Tutorials › Linear Algebra › Independence of Polynomial Lists

Basis and Dimension · Tutorial 308 of 1000

Independence of Polynomial Lists

Translate polynomial relations into equations on coefficients to decide independence and identify dependence relations.

Intermediate 9 min read

What You'll Learn

  • Express polynomial independence as a homogeneous system of coefficient equations.
  • Construct coefficient vectors using a consistent ordering of powers.
  • Test independence of a list of monomials by comparing coefficients.
  • Find and verify a dependence relation among non-monomial polynomials.
  • Determine how a parameter affects independence in a bounded-degree polynomial space.
  • Prove that nonzero polynomials of distinct degrees are linearly independent.

Polynomial Relations Are Coefficient Equations

Polynomials can be treated as vectors, but their entries are coefficients rather than coordinates written in a column. This gives a direct way to test a list for linear independence: form a linear combination with unknown scalar coefficients, collect like powers of \(x\), and set each polynomial coefficient equal to zero. The resulting scalar equations decide whether the only relation is the trivial one.

We work over \(\mathbb{R}\). A polynomial in \(\mathbb{R}[x]\) is a finite expression \(a_0+a_1x+\cdots+a_dx^d\), where each \(a_j\in\mathbb{R}\). Its coefficients are uniquely determined; in particular, the zero polynomial has every coefficient equal to zero. For a nonnegative integer \(d\), write \(P_d\) for the set of polynomials of degree at most \(d\), including the zero polynomial. Every polynomial in \(P_d\) can be written with coefficients from the constant term through the \(x^d\) term, inserting zero coefficients when a power is absent.

Definition: A list \(p_1,\dots,p_k\) of polynomials is linearly independent if \(c_1p_1+\cdots+c_kp_k=0\), for \(c_1,\dots,c_k\in\mathbb{R}\), forces \(c_1=\cdots=c_k=0\). If there is a choice of coefficients not all zero for which the combination is the zero polynomial, the list is linearly dependent; that equation is a dependence relation.

To apply the definition, write each polynomial using the same ordered powers of \(x\). For polynomials in \(P_d\), the coefficient vector of \(p=a_0+a_1x+\cdots+a_dx^d\) is the column \((a_0,a_1,\dots,a_d)^{\top}\). The order matters for keeping equations organized: here the constant coefficient comes first, then the coefficient of \(x\), and so on. A missing power contributes a zero entry.

Theorem (Coefficient test): Let \(p_1,\dots,p_k\in P_d\), and form a \((d+1)\times k\) matrix whose \(j\)-th column is the coefficient vector of \(p_j\), in the order \(1,x,\dots,x^d\). The list \(p_1,\dots,p_k\) is linearly independent if and only if the columns of this matrix are linearly independent.

Proof. Write \(p_j=\sum_{i=0}^{d}a_{ij}x^i\), and let \(\mathbf{c}=(c_1,\dots,c_k)^{\top}\). The coefficient of \(x^i\) in \(c_1p_1+\cdots+c_kp_k\) is \(\sum_{j=1}^{k}a_{ij}c_j\). Therefore the polynomial combination is zero exactly when every one of these coefficients is zero. In matrix form, those \(d+1\) equations are \(\mathbf{A}\mathbf{c}=\mathbf{0}\), where \(\mathbf{A}=(a_{ij})\) is the coefficient matrix. Thus the polynomial list is independent exactly when the only solution is \(\mathbf{c}=\mathbf{0}\), which is also exactly the definition of independence of the columns of \(\mathbf{A}\). \(\square\)

This theorem justifies the method, rather than merely suggesting a convenient computation. Equality of polynomials is equality of every coefficient, so no information is lost when the polynomial equation is replaced by a system of scalar equations. After setting up that system, the familiar row-reduction test for independence applies to its coefficient matrix.

Worked Examples: Comparing Coefficients

Worked Example: A List of Monomials

Test whether \(1,x,x^2,x^3\) are linearly independent in \(P_3\).

Let \(c_0,c_1,c_2,c_3\in\mathbb{R}\), and suppose a linear combination is zero:

$$ c_0\cdot 1+c_1x+c_2x^2+c_3x^3=0. $$

The coefficient of \(1\) is \(c_0\), the coefficient of \(x\) is \(c_1\), the coefficient of \(x^2\) is \(c_2\), and the coefficient of \(x^3\) is \(c_3\). Since the zero polynomial has every coefficient equal to zero,

$$ c_0=0,\qquad c_1=0,\qquad c_2=0,\qquad c_3=0. $$

Only the trivial choice of coefficients gives the zero polynomial, so the list is independent.

Check: Substituting these values into the proposed combination gives \(0\cdot1+0\cdot x+0\cdot x^2+0\cdot x^3=0\). The coefficient equations allow no other solution.

Worked Example: Finding a Dependence Relation

Test whether \(p_1=1+x\), \(p_2=x+x^2\), and \(p_3=1+2x+x^2\) are linearly independent. If they are dependent, find a relation.

Set \(c_1p_1+c_2p_2+c_3p_3=0\) and collect the constant, \(x\), and \(x^2\) coefficients:

$$ (c_1+c_3)+(c_1+c_2+2c_3)x+(c_2+c_3)x^2=0. $$

Equating coefficients gives \(c_1+c_3=0\), \(c_1+c_2+2c_3=0\), and \(c_2+c_3=0\). Choose \(c_3=1\). The first equation gives \(c_1=-1\), and the third gives \(c_2=-1\). The middle equation is then \(-1-1+2=0\), so all three equations hold. The coefficients are not all zero, and the list is dependent.

