Functionals as Elements of a New Vector Space
A linear functional takes a vector as input and returns a scalar. The set of all such functionals is itself a vector space: we can add two functionals by adding their outputs, or multiply one by a scalar by scaling its outputs. The important question is how large this new space is. For a finite-dimensional vector space, its dimension matches the dimension of the original space. The proof constructs enough functionals from a basis of the original space and then shows that they account for every functional.
Let \(\mathbb{F}\) denote either \(\mathbb{R}\) or \(\mathbb{C}\), and let \(V\) be a vector space over \(\mathbb{F}\). A linear functional on \(V\) is a linear map from \(V\) to \(\mathbb{F}\). Thus a function \(f:V\to\mathbb{F}\) is a linear functional when, for all \(\mathbf{u},\mathbf{v}\in V\) and \(c\in\mathbb{F}\), it satisfies \(f(\mathbf{u}+\mathbf{v})=f(\mathbf{u})+f(\mathbf{v})\) and \(f(c\mathbf{u})=cf(\mathbf{u})\). In particular, \(f(\mathbf{0})=0\), since \(f(\mathbf{0})=f(0\mathbf{v})=0f(\mathbf{v})=0\).
The elements of \(V^*\) are functions, not vectors in \(V\). The operations above are defined pointwise: to add two functionals, evaluate both at the same input and add the resulting scalars. The scalar field remains \(\mathbb{F}\).
Proof. First, the operations stay inside \(V^*\). For \(f,g\in V^*\), \((f+g)(\mathbf{u}+\mathbf{v})=f(\mathbf{u})+f(\mathbf{v})+g(\mathbf{u})+g(\mathbf{v})=(f+g)(\mathbf{u})+(f+g)(\mathbf{v})\), and \((f+g)(c\mathbf{u})=cf(\mathbf{u})+cg(\mathbf{u})=c(f+g)(\mathbf{u})\). For \(a\in\mathbb{F}\), \((af)(\mathbf{u}+\mathbf{v})=a f(\mathbf{u})+a f(\mathbf{v})\) and \((af)(c\mathbf{u})=ac f(\mathbf{u})=c(af)(\mathbf{u})\), so \(f+g\) and \(af\) are also linear functionals. The zero function \(0_{V^*}\), given by \(0_{V^*}(\mathbf{v})=0\) for every \(\mathbf{v}\), is a linear functional and is the additive identity. The additive inverse of \(f\) is the functional \((-f)(\mathbf{v})=-f(\mathbf{v})\). Finally, each remaining vector-space axiom follows by evaluating both sides at an arbitrary \(\mathbf{v}\in V\): for example, \(((f+g)+h)(\mathbf{v})=(f(\mathbf{v})+g(\mathbf{v}))+h(\mathbf{v})=(f(\mathbf{v})+(g(\mathbf{v})+h(\mathbf{v})))=(f+(g+h))(\mathbf{v})\), and \((a(f+g))(\mathbf{v})=a(f(\mathbf{v})+g(\mathbf{v}))=(af+ag)(\mathbf{v})\). The zero and identity scalar laws reduce in the same way to the corresponding scalar laws in \(\mathbb{F}\). Therefore the stated operations satisfy all vector-space axioms. \(\square\)
Examples of Functionals and Their Operations
Worked Example: Adding and Scaling Functionals on \(\mathbb{R}^2\)
Define \(f,g:\mathbb{R}^2\to\mathbb{R}\) by \(f(x,y)=2x-y\) and \(g(x,y)=x+3y\). For real \(a,b\), the pointwise combination \(af+bg\) is the function
This is a linear functional because it is a linear expression in \(x\) and \(y\). For example, taking \(a=2\) and \(b=-1\) gives \((2f-g)(x,y)=3x-5y\). At \((x,y)=(1,-2)\), the original functionals give \(f(1,-2)=2+2=4\) and \(g(1,-2)=1-6=-5\), so \((2f-g)(1,-2)=2(4)-(-5)=13\). The formula gives \(3(1)-5(-2)=13\) as well, checking the pointwise calculation.
