From a Transformation to a Functional on Its Domain
A linear transformation sends vectors in one space to vectors in another. A linear functional assigns a scalar to each vector in its space. Given a transformation \(T:V\to W\) and a functional on \(W\), we can first transform a vector in \(V\) and then evaluate the functional on the result. This produces a scalar-valued rule on \(V\), and linearity of both operations ensures that the rule is itself a functional.
Let \(V\) and \(W\) be vector spaces over the same field \(\mathbb{F}\), where \(\mathbb{F}\) is \(\mathbb{R}\) or \(\mathbb{C}\). A linear functional on \(V\) is a linear map \(\varphi:V\to\mathbb{F}\). The dual space \(V^*\) is the vector space of all linear functionals on \(V\). Linearity here means \(\varphi(a\mathbf{v}+b\mathbf{u})=a\varphi(\mathbf{v})+b\varphi(\mathbf{u})\) for all \(\mathbf{u},\mathbf{v}\in V\) and \(a,b\in\mathbb{F}\). Over \(\mathbb{C}\), this is still ordinary complex linearity, not conjugate-linearity.
The order of operations follows directly from the types. For \(\mathbf{v}\in V\), first \(T\) sends \(\mathbf{v}\) to \(T(\mathbf{v})\in W\), then \(\varphi\) sends that vector to a scalar. The composition \(\varphi\circ T:V\to\mathbb{F}\) is therefore defined. The dual map acts on functionals, not on vectors: its input is \(\varphi\in W^*\), and its output is a functional in \(V^*\).
Why the Definition Produces a Functional
Proof. Fix \(\varphi\in W^*\). To show \(T^*(\varphi)\) is in \(V^*\), take \(\mathbf{u},\mathbf{v}\in V\) and \(a,b\in\mathbb{F}\). Using the definition of composition, the linearity of \(\varphi\), and the linearity of \(T\), we obtain \(\bigl(T^*(\varphi)\bigr)(a\mathbf{u}+b\mathbf{v})=\varphi(T(a\mathbf{u}+b\mathbf{v}))=\varphi(aT(\mathbf{u})+bT(\mathbf{v}))=a\varphi(T(\mathbf{u}))+b\varphi(T(\mathbf{v}))\). This equals \(a\bigl(T^*(\varphi)\bigr)(\mathbf{u})+b\bigl(T^*(\varphi)\bigr)(\mathbf{v})\), so \(T^*(\varphi)\) is linear on \(V\). Now take \(\varphi,\psi\in W^*\) and \(a,b\in\mathbb{F}\). For each \(\mathbf{v}\in V\), \(\bigl(T^*(a\varphi+b\psi)\bigr)(\mathbf{v})=(a\varphi+b\psi)(T(\mathbf{v}))=a\varphi(T(\mathbf{v}))+b\psi(T(\mathbf{v}))=\bigl(aT^*(\varphi)+bT^*(\psi)\bigr)(\mathbf{v})\). The two functionals therefore agree on every \(\mathbf{v}\), proving \(T^*(a\varphi+b\psi)=aT^*(\varphi)+bT^*(\psi)\). Hence \(T^*\) is linear. \(\square\)
This construction does not require \(V\) or \(W\) to be finite-dimensional. Finite-dimensional coordinates are useful for computing examples, but the definition itself only uses the vector spaces, the linear map, and composition.
Worked Example: Pulling Back a Functional from R³
Let \(T:\mathbb{R}^2\to\mathbb{R}^3\) be defined by \(T(x,y)=(x+2y,\,3x-y,\,y)\), and let \(\varphi\in(\mathbb{R}^3)^*\) be \(\varphi(a,b,c)=2a-b+4c\). Compute \(T^*(\varphi)\).
Take an arbitrary input \((x,y)\in\mathbb{R}^2\). Apply \(T\) first, then evaluate \(\varphi\): \(\bigl(T^*(\varphi)\bigr)(x,y)=\varphi(T(x,y))=\varphi(x+2y,\,3x-y,\,y)\). Substituting the three coordinates into \(\varphi\) gives \(2(x+2y)-(3x-y)+4y=2x+4y-3x+y+4y=-x+9y\). Therefore \(T^*(\varphi)\) is the functional on \(\mathbb{R}^2\) given by \((x,y)\mapsto -x+9y\).
Check: at \((1,0)\), the transformed vector is \(T(1,0)=(1,3,0)\), and \(\varphi(1,3,0)=2-3=-1\), matching \(-1+9(0)=-1\). At \((0,1)\), \(T(0,1)=(2,-1,1)\), so \(\varphi(2,-1,1)=4+1+4=9\), matching \(-0+9(1)=9\).