Check: Substituting the coefficients into the original polynomial combination gives \((-1)(1+x)+(-1)(x+x^2)+(1)(1+2x+x^2)\). Its constant coefficient is \(-1+1=0\), its \(x\) coefficient is \(-1-1+2=0\), and its \(x^2\) coefficient is \(-1+1=0\). Thus the combination is the zero polynomial, as required.

Worked Example: A Parameter Changes Independence

For which real values of \(t\) is the list \(p_1=1+x\), \(p_2=x+x^2\), \(p_3=1+tx+x^2\) linearly independent?

Suppose \(c_1p_1+c_2p_2+c_3p_3=0\). Equating the coefficients of \(1,x,x^2\), respectively, gives

$$ \begin{aligned} c_1+c_3&=0,\\ c_1+c_2+tc_3&=0,\\ c_2+c_3&=0. \end{aligned} $$

The first and third equations imply \(c_1=-c_3\) and \(c_2=-c_3\). Substitute both into the middle equation:

$$ -c_3-c_3+tc_3=(t-2)c_3=0. $$

If \(t\ne2\), then \(c_3=0\), and consequently \(c_1=c_2=0\). The list is independent. If \(t=2\), take \(c_3=1\); then \(c_1=c_2=-1\), giving a nontrivial relation. Therefore the list is independent exactly when \(t\ne2\).

Check: At \(t=2\), \(p_3=1+2x+x^2=(1+x)+(x+x^2)=p_1+p_2\). Hence \(-p_1-p_2+p_3=0\), confirming dependence at the exceptional value. For \(t\ne2\), the coefficient equations force all three coefficients to be zero, confirming independence.

Distinct Degrees Give Independence

Coefficient comparison also gives a useful result that avoids solving a full system in some cases. The degree of a nonzero polynomial is the largest power of \(x\) with a nonzero coefficient. If the degrees in a list are all different, the polynomial of largest degree contributes a power that none of the other polynomials can contribute. Its coefficient in any zero combination must therefore vanish. Repeating this reasoning rules out every nonzero coefficient in the relation.

Theorem: Any finite list of nonzero polynomials in \(\mathbb{R}[x]\) with pairwise distinct degrees is linearly independent.

Proof. Let \(p_1,\dots,p_k\) be nonzero polynomials with pairwise distinct degrees, and suppose \(c_1p_1+\cdots+c_kp_k=0\). Choose the index \(j\) for which \(p_j\) has the greatest degree, and call that degree \(m\). Let \(a\ne0\) be the coefficient of \(x^m\) in \(p_j\). Every other polynomial in the list has degree less than \(m\), so its coefficient of \(x^m\) is zero. The coefficient of \(x^m\) in the proposed zero combination is therefore \(c_ja\). Since the combination is the zero polynomial, \(c_ja=0\), and \(a\ne0\) implies \(c_j=0\). Remove this term from the relation. Among the remaining polynomials, the degrees are still pairwise distinct; the same argument forces the coefficient of the one with greatest remaining degree to be zero. Repeating this finite argument gives \(c_i=0\) for every index \(i\). Thus the list is independent. \(\square\)

The nonzero hypothesis matters: the zero polynomial has no degree under the definition above and is dependent in any list that contains it. The theorem is a sufficient test, not a necessary one. For instance, two independent polynomials can have the same degree; distinct degrees are one useful reason for independence, not the only possible reason.

Common Pitfall: Matching Only the Highest Powers

Comparing only the highest-degree terms is not enough to establish a dependence relation. Consider \(p_1=1+x^2\) and \(p_2=x+x^2\). A tempting but incorrect calculation is to note that both have \(x^2\) coefficient \(1\), choose coefficients \(1\) and \(-1\) to cancel those terms, and conclude that \(p_1-p_2=0\). The actual calculation is

$$ (1+x^2)-(x+x^2)=1-x\ne0. $$

The cancellation of the \(x^2\) terms is real, but the constant and \(x\) coefficients must also be checked. In full, \(c_1p_1+c_2p_2=0\) requires \(c_1=0\) from the constant term, \(c_2=0\) from the \(x\) term, and \(c_1+c_2=0\) from the \(x^2\) term. These equations force the trivial choice, so the two polynomials are independent. Every power from the constant through the largest degree must be included, even when a polynomial has a zero coefficient at one of those powers.

The coefficient method is useful whenever a finite list of polynomials has a common degree bound: it turns an abstract independence question into a finite system of scalar equations. The chosen degree bound must be at least as large as every degree in the list, and the coefficient vectors must all use the same ordering. These details prevent omitted terms or mismatched coordinates from producing a false conclusion.

Check Your Understanding

Use coefficient equations in each question, and verify any dependence relation by substitution.

  1. Test whether \(1-x\), \(x+x^2\), and \(1+x^2\) are linearly independent in \(P_2\).
  2. Find a nontrivial dependence relation, or show independence, for \(1+x\), \(2+2x\).
  3. For which real values of \(t\) are \(1+x\) and \(1+tx\) linearly independent?
  4. Explain why the coefficient vector of \(2-x+3x^3\), using powers \(1,x,x^2,x^3\), must include a zero entry.
  5. State the hypotheses of the theorem about polynomials of distinct degrees, and explain why a list containing the zero polynomial is not covered.

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