Coordinates in Polynomial Bases

Polynomial Coordinates Record a Choice of Basis

A polynomial can be written in more than one useful collection of building blocks. In the usual powers of \(x\), the polynomial \(x^2\) has coefficient list \((0,0,1)\). In powers of \(x-1\), the same polynomial uses three terms: \(x^2=1+2(x-1)+(x-1)^2\). These coefficient lists differ because coordinates describe coefficients relative to a particular ordered basis, not coefficients that belong to the polynomial independently of any basis.

Throughout, scalars are real. Let \(P_2(\mathbb{R})\) denote the vector space of real polynomials of degree at most two, including the zero polynomial. Every element has the form \(a+bx+cx^2\), where \(a,b,c\in\mathbb{R}\). Two such polynomials are equal exactly when their coefficients of \(1,x,x^2\) agree. An ordered basis of this space is an ordered list of polynomials that is linearly independent and spans \(P_2(\mathbb{R})\).

The standard ordered basis is \(\mathcal{B}_0=(1,x,x^2)\). Another basis used throughout this tutorial is the shifted basis \(\mathcal{B}_1=(1,x-1,(x-1)^2)\). For either basis, coordinates tell us precisely how to combine its listed polynomials to obtain the desired polynomial.

Existence and Uniqueness of Polynomial Coordinates

Proof. Because \(\mathcal{B}\) spans \(P_2(\mathbb{R})\), there are scalars \(c_0,c_1,c_2\) such that \(p=c_0q_0+c_1q_1+c_2q_2\). Thus at least one coordinate vector exists. To show uniqueness, suppose two representations give the same polynomial: \(p=c_0q_0+c_1q_1+c_2q_2=d_0q_0+d_1q_1+d_2q_2\). Subtracting the second expression from the first gives \((c_0-d_0)q_0+(c_1-d_1)q_1+(c_2-d_2)q_2=0\). Since the basis is linearly independent, the only coefficients that produce the zero polynomial are all zero. Therefore \(c_i-d_i=0\) for each \(i\), so \(c_i=d_i\) for each \(i\), and the coordinate vector is unique. \(\square\)

This result justifies two complementary tasks. To find coordinates, express a polynomial as a combination of the basis elements. To reconstruct a polynomial, use a given coordinate vector as the coefficients in that combination. In either task, expanding both sides into powers of \(x\) gives a reliable way to compare coefficients and verify the answer.

Coordinates in the Standard Polynomial Basis

For \(\mathcal{B}_0=(1,x,x^2)\), if \(p=a+bx+cx^2\), then its coordinate vector is \((a,b,c)^{\top}\). This follows directly from the representation \(p=a\cdot1+b\cdot x+c\cdot x^2\). The statement is specific to this basis: the same list of numbers used with another basis generally produces another polynomial.

Finding Coordinates in a Shifted Basis

In the shifted basis \(\mathcal{B}_1=(1,x-1,(x-1)^2)\), the coefficient of a basis element is not generally the coefficient of the corresponding power of \(x\). Suppose the unknown coordinate vector is \((c_0,c_1,c_2)^{\top}\). Expand the basis representation using \((x-1)^2=x^2-2x+1\):

Matching the constant, \(x\), and \(x^2\) coefficients with a given polynomial \(a+bx+cx^2\) produces three equations. The \(x^2\) coefficient determines the last coordinate first; the \(x\) coefficient then determines the middle coordinate; the constant coefficient determines the first. This order makes the calculation direct.

Proof. Expanding \(c_0+c_1(x-1)+c_2(x-1)^2\) gives constant coefficient \(c_0-c_1+c_2\), coefficient of \(x\) equal to \(c_1-2c_2\), and coefficient of \(x^2\) equal to \(c_2\). For this expression to equal \(a+bx+cx^2\), equality of polynomials requires \(c_2=c\), \(c_1-2c_2=b\), and \(c_0-c_1+c_2=a\). The first equation gives \(c_2=c\). Substituting this into the second gives \(c_1=b+2c\). Substituting both values into the third gives \(c_0-(b+2c)+c=a\), hence \(c_0=a+b+c\). These values produce the required polynomial representation, and the theorem on existence and uniqueness of coordinates ensures that this representation gives the coordinate vector. \(\square\)

Reconstructing from Shifted Coordinates

Reconstruction uses the same defining equation as finding coordinates, but starts with the coefficient list. Given \((c_0,c_1,c_2)^{\top}\) in \(\mathcal{B}_1\), form \(c_0+c_1(x-1)+c_2(x-1)^2\) and expand if the answer is wanted in powers of \(x\). Keeping the basis visible until the last step helps prevent the coordinates from being mistaken for ordinary polynomial coefficients.

Common Mistake: Reusing Coefficients across Bases

A common error is to read the coefficients of \(1,x,x^2\) as though they were automatically the coordinates in \(\mathcal{B}_1\). For example, the standard-basis coefficients of \(x^2\) are \((0,0,1)^{\top}\). Incorrectly copying them into the shifted basis would claim \(x^2=0\cdot1+0\cdot(x-1)+1\cdot(x-1)^2\), or \(x^2=(x-1)^2\). But \((x-1)^2=x^2-2x+1\), which is not \(x^2\).

The correction is to solve for the coordinates in the basis actually named. For \(x^2\), the shifted-basis formula uses \(a=0\), \(b=0\), \(c=1\), and gives \((1,2,1)^{\top}\). Substitution verifies the correction:

Order matters as well as choice of basis. A coordinate vector is attached to an ordered list: its first entry multiplies the first basis polynomial, its second entry multiplies the second, and so on. When checking an answer, write out that ordered linear combination and expand it. Equality of the resulting coefficients is a direct check that the coordinates reconstruct the intended polynomial.