Why L(V,W) Is a Vector Space

From Operations on Vectors to Operations on Maps

Linear maps can themselves be added and multiplied by scalars. The resulting operations are defined by applying the maps to each input and then carrying out the corresponding operation in the codomain. The key question is whether these operations do more than produce functions: do they keep us inside the collection of linear maps, and do they satisfy all the vector-space laws?

Let \(\mathbb{F}\) be either \(\mathbb{R}\) or \(\mathbb{C}\), and let \(V\) and \(W\) be vector spaces over the same field \(\mathbb{F}\). A map \(T:V\to W\) is linear if, for all \(\mathbf{u},\mathbf{v}\in V\) and \(a\in\mathbb{F}\), it satisfies \(T(\mathbf{u}+\mathbf{v})=T(\mathbf{u})+T(\mathbf{v})\) and \(T(a\mathbf{u})=aT(\mathbf{u})\). Write \(L(V,W)\) for the set of all such linear maps.

These formulas define functions from \(V\) to \(W\). But the definition of \(L(V,W)\) requires more: those functions must be linear. Closure under the operations is therefore the first point to establish. In what follows, the zero map, denoted \(\mathbf{0}_{L(V,W)}\), is the function that sends every vector in \(V\) to the zero vector \(\mathbf{0}\) in \(W\). The zero map and the zero vector are different objects.

Closure: The Operations Produce Linear Maps

Proof. Take \(\mathbf{u},\mathbf{v}\in V\) and \(a\in\mathbb{F}\). Since \(S\) and \(T\) are linear, \((S+T)(\mathbf{u}+\mathbf{v})=S(\mathbf{u}+\mathbf{v})+T(\mathbf{u}+\mathbf{v})=S(\mathbf{u})+S(\mathbf{v})+T(\mathbf{u})+T(\mathbf{v})\). Addition in \(W\) is commutative and associative, so this equals \((S+T)(\mathbf{u})+(S+T)(\mathbf{v})\). Also, \((S+T)(a\mathbf{u})=S(a\mathbf{u})+T(a\mathbf{u})=aS(\mathbf{u})+aT(\mathbf{u})=a(S+T)(\mathbf{u})\). Thus \(S+T\) is linear. For the scalar multiple, \((aT)(\mathbf{u}+\mathbf{v})=a\bigl(T(\mathbf{u})+T(\mathbf{v})\bigr)=aT(\mathbf{u})+aT(\mathbf{v})\), and \((aT)(b\mathbf{u})=aT(b\mathbf{u})=a\bigl(bT(\mathbf{u})\bigr)=b\bigl(aT(\mathbf{u})\bigr)\) for every \(b\in\mathbb{F}\). Hence \(aT\) is linear. The zero map is linear because it sends both sides of each defining equation to \(\mathbf{0}\). Finally, \((-T)(\mathbf{u}+\mathbf{v})=-T(\mathbf{u})-T(\mathbf{v})\) and \((-T)(a\mathbf{u})=a(-T)(\mathbf{u})\), so \(-T\) is linear as well. \(\square\)

Why Every Vector-Space Law Holds

Closure alone is not enough: a vector space must also satisfy the usual laws for addition and scalar multiplication. Those laws can be checked directly, without assuming anything special about the form of \(V\) or \(W\). The reason is that two functions are equal exactly when they give equal outputs at every input. So we can take an arbitrary \(\mathbf{v}\in V\), evaluate each side of a proposed law there, and use the corresponding vector-space law in \(W\).

Proof. The proposition proves that \(L(V,W)\) is closed under addition and scalar multiplication, contains the zero map, and contains the pointwise negative of each of its elements. It remains to check the vector-space laws. Let \(S,T,U\in L(V,W)\), \(a,b\in\mathbb{F}\), and \(\mathbf{v}\in V\). Using the definitions and the laws in \(W\), we obtain \(((S+T)+U)(\mathbf{v})=(S(\mathbf{v})+T(\mathbf{v}))+U(\mathbf{v})=S(\mathbf{v})+(T(\mathbf{v})+U(\mathbf{v}))=(S+(T+U))(\mathbf{v})\). Thus addition is associative. Likewise, \((S+T)(\mathbf{v})=S(\mathbf{v})+T(\mathbf{v})=T(\mathbf{v})+S(\mathbf{v})=(T+S)(\mathbf{v})\), so it is commutative.

