Pointwise Operations on Linear Maps

Combining Maps at Each Input

Linear maps can be combined in much the same way as vectors: add their outputs or multiply their outputs by a scalar. The important point is that these operations are defined input by input. To add two maps, they must accept the same inputs and produce outputs in the same space. The same requirement lets us form scalar multiples of a map.

Let \(V\) and \(W\) be vector spaces over the same field \(\mathbb{F}\), where \(\mathbb{F}\) is \(\mathbb{R}\) or \(\mathbb{C}\). Write \(\mathcal{L}(V,W)\) for the set of linear maps from \(V\) to \(W\). Recall that a map \(T:V\to W\) is linear if \(T(a\mathbf{v}+b\mathbf{w})=aT(\mathbf{v})+bT(\mathbf{w})\) for all \(\mathbf{v},\mathbf{w}\in V\) and scalars \(a,b\in\mathbb{F}\).

These are pointwise definitions: choose an input, evaluate each map there, and then perform the corresponding operation on the outputs. The scalar \(c\) multiplies the output of \(T\), not the map’s input as a separate operation. Since \(T\) is linear, \(cT(\mathbf{v})=T(c\mathbf{v})\) also holds, but the definition of \(cT\) is the output-side formula.

The common-space condition matters. If \(T:V\to W\) and \(S:V\to Z\) have different codomains, the expression \(T(\mathbf{v})+S(\mathbf{v})\) is not defined by these rules: the outputs need to be vectors in one common vector space. Likewise, maps with different domains cannot be added by this pointwise definition because there is no common input to use for every evaluation.

Why the Result Is Still a Linear Map

Pointwise formulas define functions, but here the intended results are again linear maps. The following check proves that addition and scalar multiplication preserve linearity; it does not require \(V\) or \(W\) to be finite-dimensional.

Proof. For any \(\mathbf{v},\mathbf{w}\in V\) and \(a,b\in\mathbb{F}\), linearity of \(T\) and \(S\) gives \((T+S)(a\mathbf{v}+b\mathbf{w})=T(a\mathbf{v}+b\mathbf{w})+S(a\mathbf{v}+b\mathbf{w})=aT(\mathbf{v})+bT(\mathbf{w})+aS(\mathbf{v})+bS(\mathbf{w})=a(T+S)(\mathbf{v})+b(T+S)(\mathbf{w})\). For the scalar multiple, \((cT)(a\mathbf{v}+b\mathbf{w})=cT(a\mathbf{v}+b\mathbf{w})=c(aT(\mathbf{v})+bT(\mathbf{w}))=a(cT)(\mathbf{v})+b(cT)(\mathbf{w})\). Thus both maps are linear and have the same domain and codomain as \(T\) and \(S\). \(\square\)

This proposition verifies that the operations stay within \(\mathcal{L}(V,W)\). It does not by itself establish all the vector-space rules for these operations; those require additional checks. In computations, one can often specify a resulting map by simplifying its output formula for a general input.

Computing with Coordinate-Space Maps

For coordinate-space maps, a formula such as the one above computes every output at once. When bases are chosen, the corresponding matrix of a sum or scalar multiple can also be obtained by the same operation on the matrices; that is the coordinate representation of these pointwise definitions.

Operations on Polynomial Maps

Polynomial spaces provide examples where the input is a polynomial rather than a coordinate column. Let \(P_n\) denote the real vector space of polynomials of degree at most \(n\), including the zero polynomial. The map’s domain and codomain must still match before a pointwise sum is formed.

In this example, evaluation at zero produces a scalar, and that scalar multiplies the polynomial \(1+x\) to give an output in the codomain. A scalar-valued evaluation alone would have codomain \(\mathbb{R}\), so it could not be added directly to the derivative map \(P_2\to P_1\). The factor \(1+x\) makes the outputs compatible.

Values on a Spanning Set Determine a Map

To compare two linear maps, it is not always necessary to calculate their outputs on every vector. A set \(G\subseteq V\) spans \(V\) if every vector in \(V\) is a finite linear combination of elements of \(G\). Agreement on such a set determines the map everywhere.

Proof. If \(T=S\), then their values agree at every input, including each \(\mathbf{g}\in G\). Conversely, suppose they agree on every element of \(G\). If \(G\) is nonempty, any \(\mathbf{v}\in V\) can be written as \(\mathbf{v}=a_1\mathbf{g}_1+\cdots+a_k\mathbf{g}_k\), with \(\mathbf{g}_i\in G\) and \(a_i\in\mathbb{F}\). By linearity and the assumed agreement, \(T(\mathbf{v})=\sum_{i=1}^{k}a_iT(\mathbf{g}_i)=\sum_{i=1}^{k}a_iS(\mathbf{g}_i)=S(\mathbf{v})\). If \(G\) is empty and spans \(V\), then \(V=\{\mathbf{0}\}\). For any linear map \(T\), linearity gives \(T(\mathbf{0})=T(\mathbf{0}+\mathbf{0})=T(\mathbf{0})+T(\mathbf{0})\), so \(T(\mathbf{0})=\mathbf{0}\); the same holds for \(S\). Therefore the maps agree on all of \(V\) in this case as well. Hence \(T=S\). \(\square\)

This result can make verification efficient. For example, when a basis is available, checking two linear maps on its basis vectors is enough to establish that they agree on the whole space. This applies to pointwise sums too: to verify a proposed formula for \(T+S\), one may check it on a spanning set, provided the proposed map is linear.

Common Pitfall: Confusing Scalar Multiplication with Composition

Scalar multiplication of a map is not the same operation as composing a map with itself. Let \(T:\mathbb{R}^2\to\mathbb{R}^2\) be \(T(x,y)=(x,2y)\). The pointwise scalar multiple is \((2T)(x,y)=2T(x,y)=(2x,4y)\). A mistaken calculation might instead write \(2T=T\circ T\), but composition gives \((T\circ T)(x,y)=T(x,2y)=(x,4y)\). At \((1,0)\), these outputs are \((2,0)\) and \((1,0)\), respectively, so the operations are different. The correction is to apply the scalar to the output of one evaluation: \((2T)(\mathbf{v})=2\,T(\mathbf{v})\).

Before combining maps, check that they have the same domain and codomain and use the same scalar field. Then evaluate each map at a general input, combine the outputs, and simplify. For polynomial maps, retaining general coefficients until the final substitution helps expose the resulting rule and reduces arithmetic errors.