Constructing a functional from a basis requires knowing that every vector has a unique expansion in that basis. If \(\mathcal{B}=(\mathbf{b}_1,\dots,\mathbf{b}_n)\) is a basis of \(V\), its vectors span \(V\), so each \(\mathbf{v}\in V\) has at least one expression \(\mathbf{v}=\sum_{i=1}^n c_i\mathbf{b}_i\). It has at most one: if \(\sum_{i=1}^n c_i\mathbf{b}_i=\sum_{i=1}^n d_i\mathbf{b}_i\), then \(\sum_{i=1}^n(c_i-d_i)\mathbf{b}_i=\mathbf{0}\), and independence of the basis forces \(c_i-d_i=0\) for every \(i\).
Worked Example: Defining a Functional from a Nonstandard Basis
Let \(\mathbf{b}_1=(1,1)\) and \(\mathbf{b}_2=(1,-1)\) in \(\mathbb{R}^2\). For \((x,y)=a\mathbf{b}_1+b\mathbf{b}_2\), the coordinate equations are \(x=a+b\) and \(y=a-b\). Adding and subtracting them gives \(a=(x+y)/2\) and \(b=(x-y)/2\), so this expression exists uniquely for every \((x,y)\).
Define \(f\) by assigning \(f(\mathbf{b}_1)=2\) and \(f(\mathbf{b}_2)=-1\), then extending linearly to vectors expressed in this basis. For \((x,y)=a\mathbf{b}_1+b\mathbf{b}_2\), this means \(f(x,y)=2a-b\). Substituting the formulas for \(a\) and \(b\),
The formula is linear, and its values on the basis are indeed \(f(1,1)=(1+3)/2=2\) and \(f(1,-1)=(1-3)/2=-1\). Thus it has exactly the prescribed values.
Functionals Constructed from a Basis
Now suppose \(V\) is finite-dimensional over \(\mathbb{F}\), with basis \(\mathcal{B}=(\mathbf{b}_1,\dots,\mathbf{b}_n)\). For each index \(i\), define \(f_i:V\to\mathbb{F}\) by taking the coefficient of \(\mathbf{b}_i\) in the unique basis expansion. Explicitly, if \(\mathbf{v}=\sum_{j=1}^n c_j\mathbf{b}_j\), set \(f_i(\mathbf{v})=c_i\). Uniqueness makes this definition unambiguous. The functions \(f_1,\dots,f_n\) are the tools for comparing the size of \(V^*\) with that of \(V\).
To check that \(f_i\) is linear, let \(\mathbf{u}=\sum_{j=1}^n a_j\mathbf{b}_j\) and \(\mathbf{v}=\sum_{j=1}^n c_j\mathbf{b}_j\). Then \(\mathbf{u}+\mathbf{v}=\sum_{j=1}^n(a_j+c_j)\mathbf{b}_j\), so \(f_i(\mathbf{u}+\mathbf{v})=a_i+c_i=f_i(\mathbf{u})+f_i(\mathbf{v})\). For \(a\in\mathbb{F}\), \(a\mathbf{u}=\sum_{j=1}^n(aa_j)\mathbf{b}_j\), so \(f_i(a\mathbf{u})=aa_i=af_i(\mathbf{u})\). Hence every \(f_i\) belongs to \(V^*\).
Proof. Let \(\mathcal{B}=(\mathbf{b}_1,\dots,\mathbf{b}_n)\) be a basis of \(V\), and construct \(f_1,\dots,f_n\) as above. First they are linearly independent. Suppose \(\sum_{i=1}^n a_i f_i=0_{V^*}\). Evaluate both sides at \(\mathbf{b}_j\). The coefficient of \(\mathbf{b}_i\) in the basis expansion of \(\mathbf{b}_j\) is \(1\) when \(i=j\) and \(0\) otherwise, so \(f_i(\mathbf{b}_j)\) is \(1\) if \(i=j\) and \(0\) if \(i\ne j\). It follows that \(a_j=\sum_{i=1}^n a_i f_i(\mathbf{b}_j)=0\). This holds for every \(j\), proving independence.
Next they span \(V^*\). Take any \(f\in V^*\) and any \(\mathbf{v}=\sum_{i=1}^n c_i\mathbf{b}_i\). Linearity gives \(f(\mathbf{v})=\sum_{i=1}^n c_i f(\mathbf{b}_i)\), while the definition of \(f_i\) gives \(f_i(\mathbf{v})=c_i\). Therefore \(\left(\sum_{i=1}^n f(\mathbf{b}_i)f_i\right)(\mathbf{v})=\sum_{i=1}^n f(\mathbf{b}_i)c_i=f(\mathbf{v})\). Since this equality holds for every \(\mathbf{v}\in V\), the two functionals are equal, so \(f=\sum_{i=1}^n f(\mathbf{b}_i)f_i\). Thus the \(n\) independent functionals span \(V^*\), and form a basis of it. Consequently \(\dim V^*=n=\dim V\).
