From Values on Points to Coordinate Vectors
A real-valued function on a finite set is completely determined by its value at each point. If the set has \(n\) points, collecting those values in a specified order produces a vector in \(\mathbb{R}^n\). The order matters: coordinates are not an unordered collection of values, but a list whose positions are assigned to particular points. Once an ordering is fixed, the correspondence between functions and vectors is exact, and it respects addition and scalar multiplication.
Let \(S\) be a finite set with \(n\geq 1\) elements. Write \(\mathcal{F}(S,\mathbb{R})\) for the set of all functions from \(S\) to \(\mathbb{R}\). We give this set the pointwise operations \( (f+g)(s)=f(s)+g(s) \) and \( (cf)(s)=cf(s) \) for every \(s\in S\), where \(f,g\in\mathcal{F}(S,\mathbb{R})\) and \(c\in\mathbb{R}\). Its zero vector is the function with value \(0\) at every point. These operations make \(\mathcal{F}(S,\mathbb{R})\) a real vector space: each vector-space identity holds because it holds for real numbers at every point.
Choose an ordering of the points of \(S\), written \((s_1,\dots,s_n)\). This means each point appears exactly once. Evaluating a function at the points in that order defines a map \(E:\mathcal{F}(S,\mathbb{R})\to\mathbb{R}^n\):
The vector \(E(f)\) records values, not a formula for the function. This is sufficient because every point of \(S\) occurs in the ordering: if two functions have the same ordered values, they agree at every point and hence are the same function. Conversely, any \(n\) real numbers can be assigned as values at the \(n\) respective points. The next result makes both observations precise.
The Evaluation Map Is an Isomorphism
Proof. First, for \(f,g\in\mathcal{F}(S,\mathbb{R})\) and \(c\in\mathbb{R}\), the \(i\)th entry of \(E(f+g)\) is \((f+g)(s_i)=f(s_i)+g(s_i)\), the \(i\)th entry of \(E(f)+E(g)\). The \(i\)th entry of \(E(cf)\) is \((cf)(s_i)=cf(s_i)\), the \(i\)th entry of \(cE(f)\). These equalities hold for every \(i\), so \(E(f+g)=E(f)+E(g)\) and \(E(cf)=cE(f)\); therefore \(E\) is linear.
To prove that \(E\) is one-to-one, suppose \(E(f)=E(g)\). Equality of the columns gives \(f(s_i)=g(s_i)\) for each \(i=1,\dots,n\). Since the list contains every point of \(S\), \(f(s)=g(s)\) for every \(s\in S\), so \(f=g\). To prove that \(E\) is onto, take any \(\mathbf{x}=(x_1,\dots,x_n)^{\top}\in\mathbb{R}^n\). Define \(f:S\to\mathbb{R}\) by \(f(s_i)=x_i\) for each \(i\). Each point occurs exactly once, so this prescription assigns exactly one value to every point and defines a function. It satisfies \(E(f)=\mathbf{x}\). Thus \(E\) is onto. The same prescription shows that this \(f\) is unique: any function mapping to \(\mathbf{x}\) must have value \(x_i\) at each \(s_i\). Therefore \(E\) is a linear bijection and has the stated inverse. \(\square\)
This proof also explains what “coordinates of a function” means here. The entries are indexed by points through the ordering, and recovering the function means assigning each entry back to its corresponding point. No rule connecting values at different points is required: a function on a finite set may take any real values independently at its distinct points.
Worked Example: Evaluating in a Specified Order
Let \(S=\{\alpha,\beta,\gamma\}\), ordered as \((\gamma,\alpha,\beta)\). Define \(f(\alpha)=-1\), \(f(\beta)=4\), and \(f(\gamma)=2\). Find \(E(f)\), then recover \(f\) from the column \(\begin{bmatrix}2\\-1\\4\end{bmatrix}\).
The first coordinate belongs to \(\gamma\), the second to \(\alpha\), and the third to \(\beta\). Therefore
For the reverse direction, assign the first entry to \(\gamma\), the second to \(\alpha\), and the third to \(\beta\). The recovered function has \(f(\gamma)=2\), \(f(\alpha)=-1\), and \(f(\beta)=4\).
Check: Evaluating the recovered function in the stated order gives \(\begin{bmatrix}f(\gamma)&f(\alpha)&f(\beta)\end{bmatrix}^{\top}=\begin{bmatrix}2&-1&4\end{bmatrix}^{\top}\), the original column.
Why Pointwise Operations Match Vector Operations
Linearity is more than a formal condition: it says that taking coordinates does not disrupt addition or scalar multiplication. The correspondence works because both operations on functions are defined point by point. For example, if \(E(f)=\mathbf{x}\) and \(E(g)=\mathbf{y}\), then the value of \(f+g\) at the \(i\)th point is the sum of the \(i\)th values, exactly as in vector addition.
Worked Example: Coordinates of a Linear Combination
Use \(S=\{p,q,r\}\) ordered as \((p,q,r)\). Let \(f(p)=1,\ f(q)=-2,\ f(r)=3\), and \(g(p)=4,\ g(q)=0,\ g(r)=-1\). Find the coordinates of \(2f-g\) by using the function operations, and compare with \(2E(f)-E(g)\).
Evaluate the combination at each point. At \(p\), \((2f-g)(p)=2(1)-4=-2\). At \(q\), \((2f-g)(q)=2(-2)-0=-4\). At \(r\), \((2f-g)(r)=2(3)-(-1)=7\). Hence
Now evaluate the two functions separately and perform the vector operations:
Check: The entries of this column are precisely the previously computed values of \(2f-g\) at \(p,q,r\). Thus \(E(2f-g)=2E(f)-E(g)\), as the linearity proof predicts.
