Linear Algebra · Worked Example 6
Is S = {v₁, v₂, v₃} Linearly Independent in ℝ³?
Step 1
Identify the Vectors
We are given the set S containing three vectors in ℝ³:
v₁ = (1, 2, 3),
v₂ = (0, 1, 2),
v₃ = (2, 3, 4)
Definition. A set of vectors is linearly independent if the only solution to
c₁v₁ + c₂v₂ + … + cₙvₙ = 0
is the trivial solution c₁ = c₂ = … = cₙ = 0.
Otherwise it is linearly dependent.
Step 2
Set Up the Vector Equation
Assume a linear combination equals the zero vector:
x v₁ + y v₂ + z v₃ = 0
Substituting component-wise:
x(1, 2, 3) + y(0, 1, 2) + z(2, 3, 4) = (0, 0, 0)
Equating each component gives the homogeneous system:
① x + 0y + 2z = 0
② 2x + y + 3z = 0
③ 3x + 2y + 4z = 0
Step 3
Write the Coefficient Matrix
The system in matrix form Ax = 0:
| 1 | 0 | 2 |
| 2 | 1 | 3 |
| 3 | 2 | 4 |
Key Theorem. The homogeneous system Ax = 0 has a
non-trivial solution if and only if det(A) = 0.
So we just need to check the determinant.
Step 4
Compute the Determinant
Expanding |A| along the first row:
|A| =
| 1 | 0 | 2 |
| 2 | 1 | 3 |
| 3 | 2 | 4 |
= 1 ×
− 0 ×
+ 2 ×
| 1 | 3 |
| 2 | 4 |
| 2 | 3 |
| 3 | 4 |
| 2 | 1 |
| 3 | 2 |
= 1(4 − 6) − 0(8 − 9) + 2(4 − 3)
= 1(−2) − 0(−1) + 2(1)
= −2 + 0 + 2
∴ |A| = 0
Since det(A) = 0, the matrix is singular.
The system Ax = 0 has
infinitely many solutions — a non-trivial solution exists.
The Vectors are Linearly Dependent
Since det(A) = 0, there exist scalars x, y, z (not all zero) such that x v₁ + y v₂ + z v₃ = 0. Hence S = {(1,2,3), (0,1,2), (2,3,4)} is linearly dependent in ℝ³.
Tutorial Video: Linear Span & Linear Combinations
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