Sep 5, 2026
If you studied Linear Algebra from Gilbert Strang (cite?) you learned about the vector space {\boldsymbol{{R}}^n = \{(x_1,\ldots,x_n)\mid x_i\in\boldsymbol{{R}}\}} of n-tuples of real numbers. After the first week of class it may have dawned on you that you are now doomed to spending the rest of the semester doing tedious numerical exercises involving “row” and “column” vectors.
If you found that unsatisfying you might have come across the pompously titled “Vector Spaces Done Right” by Sheldon Axler. (cite?) It takes a modern axiomatic approach that clarifies how scalar multiplication, vector addition, and their connection through distributive laws clarify the structure of vector spaces and linear operators between them. You will not learn what a determinant is until the last chapter.
I studied from (Nering 1964) when I was getting a “C” in my first linear algebra course. When we got to determinants I noped out. Who ordered those?
Banchoff Wermer
Halmos
Peano
Euclid
There are many vector spaces that are not just n-tuples of numbers.
Unlike Dicken’s tale, there is a second vector space that has the same form as the first. If \boldsymbol{{n}} = \{1,\ldots,n\} define \boldsymbol{{R}}^{\boldsymbol{{n}}} = \{x\colon\boldsymbol{{n}}\to\boldsymbol{{R}}\}. This is isomorphic to \boldsymbol{{R}}^n. Every x = (x_1,\ldots,x_n) corresponds to \boldsymbol{{x}}\colon\boldsymbol{{n}}\to\boldsymbol{{R}} where \boldsymbol{{x}}(i) = x_i, i\in\boldsymbol{{n}}.
Eventually I got a PhD in Operator Theory and a job at Brown by solving a problem about closed invariant subspaces of bounded linear operators on Hilbert spaces. (Beurling 1949) found all the invariant subspaces of the unilateral shift operator in infinite dimensions. If S(x_1, x_2, \ldots)\mapsto(0,x_1, x_2, \ldots) then S has no eigenvectors, but it has plenty of invariant subspaces. Arne found all of them. Beurling inspired generations of mathematicians to extend his result.
He also solved the Geheimschreiber teletype cipher with pencil and paper in two weeks. To this day, nobody knows how he did that.
It wasn’t until I started implementing math on computers that I realized there was an equivalent, but different, definition of a vector space. If A and B are sets, then the set exponential {B^A = \{f\colon A\to B\} = \{A\to B\}} is the set of all functions from A to B. We can identify \boldsymbol{{R}}^n with {\boldsymbol{{R}}^{\{1,\ldots,n\}}}. The tuple {x = (x_1,\ldots,x_n)} can be identified with the function {\boldsymbol{{x}}\colon\boldsymbol{{n}} \to\boldsymbol{{R}}} by {x_i = \boldsymbol{{x}}(i)} where {\boldsymbol{{n}} = \{1,\ldots,n\}} and {i\in\boldsymbol{{n}}}. We use bold font to distinguish the product from the exponential vector space definitions.
The standard basis of \boldsymbol{{R}}^n is \{e_i\}_{i=1}^n where e_i is an n-tuple of 0’s except for a 1 in the i-th index so {(x_1,\ldots,x_n) = x_1 e_1 + \cdots + x_n e_n}. The corresponding basis of {\boldsymbol{{R}}^{\boldsymbol{{n}}}} is \{\boldsymbol{{e}}_i\}_{i=1}^n where \boldsymbol{{e}}_i(j) = \delta_{ij}, the Kronecker delta, so \boldsymbol{{x}} = \sum_{i=1}^n \boldsymbol{{x}}(i)\boldsymbol{{e}}_i.
The exponential definition is more flexible. Given any set I, \boldsymbol{{R}}^I is a vector space. Define the scalar multiplication and vector addition pointwise by {(a\boldsymbol{{x}})(i) = a(\boldsymbol{{x}}(i))} for {a\in\boldsymbol{{R}}} and {(\boldsymbol{{x}} + \boldsymbol{{y}})(i) = \boldsymbol{{x}}(i) + \boldsymbol{{y}}(i)} for {\boldsymbol{{x}},\boldsymbol{{y}}\in\boldsymbol{{R}}^I}.
An abstract vector space is any abelian (commutative) group V having a scalar multiplication that satisfies the distributive laws {a(x + y) = ax + ay} and {(a + b)x = ax + bx} for a,b\in\boldsymbol{{R}} and x,y\in V. While it is true every finite-dimensional vector space is isomorphic to \boldsymbol{{R}}^n for some n, there are many abstract vector spaces that are not.
Exercise. Show continuous functions on the interval [a,b]\subseteq\boldsymbol{{R}}, C([a,b]), is a vector space.
Exercise. Show the set of solutions to the differential equation \sum_j a_j d^j x(t)/dt^j = 0 is a vector space.
Recall a function {T\colon\boldsymbol{{R}}^n\to\boldsymbol{{R}}^m} is a linear operator if {T(ax + y) = aTx + Ty} for a\in\boldsymbol{{R}} and x,y\in\boldsymbol{{R}}^n. The set of all linear operators from vector space V to vector space W is denoted in various ways. The category theoretic teminology is {\operatorname{Hom}(V,W)} for the set of all homomorphisms. The most common is \mathcal{L}(V,W). If the vectors spaces have an associated topology then \mathcal{B}(V,W) indicates all continuous operators. In finite dimensions all continuous linear operators are \mathcal{B}ounded. We will use the non-standard notation {[V\to W] = \{T:V\to W\mid T\text{ is linear}\}}.
Exercise. Show if T\colon V\to W is linear then T(ax) = a(Tx) and T(x + y) = Tx + Ty for a\in\boldsymbol{{R}} and x,y\in V.
Note the scalar multiplication and vector addition on the right-hand side of these equations is occuring in W.
Exercise. SHow linear operators are a vector space.
The dual of a vector space V is the space of all linear functionals {V^* = [V\to\boldsymbol{{R}}]}.
We can define a function {}^*\colon\boldsymbol{{R}}^n\to\boldsymbol{{R}}^{\boldsymbol{{n}}}
This can be used to show matrix multiplication is composition of linear operators.