is that because the proof is effectively the same for each vector space, we just need to prove it once in the abstract setting to know it is true for any vector space. In particular, the result and its proof does not rely on what the vectors are, only that they have the properties that vectors should have (i.e. the ones the axioms of a vector space say they should have).
This may be a helpful analogy. If you want to prove that the square of an even number is even, how do you do it? Let $n$ be an even number, then $n = 2k$ for some integer $k$. Now $n^2 = (2k)^2 = 4k^2 = 2(2k^2)$, so $n^2$ is even. Now, you could rewrite this proof for any particular choice of even number to verify it in that case, but, excuse my repetition, why would you want to? Using the 'abstract' proof, we see the result holds for any even number; that is, it doesn't rely on what the number $n$ is, only that it satisfies the property of being even.