Computing on Polynomial Spaces
A functional need not be given by a coordinate dot product. For a finite-dimensional polynomial space, it may be defined by evaluations or by combining coefficients. The same rule applies: insert the transformed object into the functional. The result must be a scalar-valued linear function of the original input.
Worked Example: Evaluation Data Becomes a Functional
Let \(P_2(\mathbb{R})\) denote the real polynomials of degree at most \(2\). Define \(S:P_2(\mathbb{R})\to\mathbb{R}^2\) by \(S(p)=(p(0),p(1))\), and let \(\psi(u,v)=3u-2v\). Find \(S^*(\psi)\).
For an arbitrary \(p(x)=a+bx+cx^2\), first compute the two evaluations: \(p(0)=a\) and \(p(1)=a+b+c\). Thus \(\bigl(S^*(\psi)\bigr)(p)=\psi(S(p))=3p(0)-2p(1)=3a-2(a+b+c)=a-2b-2c\). So \(S^*(\psi)\) is the functional \(p\mapsto a-2b-2c\), where \(a,b,c\) are the coefficients of \(p\).
Check: for \(p(x)=1\), the original evaluation is \(\psi(1,1)=3-2=1\), and the computed functional gives \(a-2b-2c=1\). For \(p(x)=x^2\), the evaluations are \(S(p)=(0,1)\), so \(\psi(S(p))=-2\); the coefficient formula gives \(0-0-2=-2\). These checks agree with the formula for arbitrary coefficients.
One can also compute the pullback by applying the map to a basis of its domain and recording the functional values. That approach is often convenient when a functional is specified through its values on a basis. The definition still governs the computation: every recorded value is \(\varphi(T(\mathbf{v}))\), not a value of \(\varphi\) on the original input unless the transformation leaves that input unchanged.
Dual Maps Reverse Composition
Suppose \(T:V\to W\) and \(S:W\to X\) are linear maps over the same field. The combined map \(S\circ T\) sends \(V\) to \(X\). Its dual map acts on functionals in \(X^*\), and the result is in \(V^*\). To obtain the same effect in two stages, first pull an \(X\)-functional back along \(S\), then pull the resulting \(W\)-functional back along \(T\).
Proof. Let \(\chi\in X^*\) and \(\mathbf{v}\in V\). By the definition of the dual map, \(\bigl((S\circ T)^*(\chi)\bigr)(\mathbf{v})=\chi((S\circ T)(\mathbf{v}))=\chi(S(T(\mathbf{v})))\). On the other hand, \(S^*(\chi)=\chi\circ S\in W^*\), so \(\bigl((T^*\circ S^*)(\chi)\bigr)(\mathbf{v})=(T^*(S^*(\chi)))(\mathbf{v})=(S^*(\chi))(T(\mathbf{v}))=\chi(S(T(\mathbf{v})))\). The two functionals have the same value on every \(\mathbf{v}\in V\), so they are equal for every \(\chi\in X^*\). For the identity map and any \(\varphi\in V^*\), \((\operatorname{id}_V)^*(\varphi)=\varphi\circ\operatorname{id}_V=\varphi\). Thus the dual of the identity map is the identity map on \(V^*\). \(\square\)
The order reversal is a consequence of the types as well as the calculation. The composition \(S\circ T\) applies \(T\) first and \(S\) second, whereas its dual applies \(S^*\) first and \(T^*\) second. This property is sometimes called contravariance: arrows reverse direction when passing to dual maps.
What the Dual Map Can Forget
Proof. By definition, \(\bigl(T^*(\varphi)\bigr)(\mathbf{v})=\varphi(T(\mathbf{v}))\) for every \(\mathbf{v}\in V\). A functional is the zero functional precisely when its value is \(0\) at every vector in its domain. Therefore \(T^*(\varphi)=0\) if and only if \(\varphi(T(\mathbf{v}))=0\) for every \(\mathbf{v}\in V\). The image of \(T\) is the set of all vectors \(T(\mathbf{v})\) with \(\mathbf{v}\in V\), so this is exactly the condition that \(\varphi\) vanishes on every vector in the image. \(\square\)
Worked Example: A Nonzero Functional Pulls Back to Zero
Let \(U:\mathbb{R}^3\to\mathbb{R}^2\) be \(U(x,y,z)=(x-y,\,y+z)\), and define \(\eta(a,b)=a+b\). Compute \(U^*(\eta)\) and check whether it is zero.