If \(n=0\), then \(V=\{\mathbf{0}\}\). Every linear functional \(f:V\to\mathbb{F}\) must satisfy \(f(\mathbf{0})=0\), so \(V^*\) consists only of the zero functional and has dimension \(0\). This also gives \(\dim V^*=\dim V\) in the zero-dimensional case. \(\square\)
Worked Example: Three Functionals from a Basis of \(\mathbb{R}^3\)
Consider \(\mathbf{b}_1=(1,0,1)\), \(\mathbf{b}_2=(0,1,1)\), and \(\mathbf{b}_3=(1,1,0)\). For \((r,s,t)=a\mathbf{b}_1+b\mathbf{b}_2+c\mathbf{b}_3\), the coordinate equations are \(a+c=r\), \(b+c=s\), and \(a+b=t\). Solving gives
These formulas give an expansion for every \((r,s,t)\), and the equations determine \(a,b,c\) uniquely, so the three vectors are a basis. Define \(f\) to take the values \(1,0,-2\) on \(\mathbf{b}_1,\mathbf{b}_2,\mathbf{b}_3\), respectively. Its value on \((r,s,t)\) is \(a-2c\), hence
Check the assigned values using the formula: \(f(1,0,1)=(-1+0+3)/2=1\), \(f(0,1,1)=(0-3+3)/2=0\), and \(f(1,1,0)=(-1-3+0)/2=-2\). This is one functional in a three-dimensional dual space; the theorem guarantees that all functionals on \(\mathbb{R}^3\) are accounted for by three independent functionals constructed from any chosen basis.
Why the Dimension Equality Matters
The equality \(\dim V^*=\dim V\) tells us that passing from vectors to scalar-valued linear measurements does not change the finite dimension. The proof also gives a practical construction: choose a basis of \(V\), prescribe a value for a functional on each basis vector, and extend to all vectors by linearity. The resulting functional is unique, because every vector has a unique basis expansion. This fact is useful whenever a problem describes a linear functional by its values on a basis rather than by a formula on all of \(V\).
The finite-dimensional hypothesis matters. The proof uses a finite basis \((\mathbf{b}_1,\dots,\mathbf{b}_n)\) and constructs exactly \(n\) functionals from it. The dimension theorem stated here makes no claim about infinite-dimensional spaces.
Common Pitfall: Conjugation Is Not Linearity
Over \(\mathbb{C}\), a functional must be linear over complex scalars, not merely additive. A common wrong choice is \(f:\mathbb{C}\to\mathbb{C}\) given by \(f(z)=\overline{z}\). It is additive, but it is not complex-linear: \(f(i\cdot1)=\overline{i}=-i\), whereas \(i f(1)=i\cdot1=i\), and \(-i\ne i\). Thus this map is not an element of the complex dual space \(\mathbb{C}^*\). By contrast, \(g(z)=(2+i)z\) is complex-linear: for \(z,w,c\in\mathbb{C}\), \(g(z+w)=(2+i)z+(2+i)w=g(z)+g(w)\) and \(g(cz)=c(2+i)z=cg(z)\).
Keep track of the scalar field when forming a dual space. A map can be linear over one field and fail to be linear over another; the definition of \(V^*\) always uses the same field as the vector space \(V\).
Check Your Understanding
Use the definitions and basis construction above to answer these questions.
- For \(f(x,y)=x-2y\) and \(g(x,y)=3x+y\) on \(\mathbb{R}^2\), compute \((f+2g)(x,y)\).
- Let \(\mathcal{B}=(\mathbf{b}_1,\mathbf{b}_2)\) be a basis and define \(f(\mathbf{b}_1)=3\), \(f(\mathbf{b}_2)=1\). Find \(f(2\mathbf{b}_1-4\mathbf{b}_2)\).
- In the dimension proof, why does evaluating \(\sum_i a_i f_i=0_{V^*}\) at \(\mathbf{b}_j\) show that \(a_j=0\)?
- Explain why a functional \(f:V\to\mathbb{F}\) is determined by its values on a basis of \(V\).
- Does \(z\mapsto\overline{z}\) belong to the complex dual of \(\mathbb{C}\)? Verify your answer using the scalar \(i\).
