Rank by using Echelon and Normal form of Matrices

Rank by using Echelon and Normal form of Matrices

1. Echelon (Triangular) Form — Definition

A matrix is in row echelon form if:

1. all rows consisting entirely of zeros lie below every non-zero row;
2. the first non-zero entry of each non-zero row (its pivot or leading entry) is strictly to the right of the pivot of the row above it.

If, in addition, every pivot equals 1 and is the only non-zero entry in its column, the matrix is in reduced row echelon form (RREF).

Then: rank = number of non-zero rows (equivalently, the number of pivots).

Echelon form is obtained with row operations only, and it is not unique — different sequences of operations give different-looking echelon matrices, but the number of pivots (hence the rank) is always the same. RREF, by contrast, is unique.

2. Normal (Canonical) Form — Definition

If, by a sequence of elementary row and column operations, a matrix A of rank r is reduced to the block form

where Iᵣ is the r × r identity matrix, then r is the rank of A, written A ~ [Iᵣ | 0 ; 0 | 0]. The normal form is unique: two matrices have the same normal form exactly when they are equivalent. Since only the rank survives in it, the normal form is the strongest statement elementary operations can make about a matrix.

3. Method — Rank by Echelon Form (step by step)

1. Take the left-most column that has a non-zero entry, and (if needed) interchange rows so that entry comes to the top.
2. Make all entries below the pivot zero with Rᵢ → Rᵢ − kR₁.
3. Move to the next column and the next row; repeat on the remaining sub-matrix.
4. Count the non-zero rows.

        [ 1  -1   1 ]
   A =  [ 2   1  -3 ]           R2 → R2 - 2R1 :  (2,1,-3) - (2,-2,2) = (0, 3, -5)
        [ 1   2  -3 ]           R3 → R3 -  R1 :  (1,2,-3) - (1,-1,1) = (0, 3, -4)

        [ 1  -1   1 ]
   ~    [ 0   3  -5 ]           R3 → R3 - R2 :  (0,3,-4) - (0,3,-5) = (0, 0, 1)
        [ 0   3  -4 ]

        [ 1  -1   1 ]
   ~    [ 0   3  -5 ]     ← echelon form, 3 non-zero rows
        [ 0   0   1 ]

   ⇒  ρ(A) = 3
   Check by determinant: only "add a multiple of another row" operations were used
   (no interchanges, no scaling), so det is unchanged:
   det A = 1 × 3 × 1 = 3  (product of the pivots) ≠ 0  ✓

7. Echelon (or RREF) vs Normal Form

Basis	Echelon / RREF	Normal form
Operations allowed	Row operations only	Row and column operations
Unique?	Echelon: no · RREF: yes	Yes
Rank read as	Number of non-zero rows	Size of the identity block Iᵣ
Use	Solving systems, quick rank	Equivalence of matrices, theory
Safe for solving systems?	Yes	No — column operations change the solution set

8. Exam Points

• Define echelon form and reduced row echelon form; find the rank by reducing to echelon form — 5-7 marks, almost certain.
• Reduce a given matrix to its normal form and hence find its rank — 5-7 marks, very frequent.
• Show that two given matrices are equivalent by reducing both to the same normal form — 5 marks.
• Why does the number of non-zero rows in echelon form equal the rank? — 3 marks.
• Differentiate echelon form and normal form (operations allowed, uniqueness) — 3-5 marks.
• Find the rank of a 4 × 5 matrix by the echelon method — 5 marks.