Functionals as Coordinates on a Space
A linear functional turns a vector into a scalar. On polynomial and matrix spaces, many useful functionals have familiar descriptions: evaluate a polynomial at a point, extract one of its coefficients, take the trace of a matrix, or select one matrix entry. The dual-basis viewpoint gives a systematic way to express each of these rules, even when the chosen basis is not the most familiar one.
Throughout, the scalar field is \(\mathbb{R}\). A linear functional on a vector space \(V\) is a linear map \(f:V\to\mathbb{R}\); that is, \(f(a\mathbf{v}+b\mathbf{w})=af(\mathbf{v})+bf(\mathbf{w})\) for all \(\mathbf{v},\mathbf{w}\in V\) and \(a,b\in\mathbb{R}\). If \(V\) is finite-dimensional with ordered basis \(\mathcal{B}=(\mathbf{b}_1,\dots,\mathbf{b}_n)\), every \(\mathbf{v}\in V\) has a unique expression \(\mathbf{v}=a_1\mathbf{b}_1+\cdots+a_n\mathbf{b}_n\).
The values of a functional on the basis determine its expression in the dual basis. This gives a practical method: evaluate the functional on each basis vector, then use those values as coefficients.
Proof. Let \(\mathbf{v}=a_1\mathbf{b}_1+\cdots+a_n\mathbf{b}_n\). By linearity, \(f(\mathbf{v})=\sum_{i=1}^n a_i f(\mathbf{b}_i)\). The proposed functional gives the same value, since \(b_i^*(\mathbf{v})=a_i\), and therefore \(\sum_{i=1}^n f(\mathbf{b}_i)b_i^*(\mathbf{v})=\sum_{i=1}^n f(\mathbf{b}_i)a_i=f(\mathbf{v})\). The two functionals agree on every \(\mathbf{v}\), so they are equal. If \(f=\sum_{i=1}^n c_i b_i^*\), applying both sides to \(\mathbf{b}_j\) gives \(f(\mathbf{b}_j)=c_j\), because \(b_i^*(\mathbf{b}_j)\) is \(1\) for \(i=j\) and \(0\) otherwise. Thus each coefficient is forced, which proves uniqueness. \(\square\)
Evaluation and Coefficients on Polynomial Spaces
Let \(P_2(\mathbb{R})\) denote the space of real polynomials of degree at most two, including the zero polynomial. For a real number \(r\), evaluation at \(r\) is the rule \(p\mapsto p(r)\). The coefficient-extraction functional \([x]p\) returns the coefficient of \(x\) in \(p\), when \(p\) is written in powers of \(x\). Both rules are linear: for evaluation, \((ap+bq)(r)=ap(r)+bq(r)\); for coefficient extraction, the coefficient of \(x\) in \(ap+bq\) is \(a[x]p+b[x]q\).
In the standard basis \((1,x,x^2)\), evaluation at \(r\) has dual-basis expression \(f_r=1\cdot b_1^*+r\cdot b_2^*+r^2\cdot b_3^*\), because the basis polynomials evaluate to \(1,r,r^2\). The coefficient functional \([x]\) is simply the second dual-basis element for this basis. A different basis changes these expressions, but not the functional itself.
Worked Example: Evaluation in a Nonstandard Polynomial Basis
In \(P_2(\mathbb{R})\), choose \(\mathcal{B}=(b_1,b_2,b_3)=(1+x,x,x^2)\). This is a basis: if \(a(1+x)+bx+cx^2=0\), then comparing the constant, \(x\), and \(x^2\) coefficients gives \(a=0\), \(a+b=0\), and \(c=0\), hence \(a=b=c=0\). Write \(b_1^*,b_2^*,b_3^*\) for its dual basis. Find the dual-basis expression for \(f(p)=p(2)\).
Evaluate \(f\) on each basis polynomial: \(f(b_1)=(1+x)|_{x=2}=3\), \(f(b_2)=x|_{x=2}=2\), and \(f(b_3)=x^2|_{x=2}=4\). The theorem gives
To check the formula, take \(p=2-x+3x^2\). Its coordinates in \(\mathcal{B}\) satisfy \(p=a(1+x)+bx+cx^2=a+(a+b)x+cx^2\). Matching coefficients gives \(a=2\), \(a+b=-1\), and \(c=3\), so \(b=-3\). Thus the dual-basis expression returns \(3(2)+2(-3)+4(3)=12\). Direct evaluation gives \(p(2)=2-2+12=12\), as required.
Worked Example: Extracting a Coefficient in the Same Basis
For \(\mathcal{B}=(1+x,x,x^2)\), express the functional \(g(p)=[x]p\) in the dual basis. Its values on the basis polynomials are \([x](1+x)=1\), \([x]x=1\), and \([x]x^2=0\). Therefore
Check this on \(p=2-x+3x^2\). The coordinates already found are \((2,-3,3)\), so \(b_1^*(p)+b_2^*(p)=2-3=-1\). In the standard power expansion the coefficient of \(x\) is \(-1\), confirming the formula. The two terms appear because the coefficient of \(x\) in \(a(1+x)+bx+cx^2\) is \(a+b\), not just \(b\).
Trace and Entry Selection on Matrix Spaces
Let \(M_{2\times2}(\mathbb{R})\) be the vector space of real \(2\times2\) matrices. For \(i,j\in\{1,2\}\), let \(\mathbf{E}_{ij}\) be the matrix with a \(1\) in row \(i\), column \(j\), and zeros elsewhere. The four matrices \(\mathbf{E}_{11},\mathbf{E}_{12},\mathbf{E}_{21},\mathbf{E}_{22}\) form the standard basis: each matrix \(\mathbf{A}=\begin{bmatrix}a&b\\c&d\end{bmatrix}\) equals \(a\mathbf{E}_{11}+b\mathbf{E}_{12}+c\mathbf{E}_{21}+d\mathbf{E}_{22}\).