For \((x,y,z)\in\mathbb{R}^3\), \(\bigl(U^*(\eta)\bigr)(x,y,z)=\eta(x-y,\,y+z)=(x-y)+(y+z)=x+z\). Thus \(U^*(\eta)\) is not the zero functional: for example, it takes the value \(1\) at \((1,0,0)\).
To see how a nonzero functional can instead pull back to zero, define \(\varphi(a,b)=a-b\). Then \(\bigl(U^*(\varphi)\bigr)(x,y,z)=\varphi(x-y,\,y+z)=(x-y)-(y+z)=x-2y-z\), which is also nonzero, so this choice does not give a zero pullback. Instead use \(\theta(a,b)=a-b\) with the map \(J(x,y,z)=(x+y,\,x+y)\). For every input, \(\bigl(J^*(\theta)\bigr)(x,y,z)=\theta(x+y,\,x+y)=(x+y)-(x+y)=0\). The functional \(\theta\) itself is nonzero, since \(\theta(1,0)=1\), but it vanishes on every output of \(J\), exactly as the proposition predicts.
Check: for any \((x,y,z)\), the output \(J(x,y,z)\) has equal coordinates, so applying \(\theta(a,b)=a-b\) gives zero. In particular, \(J^*(\theta)\) takes the value \(0\) at every input, as required for the zero functional.
Common Composition Pitfall
Composition errors are often errors of domain and codomain. In the example \(U:\mathbb{R}^3\to\mathbb{R}^2\), with \(\eta\in(\mathbb{R}^2)^*\), the composition \(\eta\circ U\) is defined and is exactly \(U^*(\eta)\): first apply \(U\), then \(\eta\). The reversed composition \(U\circ\eta\) is not defined, because \(\eta\) outputs a scalar in \(\mathbb{R}\), while \(U\) requires an input in \(\mathbb{R}^3\).
A concrete wrong calculation would write \(U\circ\eta\) and try to feed \(\eta(a,b)\), a single scalar, into \(U\) as though it were a triple \((x,y,z)\). That step has no meaning: \(U\) has no scalar input. The correction is to start with \((x,y,z)\in\mathbb{R}^3\), compute \(U(x,y,z)=(x-y,y+z)\), and then evaluate \(\eta\) on that pair. This produces the valid formula \(\bigl(U^*(\eta)\bigr)(x,y,z)=x+z\).
It is also important to keep the direction of the dual map visible. A functional \(\varphi\in W^*\) is an input to \(T^*\), and \(T^*(\varphi)\) belongs to \(V^*\). The map \(T^*\) does not apply \(T\) to a functional, nor does it send vectors in \(V\) to vectors in \(W\).
Why the Dual Map Is Useful
The dual map translates information about outputs into a functional on inputs. For example, a measurement made after applying \(T\) can be represented by \(\varphi\in W^*\); the corresponding measurement of an original input is \(T^*(\varphi)\). This is useful when transformations are composed, because the composition law lets one pull a functional backward one map at a time. The zero-pullback criterion also identifies functionals that cannot distinguish any outputs of a given transformation from zero.
For finite-dimensional spaces, choosing bases allows these functionals to be calculated by finite coordinate computations. The conceptual definition remains composition: apply the transformation to an arbitrary input and then evaluate the chosen output functional. Keeping this order explicit checks both that the computation is well-typed and that the resulting expression belongs to the correct dual space.
Check Your Understanding
For each question, keep track of the domain and codomain before simplifying any expression.
- If \(T:\mathbb{R}^2\to\mathbb{R}^3\) is \(T(x,y)=(x,y,x+y)\) and \(\varphi(a,b,c)=a+2b-c\), compute \(T^*(\varphi)\).
- Let \(S:P_2(\mathbb{R})\to\mathbb{R}\) be \(S(p)=p(1)\), and let \(c\mapsto 4c\) be a functional on \(\mathbb{R}\). Compute its pullback along \(S\).
- For maps \(T:V\to W\) and \(S:W\to X\), explain why \((S\circ T)^*=T^*\circ S^*\), including the space in which each intermediate functional lies.
- Give a nonzero functional on \(\mathbb{R}^2\) whose pullback along \(J(x,y)=(x,x)\) is zero, and verify the claim on an arbitrary input.
- For \(U:\mathbb{R}^3\to\mathbb{R}^2\), state whether \(\eta\circ U\) or \(U\circ\eta\) is the dual-map composition when \(\eta\in(\mathbb{R}^2)^*\), and justify using the input and output types.
