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Friday, June 5, 2026

MA25C02| Example5 | Linear Combination of Polynomials 🚫 Yes or No? | Vector Spaces | Linear Algebra 🧮

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Can p₁ be written as a Linear Combination of p₂ and p₃?

Example Problem Space P₃(ℝ) Coefficient Comparison Method

What is a Linear Combination?

Definition

A vector (or polynomial) p₁ is a linear combination of p₂ and p₃ if there exist scalars a, b ∈ ℝ such that:

p₁ = a · p₂ + b · p₃

In the polynomial space P₃(ℝ), polynomials behave like vectors: we can add them and scale them. Checking for a linear combination reduces to asking whether a certain system of linear equations has a consistent solution.

Given Polynomials

Exam Problem ·5

Verify whether the first polynomial can be expressed as a linear combination of the other two in P₃(ℝ):

p₁(x) = x³ − 8x² + 4x p₂(x) = x³ − 2x² + 3x − 1 p₃(x) = x³ − 2x + 3

Step-by-Step Verification

1

Set Up the Equation

We want scalars a, b ∈ ℝ such that:

a · p₂(x) + b · p₃(x) = p₁(x)

Substituting each polynomial:

a(x³ − 2x² + 3x − 1) + b(x³ − 2x + 3) = x³ − 8x² + 4x
2

Expand & Collect Like Terms

ax³ − 2ax² + 3ax − a + bx³ − 2bx + 3b = x³ − 8x² + 4x (a + b)x³ + (−2a)x² + (3a − 2b)x + (−a + 3b) = x³ − 8x² + 4x
3

Compare Coefficients

Since two polynomials are equal if and only if all corresponding coefficients match, we extract one equation per degree:

(1) a + b = 1 coeff. of x³
(2) −2a = −8 coeff. of x²
(3) 3a − 2b = 4 coeff. of x
(4) −a + 3b = 0 constant term
4

Solve Equations (1) & (2)

From equation (2):

−2a = −8 a = 4

Substituting a = 4 into equation (1):

4 + b = 1 b = −3
5

Check Equations (3) & (4)

We must verify that a = 4, b = −3 satisfies the remaining equations.

Equation (3): 3a − 2b = 4 ?
3(4) − 2(−3) = 12 + 6 = 18 ≠ 4
✗ Fails
Equation (4): −a + 3b = 0 ?
−(4) + 3(−3) = −4 − 9 = −13 ≠ 0
✗ Fails

Both remaining equations are violated. The system is inconsistent — no pair (a, b) can satisfy all four equations simultaneously.

Final Answer

No — p₁(x) = x³ − 8x² + 4x cannot be expressed as a linear combination of p₂ and p₃ in P₃(ℝ), because the resulting system of equations is inconsistent.

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