What a Linear Functional Does
A linear functional takes vectors as inputs and real numbers as outputs while preserving linear combinations. Although its output is only one number, the equation that says this number is zero can describe a substantial subspace of the input space. In this tutorial, \(V\) denotes a vector space over \(\mathbb{R}\); when a result requires finite dimension, that condition is stated. The main examples use \(\mathbb{R}^n\) and the space of real polynomials of degree at most two.
The two linearity conditions express preservation of addition and scalar multiplication. They imply \(f(\mathbf{0})=0\): taking \(c=0\) gives \(f(\mathbf{0})=0f(\mathbf{v})=0\). Thus a proposed functional with a nonzero value at the zero vector cannot be linear. This is a quick test, but passing it alone does not prove linearity; the defining conditions must still hold for every input.
In a finite-dimensional real vector space, a basis \(\mathcal{B}=(\mathbf{b}_1,\ldots,\mathbf{b}_n)\) gives each vector a unique expression \(\mathbf{v}=c_1\mathbf{b}_1+\cdots+c_n\mathbf{b}_n\). A functional’s values on those basis vectors determine its value everywhere.
Proof. The vector \(\mathbf{v}\) is a finite sum of scalar multiples of basis vectors. Repeatedly applying preservation of addition gives \(f(\mathbf{v})=f(c_1\mathbf{b}_1)+\cdots+f(c_n\mathbf{b}_n)\); applying preservation of scalar multiplication to each term gives the stated formula. If two linear functionals agree on every \(\mathbf{b}_i\), the formula gives the same value for both on every vector \(\mathbf{v}\), because every vector has such an expansion. Therefore the values on the basis determine the functional. \(\square\)
Testing Linearity and Finding a Kernel
For a formula in coordinates, the addition and scalar conditions can often be checked by direct substitution. For example, if \(f(x,y)=ax+by\) for fixed real numbers \(a,b\), then
and \(f(c(x,y))=a(cx)+b(cy)=cf(x,y)\). The coefficients \(a,b\) are fixed; they are not additional inputs to the function. Once linearity is established, finding the kernel means solving the equation \(f(\mathbf{v})=0\).
Worked Example: A Functional on R³
Define \(f:\mathbb{R}^3\to\mathbb{R}\) by \(f(x,y,z)=2x-y+3z\). Verify linearity and describe its kernel.
For \(\mathbf{u}=(x_1,y_1,z_1)\), \(\mathbf{v}=(x_2,y_2,z_2)\), and \(c\in\mathbb{R}\), substitution gives
So \(f\) is linear. A vector belongs to its kernel exactly when \(2x-y+3z=0\), or \(y=2x+3z\). Let \(x=s\) and \(z=t\), where \(s,t\in\mathbb{R}\). Then every vector in the kernel has the form
Conversely, every vector of this form satisfies the kernel equation. Indeed, \(f(s,2s+3t,t)=2s-(2s+3t)+3t=0\). The two displayed vectors also each have functional value zero: \(2-2+0=0\) and \(0-3+3=0\). This checks the description directly.
This example shows how a kernel calculation translates a single linear equation into a parametrization. The parameters must be allowed to vary over all real numbers, and the resulting expression must describe every solution, not merely provide some vectors that happen to satisfy the equation.
Worked Example: A Formula That Is Not Linear
Consider \(q:\mathbb{R}^2\to\mathbb{R}\), \(q(x,y)=x^2+y\). Test whether \(q\) is linear.
Scalar multiplication already gives a counterexample. Take \(\mathbf{u}=(1,0)\) and \(c=2\). Then
Since \(4\ne2\), the required identity \(q(c\mathbf{u})=cq(\mathbf{u})\) fails. Therefore \(q\) is not a linear functional. The fact that this formula has one real output for each vector does not make it linear; the preservation conditions are essential.
Coordinate Formulas in Different Spaces
The coordinate formula theorem applies to any basis, not just the standard basis of \(\mathbb{R}^n\). In standard coordinates, a linear functional on \(\mathbb{R}^n\) has the form \(f(x_1,\ldots,x_n)=a_1x_1+\cdots+a_nx_n\), where \(a_i=f(\mathbf{e}_i)\). On a polynomial space, the coefficients of the polynomial play the role of coordinates, so a functional can be calculated from those coefficients as well.
Worked Example: A Functional in a Nonstandard Basis
Let \(\mathcal{B}=((1,1),(1,-1))\) in \(\mathbb{R}^2\), and define \(f(x,y)=3x+y\). Find the coordinate formula for \(f\) using \(\mathcal{B}\).
First, for \((x,y)=s(1,1)+t(1,-1)\), its coordinates satisfy \(x=s+t\) and \(y=s-t\). Adding these equations gives \(s=(x+y)/2\); subtracting the second from the first gives \(t=(x-y)/2\). Thus every \((x,y)\) has the unique such representation, so the two vectors form a basis. The values of \(f\) on its basis vectors are
By the coordinate formula theorem, if \([(x,y)]_{\mathcal{B}}=(s,t)\), then \(f(x,y)=4s+2t\). Substituting the expressions for \(s,t\) gives
For a direct check, \((2,0)=1(1,1)+1(1,-1)\), so its basis-coordinate formula gives \(4(1)+2(1)=6\). The original formula gives \(f(2,0)=3(2)+0=6\), as required.
