Linear Algebra
Polynomial Spaces
P₄(ℝ)
RREF
[P₄ Example]
Are Four Degree-4 Polynomials Linearly Dependent? Testing in P₄(ℝ) Using Rank
Linear independence isn't just for geometric vectors — it applies to entire spaces of
polynomials. In this post, we encode four degree-4 polynomials as coefficient vectors
in ℝ⁵, form a 5×4 matrix, and row-reduce it to RREF. When rank < number of
polynomials, the set is dependent — and that's exactly what we find here.
The polynomial space P₄(ℝ) consists of all real polynomials of degree at most 4.
Its standard basis is {x⁴, x³, x², x, 1} — five basis elements,
making it a 5-dimensional vector space. Every polynomial in P₄(ℝ) is fully described
by its five coefficients, so testing linear dependence reduces to a familiar matrix problem.
The Problem5
Example 5 — P₄(ℝ) Linear Dependence Test
Determine whether the following four polynomials in P₄(ℝ) are linearly independent or linearly dependent.
p₁
x⁴ − x³ + 5x² − 8x + 6
p₂
−x⁴ + x³ − 5x² + 5x − 3
p₃
x⁴ + 0x³ + 3x² − 3x + 5
p₄
2x⁴ + x³ + 4x² + 8x + 0
Key Idea — Encoding Polynomials as Vectors
Standard basis of P₄(ℝ): { x⁴, x³, x², x, 1 }
Each polynomial is represented by its ordered list of coefficients. For example,
p₁ = x⁴ − x³ + 5x² − 8x + 6 becomes the vector (1, −1, 5, −8, 6).
Linear dependence of the polynomials is equivalent to linear dependence of these
coefficient vectors.
Setting Up the Linear Combination
Assume scalars a, b, c, d ∈ ℝ such that:
a·p₁ + b·p₂ + c·p₃ + d·p₄ = 0 (zero polynomial)
Expanding and collecting by basis element:
a − b + c + 2d = 0 ← coefficient of x⁴
−a + b + 0c + d = 0 ← coefficient of x³
5a − 5b + 3c + 4d = 0 ← coefficient of x²
−8a + 5b − 3c + 8d = 0 ← coefficient of x
6a − 3b + 5c + 0d = 0 ← constant term
Matrix Form AX = 0
Arranging the coefficients into a 5×4 matrix (one row per basis element, one column per polynomial):
1 -1 1 2
-1 1 0 1
5 -5 3 4
-8 5 -3 8
6 -3 5 0
a
b
c
d
=
0
0
0
0
0
Row Reduction — Step by Step
Initial Matrix A
1 -1 1 2
-1 1 0 1
5 -5 3 4
-8 5 -3 8
6 -3 5 0
R₂ ← R₂ + R₁
R₃ ← R₃ − 5R₁
R₄ ← R₄ + 8R₁
R₅ ← R₅ − 6R₁
↓ Eliminate column 1 below pivot
After eliminating column 1
1 -1 1 2
0 0 1 3
0 0 -2 -6
0 -3 5 24
0 3 -1 -12
R₂ ↔ R₅
(swap to get pivot in column 2)
↓ Swap R₂ and R₅ for column 2 pivot
After R₂ ↔ R₅
1 -1 1 2
0 3 -1 -12
0 0 -2 -6
0 -3 5 24
0 0 1 3
R₂ ← (1/3)·R₂
↓ Scale pivot row 2
After scaling R₂
1 -1 1 2
0 1 -1/3 -4
0 0 -2 -6
0 -3 5 24
0 0 1 3
R₁ ← R₁ + R₂
R₄ ← R₄ + 3R₂
↓ Eliminate column 2 above and below
After eliminating column 2
1 0 2/3 -2
0 1 -1/3 -4
0 0 -2 -6
0 0 4 -12
0 0 1 3
R₃ ← (1/−2)·R₃
↓ Scale pivot row 3
After scaling R₃
1 0 2/3 -2
0 1 -1/3 -4
0 0 1 3
0 0 4 -12
0 0 1 3
R₁ ← R₁ − (2/3)R₃
R₂ ← R₂ + (1/3)R₃
R₄ ← R₄ − 4R₃
R₅ ← R₅ − R₃
↓ Eliminate column 3 above and below
Reduced Row Echelon Form (RREF)
1 0 0 -4
0 1 0 -3
0 0 1 3
0 0 0 0
0 0 0 0
← Pivot row 1 (rank +1)
← Pivot row 2 (rank +1)
← Pivot row 3 (rank +1)
← zero row
← zero row
Rank Analysis
1
Free variables (n − rank)
Interpreting the Result
The RREF has 3 pivot rows and 2 zero rows, confirming
rank(A) = 3. With 4 unknowns (a, b, c, d) and rank 3,
there is 1 free variable — meaning infinitely many non-trivial
solutions exist.
Concretely: we can find a, b, c, d not all zero such that
a·p₁ + b·p₂ + c·p₃ + d·p₄ = 0. One of the polynomials is a linear combination
of the other three — the fourth polynomial is "hiding" inside the span of the rest.
ρ(A) = 3 < n = 4 → Non-trivial solution exists → 1 free variable
∴ The set {p₁, p₂, p₃, p₄} is Linearly Dependent in P₄(ℝ)
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