For the zero map, \((S+\mathbf{0}_{L(V,W)})(\mathbf{v})=S(\mathbf{v})+\mathbf{0}=S(\mathbf{v})\), so \(S+\mathbf{0}_{L(V,W)}=S\). For the additive inverse, \((S+(-S))(\mathbf{v})=S(\mathbf{v})-S(\mathbf{v})=\mathbf{0}=\mathbf{0}_{L(V,W)}(\mathbf{v})\), giving \(S+(-S)=\mathbf{0}_{L(V,W)}\).

For distributivity over addition of scalars, \(((a+b)S)(\mathbf{v})=(a+b)S(\mathbf{v})=aS(\mathbf{v})+bS(\mathbf{v})=(aS+bS)(\mathbf{v})\). For distributivity over addition of maps, \((a(S+T))(\mathbf{v})=a(S(\mathbf{v})+T(\mathbf{v}))=aS(\mathbf{v})+aT(\mathbf{v})=(aS+aT)(\mathbf{v})\). For compatibility of scalar multiplication, \((a(bS))(\mathbf{v})=a(bS(\mathbf{v}))=(ab)S(\mathbf{v})=((ab)S)(\mathbf{v})\). Finally, \((1S)(\mathbf{v})=1S(\mathbf{v})=S(\mathbf{v})\). Since \(\mathbf{v}\) was arbitrary, each equality at inputs proves the corresponding equality of functions. These are all the vector-space laws, so \(L(V,W)\) is a vector space over \(\mathbb{F}\). \(\square\)

What the Result Does—and Does Not—Depend On

The proof uses only the vector-space laws in the codomain \(W\) and the fact that the maps are linear. It does not require \(V\) or \(W\) to have finite dimension. In particular, it does not need coordinates or matrices: those can be useful representations, but the definition and proof work directly with functions.

The common scalar field is essential to the setup. The expressions \(aT(\mathbf{v})\) and \(T(a\mathbf{v})\) both use the scalar \(a\), so the domain and codomain must be vector spaces over the same field for the stated linearity and pointwise operations to apply. Also, \(S+T\) is formed by adding outputs in one codomain \(W\); if two maps have different codomains, this pointwise sum has not been defined by the operations used here.

A useful distinction is that addition of maps is not composition. Both operations can produce maps, but they have different definitions and purposes. Under pointwise addition, evaluate both maps at the same input and add their outputs. Composition instead feeds the output of one map into another, and requires compatible domains and codomains.

Common Pitfall: Confusing a Sum with a Composition

Let \(f,g:\mathbb{R}^2\to\mathbb{R}^2\) be \(f(x,y)=(x,0)\) and \(g(x,y)=(0,y)\). A mistaken calculation might treat their sum as a composition and claim \((f+g)(1,1)=f(g(1,1))=f(0,1)=(0,0)\). The first equality is wrong: the definition of pointwise addition is \((f+g)(1,1)=f(1,1)+g(1,1)\). In fact, \(f(1,1)=(1,0)\), \(g(1,1)=(0,1)\), and therefore \((f+g)(1,1)=(1,0)+(0,1)=(1,1)\). The composition \(f\circ g\) is a different map; here \((f\circ g)(1,1)=(0,0)\). The different outputs make the distinction concrete.

The central proof strategy is reusable: establish that the proposed operations keep functions inside the set, then check each vector-space law by evaluating both sides at an arbitrary input. For \(L(V,W)\), every check reduces to a law already satisfied by vectors in \(W\). This explains why the collection of linear maps has its own vector-space structure, rather than merely being a set of functions with convenient operations.