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Matrices

1.8 Rank of a Matrix

Rank of a matrix via row echelon form, full rank, the rank-nullity theorem, null space and homogeneous systems, and solvability of AX = B.

1.8 Rank of a Matrix

Definition 1.32: The rank of A, Rank(A), is the number of nonzero rows after reducing A to row echelon form. It equals the number of linearly independent rows.

Example 1.30: Find Rank(M) for M = [[−1,6,−2],[3,1,−1],[2,−10,3]]

StepOperationResult
Start[−1,6,−2 | 3,1,−1 | 2,−10,3]
1R₂→R₂+3R₁, R₃→R₃+2R₁[1,−6,2 | 0,19,−7 | 0,2,−1]
2R₃⇄R₂/2, further ops[1,0,−1 | 0,1,−½ | 0,0,5/2]
3Scale R₃[1,0,0 | 0,1,0 | 0,0,1] = I₃
All 3 rows nonzero → Rank(M) = 3 (full rank)

Example 1.31: M = [[−1,6,−6],[3,1,−20],[2,−10,8]] → reduces to [[−1,6,−6],[0,1,−2],[0,0,0]] → Rank = 2 (one zero row)

Fundamental Theorem

Theorem 1.3: Rank(A) = Rank(AT) and Rank(A) ≤ min(m,n).
If Rank(A) = min(m,n), then A has full rank.

1.9 Null Space

Definition 1.34: The null space (kernel) of A is:
Null(A) = {X | AX = 0}
The nullity = dimension of the null space = number of free parameters in the solution.

Rank-Nullity Theorem

nullity(A) + Rank(A) = n  (number of columns)

Example — full rank: For M from Ex 1.30 (Rank = 3, n = 3):

AX = 0 → X = 0 (trivial only). nullity = 0. Check: 0 + 3 = 3 ✓

Example — rank 2: For M from Ex 1.31 (Rank = 2, n = 3):

Reduced: [−1,6,−6 | 0,1,−2 | 0,0,0] → x = 6t, y = 2t, z = t
6t
2t
t
X = t·[6,2,1]ᵀ — 1 free param → nullity = 1
Check: nullity(1) + Rank(2) = 3 = n ✓

Properties of AX = 0 (Homogeneous)

  • Always consistent (X = 0 is always a solution)
  • Either unique solution X = 0, or infinitely many
  • If A is n×n: unique solution iff det(A) ≠ 0
  • If X₁ and X₂ are solutions, so is αX₁ + βX₂

1.10 Solvability of AX = B

The inhomogeneous system AX = B is consistent iff Rank(A) = Rank([A|B]).
The general solution = particular solution Y + any solution X of AX = 0.