Linear Algebra
ℝ⁴ Vectors
Row Reduction
Proving Three Vectors in ℝ⁴ Are Linearly Dependent — A Full Row Reduction Walkthrough
Four-dimensional vectors can't be visualised, but they can absolutely be tested for linear
dependence. In this post, we take three vectors in ℝ⁴, form a 4×3 coefficient matrix,
and row-reduce it to RREF — revealing a rank of 2, which is all we need to confirm
dependence. A complete worked example from the N/D-25 exam.
When we say vectors are linearly dependent, we mean at least one of them can be
written as a linear combination of the others — it adds no new "direction" to the set.
The cleanest way to verify this is to set up a homogeneous system and check whether it
admits a non-trivial solution. If rank < number of unknowns, dependence is confirmed.
The Problem 4
Example 4
Show that the vectors u, v, and w
form a linearly dependent set in ℝ⁴.
Setting Up the Linear Combination
Assume scalars x, y, z ∈ ℝ satisfy:
x·u + y·v + z·w = 0
x(0, 3, 1, −1) + y(6, 0, 5, 1) + z(4, −7, 1, 3) = (0, 0, 0, 0)
System of Equations
Equating each of the four components gives one equation per row:
0x + 6y + 4z = 0 ← component 1
3x + 0y − 7z = 0 ← component 2
x + 5y + z = 0 ← component 3
−x + y + 3z = 0 ← component 4
In matrix form AX = 0:
0 6 4
3 0 -7
1 5 1
-1 1 3
x
y
z
=
0
0
0
0
Row Reduction — Step by Step
Initial matrix A
0 6 4
3 0 -7
1 5 1
-1 1 3
R₁ ↔ R₃
(swap for a leading 1)
↓ Swap R₁ and R₃
After R₁ ↔ R₃
1 5 1
3 0 -7
0 6 4
-1 1 3
R₂ ← R₂ − 3R₁
R₄ ← R₄ + R₁
↓ Eliminate column 1 below pivot
After eliminating column 1
1 5 1
0 -15 -10
0 6 4
0 6 4
R₂ ← −(1/15)·R₂
↓ Scale pivot row 2
After scaling R₂
1 5 1
0 1 2/3
0 6 4
0 6 4
R₁ ← R₁ − 5R₂
R₃ ← R₃ − 6R₂
R₄ ← R₄ − 6R₂
↓ Eliminate column 2 above and below
Reduced Row Echelon Form (RREF)
1 0 -1/3
0 1 2/3
0 0 0
0 0 0
← Pivot (rank contributes 1)
← Pivot (rank contributes 1)
← zero row
← zero row
Interpreting the Result
The RREF has exactly 2 non-zero rows, giving rank(A) = 2.
Since we have 3 unknowns (x, y, z) and rank 2, the system has
3 − 2 = 1 free variable, meaning infinitely many non-trivial solutions exist.
That is, we can find scalars x, y, z — not all zero — such that
x·u + y·v + z·w = 0. One vector is expressible as a combination of the other two.
rank(A) = 2 < 3 (number of unknowns) → Non-trivial solution exists
∴ The vectors u, v, and w are Linearly Dependent in ℝ⁴
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