One can also describe individual coordinates using functions that isolate one point. For each \(i\), define \(\delta_i:S\to\mathbb{R}\) by \(\delta_i(s_i)=1\) and \(\delta_i(s_j)=0\) whenever \(j\ne i\). This is the indicator function of the single point \(s_i\). Its evaluation column is the standard unit vector \(\mathbf{e}_i\): it has a \(1\) in position \(i\) and \(0\) in every other position.
Proof. Fix any \(f\in\mathcal{F}(S,\mathbb{R})\) and consider \(h=\sum_{i=1}^{n}f(s_i)\delta_i\). At a point \(s_j\), every term with \(i\ne j\) is zero because \(\delta_i(s_j)=0\), while the term with \(i=j\) equals \(f(s_j)\delta_j(s_j)=f(s_j)\). Thus \(h(s_j)=f(s_j)\) for each \(j\). The points \(s_1,\dots,s_n\) exhaust \(S\), so \(h=f\); hence the list spans the function space.
For uniqueness, suppose \(\sum_{i=1}^{n}c_i\delta_i\) is the zero function. Evaluating at \(s_j\) gives \(c_j=0\), because \(\delta_j(s_j)=1\) and \(\delta_i(s_j)=0\) for \(i\ne j\). This holds for every \(j\), so all coefficients vanish. The list is linearly independent and spans, and therefore is a basis. The same evaluation argument shows that the coefficients in the displayed expression must be \(f(s_i)\), so the expression is unique. \(\square\)
The indicator functions make the coordinate correspondence concrete: each \(\delta_i\) contributes only to the value at \(s_i\). Under \(E\), they correspond respectively to \(\mathbf{e}_1,\dots,\mathbf{e}_n\). In particular, the vector of values of \(f\) lists exactly the coefficients needed to combine these indicator functions.
Worked Example: Reconstructing a Function with Indicators
Let \(S=\{u,v,w,z\}\) have ordering \((u,v,w,z)\), and let \(f(u)=3,\ f(v)=0,\ f(w)=-2,\ f(z)=5\). Write \(f\) as a combination of the four indicator functions and check the values.
By the formula just proved, the coefficient of each indicator function is the value of \(f\) at its point:
At \(u\), the right side has value \(3(1)+0(0)-2(0)+5(0)=3\). At \(v\), its value is \(3(0)+0(1)-2(0)+5(0)=0\). At \(w\), its value is \(3(0)+0(0)-2(1)+5(0)=-2\). At \(z\), its value is \(3(0)+0(0)-2(0)+5(1)=5\).
Check: The reconstructed function has the prescribed values at all four points. Its coordinate column is \(\begin{bmatrix}3&0&-2&5\end{bmatrix}^{\top}\), which also equals \(3\mathbf{e}_1+0\mathbf{e}_2-2\mathbf{e}_3+5\mathbf{e}_4\).
The Ordering Is Part of the Coordinate Convention
A finite set does not necessarily come with a preferred first, second, or third point. The evaluation-coordinate map therefore depends on the ordering chosen. Choosing a different ordering does not change the function space or the values of the function; it changes which value occupies each coordinate position. Both choices give isomorphisms with \(\mathbb{R}^n\), but their coordinate columns may differ.
For example, on \(S=\{a,b,c\}\), suppose \(f(a)=7,\ f(b)=1,\ f(c)=-3\). In the order \((a,b,c)\), the coordinate column is \(\begin{bmatrix}7&1&-3\end{bmatrix}^{\top}\). In the order \((c,a,b)\), it is \(\begin{bmatrix}-3&7&1\end{bmatrix}^{\top}\). These columns encode the same function under different point-to-position assignments.
Common Mistake: Mixing the Order and the Values
Suppose \(S=\{a,b,c\}\) is ordered as \((a,b,c)\), and a function has \(f(a)=7,\ f(b)=1,\ f(c)=-3\). A wrong calculation might place the values in the order \((c,a,b)\) and report \(\begin{bmatrix}-3&7&1\end{bmatrix}^{\top}\) as the coordinate column for \((a,b,c)\). That column instead corresponds to the function whose values are \(-3\) at \(a\), \(7\) at \(b\), and \(1\) at \(c\). The correction is to match each coordinate position to the point named in the ordering: for \((a,b,c)\), the column is \(\begin{bmatrix}f(a)&f(b)&f(c)\end{bmatrix}^{\top}=\begin{bmatrix}7&1&-3\end{bmatrix}^{\top}\).
Always write down the point order before forming the column. When recovering a function from coordinates, use that same order in reverse. This prevents a coordinate vector from being mistaken for an unordered list of function values.
Check Your Understanding
Use the specified ordering whenever a question asks for coordinates, and justify your conclusions from the definitions.
- Let \(S=\{x,y,z\}\) be ordered as \((z,x,y)\). If \(f(x)=2,\ f(y)=-1,\ f(z)=4\), find \(E(f)\).
- For a finite set \(S=\{s_1,\dots,s_n\}\), explain why equality \(E(f)=E(g)\) implies \(f=g\).
- Given an arbitrary column \(\mathbf{x}=(x_1,\dots,x_n)^{\top}\), define the function that maps to \(\mathbf{x}\) under the ordering \((s_1,\dots,s_n)\).
- For the ordering \((s_1,\dots,s_n)\), state the values of the indicator function \(\delta_j\) at \(s_i\) when \(i=j\) and when \(i\ne j\).
- On \(S=\{a,b\}\), compare the coordinate columns of a function with values \(5\) at \(a\) and \(-2\) at \(b\), using orders \((a,b)\) and \((b,a)\).
