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I often stumble upon books using the term "linear space" (outside of Linear Algebra) and I have never been totally comfortable with this. Perhaps I am over complicating this, but my intuition says that a "linear space" is one which has some sort of basis, say $\{a, b\}$, and all the other elements can be thought of as linear combinations of the elements in the basis like $j=\alpha a + \beta b$ (for some $\alpha, \beta$). This would further exclude the possibility of some sort of non-linear combinations like $j=\alpha a + \gamma a^2 + \beta b$ from our linear space.

Is my intuition on the right track? Should I even bother trying to define "linear space" to myself? Is "linear space" synonymous with "vector space"?

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2 Answers

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Formally, you need an underlying field $F$ under your linear space. The space (let's say $L$) is then whatever set of whatever object, such that you have two maps: Sum as a map from $L\times L\mapsto L$ and multiplication as a mup from $L\times F\mapsto L$. They have to satisfy some axioms of course (sum is commutative, both are associative, the distibutive law is valid).

So for example take the set of real functions $\mathbb{R}\to\{0,\dots,6\}$, and define $(a+b)(x)=a(x)+b(x)\mod 7$ and $(ta)(x)=ta(x) \mod 7$. Since $\mathbb{Z}_7$ is a field, this is a linear space, by definition. No intuition about lines and bases works here.

The elements of your $L$ can be potatoes, as long as you're able to find a field and two operations that satisfy the definition.

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However, in general if the literature says that something "is a linear space", they in general mean that it's a set closed under addition and multiplication by numbers from some field, where the defition of the operations as well as the underlying field are usually clear from the context.

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Lots of things can be the basis of a linear space, not just the $(1,2,6)$-type vectors you are used to.

For example, the set of polynomials of degree 2 or less - so quadratic, linear and constant polynomials - is a three-dimensional vector space. In this case, the basis is $\{1,x,x^2\}$. So although you do have the non-linear $x^2$ present, every polynomial is a linear combination of the three basis polynomials: $y=2x+3 = 2(1)+3(x)+0(x^2)$.

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