Worked Example: A Functional on Polynomials
Let \(P_2(\mathbb{R})\) be the real vector space of polynomials of degree at most two, including the zero polynomial. Define \(L:P_2(\mathbb{R})\to\mathbb{R}\) by \(L(p)=p'(1)-p(0)\). Find its coordinate formula and kernel.
Write an arbitrary polynomial as \(p(x)=a+bx+cx^2\), with \(a,b,c\in\mathbb{R}\). Its derivative is \(p'(x)=b+2cx\), so \(p'(1)=b+2c\), while \(p(0)=a\). Therefore
This formula also verifies linearity: if \(q(x)=d+ex+gx^2\) and \(r\in\mathbb{R}\), then the coefficients of \(p+q\) are \(a+d,b+e,c+g\), and those of \(rp\) are \(ra,rb,rc\). Consequently, \(L(p+q)=-(a+d)+(b+e)+2(c+g)=L(p)+L(q)\), and \(L(rp)=-ra+rb+2rc=rL(p)\).
The kernel equation is \(-a+b+2c=0\), so \(a=b+2c\). Hence each polynomial in the kernel is
To check, \(L(1+x)=1-1=0\) and \(L(2+x^2)=2-2=0\). By linearity, every displayed combination is in the kernel. Conversely, the kernel equation forced \(a=b+2c\), so every polynomial in the kernel has exactly the displayed form. For example, taking \(b=1,c=-1\) gives \(p(x)=-1+x-x^2\). Its derivative at \(1\) is \(1-2=-1\), its value at \(0\) is \(-1\), and \(L(p)=-1-(-1)=0\).
How the Kernel Relates to Other Values
The kernel is the set of inputs mapped to zero. More generally, the inputs that a functional maps to a fixed real number form either an empty set or a translate of its kernel. A translate of a set \(S\) by \(\mathbf{v}_0\) means the set \(\mathbf{v}_0+S=\{\mathbf{v}_0+\mathbf{w}:\mathbf{w}\in S\}\).
Proof. Assume first that \(f(\mathbf{v}_0)=r\). If \(f(\mathbf{v})=r\), linearity gives \(f(\mathbf{v}-\mathbf{v}_0)=f(\mathbf{v})-f(\mathbf{v}_0)=r-r=0\). Thus \(\mathbf{v}-\mathbf{v}_0\in\ker f\), so \(\mathbf{v}\in\mathbf{v}_0+\ker f\). Conversely, if \(\mathbf{v}=\mathbf{v}_0+\mathbf{w}\) with \(\mathbf{w}\in\ker f\), then \(f(\mathbf{v})=f(\mathbf{v}_0)+f(\mathbf{w})=r+0=r\). This proves the equality. Now suppose \(f\) is not the zero functional. There is a vector \(\mathbf{u}\) with \(f(\mathbf{u})\ne0\). For any \(r\in\mathbb{R}\), set \(\mathbf{v}_0=(r/f(\mathbf{u}))\mathbf{u}\). Then \(f(\mathbf{v}_0)=(r/f(\mathbf{u}))f(\mathbf{u})=r\), so the set is nonempty. \(\square\)
In particular, the theorem’s first statement at \(r=0\) is consistent with the definition: the zero level set is \(\ker f\) itself. For a nonzero functional, every other level set is a translate of that same kernel, so the kernel describes how the different level sets are related.
Common Pitfall: A Formula Must Preserve Scaling
A frequent mistake is to treat any simple-looking expression as linear without checking how it behaves under scalar multiplication. For \(q(x,y)=x^2+y\), a tempting but incorrect calculation is \(q(2,0)=2q(1,0)\). The left side is \(4\), while the right side is \(2\), so the claimed equality is false. The correction is to test the defining condition using the actual input vector: with \(\mathbf{u}=(1,0)\), \(q(2\mathbf{u})=4\ne2=2q(\mathbf{u})\). This single counterexample is enough to disprove linearity.
For kernel calculations, a different error is to stop after finding some vectors with functional value zero. A complete description must also show that every vector in the kernel has been included. In the polynomial example, substitution verifies that \(b(1+x)+c(2+x^2)\) always lies in the kernel; solving \(-a+b+2c=0\) and obtaining \(a=b+2c\) proves the converse.
Check Your Understanding
For each question, use the defining conditions or substitute into the claimed kernel equation to check your result.
- For \(f(x,y)=5x-2y\), verify linearity and find \(\ker f\).
- Does \(g(x,y)=x+3y+1\) define a linear functional? Give a specific failed condition or explain why it passes the definition.
- Let \(L(p)=p(1)\) on \(P_2(\mathbb{R})\). Write \(L\) in terms of the coefficients of \(p(x)=a+bx+cx^2\), and describe its kernel.
- Suppose \(f:V\to\mathbb{R}\) is linear and \(f(\mathbf{v}_0)=4\). Describe all \(\mathbf{v}\) such that \(f(\mathbf{v})=4\).
- Explain why a functional’s values on a basis determine its values on every vector in a finite-dimensional vector space.