The trace functional is \(\operatorname{tr}:M_{2\times2}(\mathbb{R})\to\mathbb{R}\), defined by \(\operatorname{tr}\mathbf{A}=a_{11}+a_{22}\). The entry-selection functional \(s_{ij}\) is defined by \(s_{ij}(\mathbf{A})=a_{ij}\). Both are linear: the diagonal entries of a sum or scalar multiple add or scale entrywise, and the same is true of any fixed entry. Thus \(\operatorname{tr}(a\mathbf{A}+b\mathbf{B})=a\operatorname{tr}\mathbf{A}+b\operatorname{tr}\mathbf{B}\), and \(s_{ij}(a\mathbf{A}+b\mathbf{B})=a s_{ij}(\mathbf{A})+b s_{ij}(\mathbf{B})\).
Worked Example: Trace in a Nonstandard Matrix Basis
Use the ordered list \(\mathcal{B}=(\mathbf{B}_1,\mathbf{B}_2,\mathbf{B}_3,\mathbf{B}_4)\), where \(\mathbf{B}_1=\mathbf{E}_{11}+\mathbf{E}_{12}\), \(\mathbf{B}_2=\mathbf{E}_{12}\), \(\mathbf{B}_3=\mathbf{E}_{21}\), and \(\mathbf{B}_4=\mathbf{E}_{22}\). A combination with coordinates \(a,b,c,d\) is
This list is a basis: any \(\begin{bmatrix}u&v\\w&z\end{bmatrix}\) is obtained uniquely by choosing \(a=u\), \(b=v-u\), \(c=w\), and \(d=z\). The trace values on the four basis matrices are \(1,0,0,1\), respectively. If \(\mathcal{B}^*=(B_1^*,B_2^*,B_3^*,B_4^*)\) is the dual basis, it follows that
For \(\mathbf{A}=\begin{bmatrix}5&7\\2&-1\end{bmatrix}\), its coordinates are \(a=5\), \(b=2\), \(c=2\), and \(d=-1\). The dual-basis expression gives \(B_1^*(\mathbf{A})+B_4^*(\mathbf{A})=5-1=4\). Directly, \(\operatorname{tr}\mathbf{A}=5+(-1)=4\), so the results agree.
Worked Example: Selecting an Entry in a Nonstandard Matrix Basis
For the same basis, find the expression for the functional \(s_{12}(\mathbf{A})=a_{12}\). The \((1,2)\) entries of \(\mathbf{B}_1,\mathbf{B}_2,\mathbf{B}_3,\mathbf{B}_4\) are \(1,1,0,0\), respectively. Hence
For \(\mathbf{A}=\begin{bmatrix}5&7\\2&-1\end{bmatrix}\), the coordinates are again \((5,2,2,-1)\), so the expression gives \(5+2=7\). The selected matrix entry is \(a_{12}=7\), which checks the result. In general, the \((1,2)\) entry of \(\begin{bmatrix}a&a+b\\c&d\end{bmatrix}\) is \(a+b\), explaining why both dual-basis terms are needed.
Common Pitfalls and a Reliable Procedure
A common error is to confuse coefficient extraction with evaluation. For \(p(x)=1+2x+x^2\), the coefficient of \(x\) is \([x]p=2\), but \(p(1)=1+2+1=4\). These are different functionals, and their values on a basis must be computed according to their definitions. Another frequent error is to treat trace as the sum of all entries. For \(\mathbf{A}=\begin{bmatrix}1&2\\3&4\end{bmatrix}\), the sum of all entries is \(10\), whereas \(\operatorname{tr}\mathbf{A}=1+4=5\). Trace uses only the main diagonal.
For any finite-dimensional space and any chosen basis, the method is the same:
Write down the ordered basis and use its dual basis; do not assume the standard coordinates remain valid for a different basis.
Compute \(f(\mathbf{b}_i)\) directly from the definition of \(f\).
Write \(f=\sum_i f(\mathbf{b}_i)b_i^*\), then test the expression on a general element or a concrete example.
This method matters because a functional can have a simple definition but a basis-dependent coordinate expression. Polynomial evaluation remains evaluation at the same number, and trace remains the sum of diagonal entries, regardless of basis. What changes is how their outputs depend on the coordinates associated with the chosen basis. In particular, the dual-basis formula provides an explicit way to recover the functional from its basis values, without needing to guess a coordinate rule.
Check Your Understanding
For each question, identify the chosen basis before computing the functional’s values on its basis elements.
- In \(P_2(\mathbb{R})\) with basis \((1,x,x^2)\), write evaluation at \(x=3\) as a combination of dual-basis elements.
- For the basis \((1+x,x,x^2)\), express the functional \([x^2]p\) in the dual basis.
- For the matrix basis \((\mathbf{E}_{11},\mathbf{E}_{12},\mathbf{E}_{21},\mathbf{E}_{22})\), express \(s_{21}(\mathbf{A})=a_{21}\) in the dual basis.
- Verify that the trace functional on \(M_{2\times2}(\mathbb{R})\) is linear by expanding \(\operatorname{tr}(a\mathbf{A}+b\mathbf{C})\) entrywise.
- For the nonstandard matrix basis in this tutorial, determine the dual-basis expression for \(s_{22}(\mathbf{A})=a_{22}\).
