Rank of a Matrix

The Rank of a Matrix is the maximum number of linearly independent rows or columns in a matrix. It essentially determines the dimensionality of the vector space formed by the rows or columns of the matrix, i.e. the unique information the matrix contains.

It helps determine:

• If a system of linear equations has solutions.
• How many rows or columns contain unique information in the matrix.

The rank of a matrix is denoted using ρ(A), where A is any matrix. Thus, the number of rows of a matrix is a limit on the rank of the matrix, which means the rank of the matrix cannot exceed the total number of rows in a matrix.

For example, if a matrix is of the order 3×3, then the maximum rank of a matrix can be 3.

Nullity of Matrix

Every column either contributes to the rank (it's linearly independent) or falls into the null space (it's a combination of others).

The number of vectors in the null space is called the nullity of the matrix.

Total columns in a matrix = Rank + Nullity

This relation is known as the Rank Nullity theorem.

Finding Rank of a Matrix

3 methods can be used to get the rank of any given matrix. These methods are as follows:

• Minor Method
• Using Elementary Operations
  • Row Echelon Form
  • Normal Form

1. Minor Method

To find the rank of a matrix using the minor method, the following steps are followed:

• Calculate the determinant of the matrix (say A).
• If det(A) ≠ 0, then the rank of matrix A = order of matrix A.
• If det(A) = 0, then the rank of the matrix is equal to the order of the maximum possible nonzero minor of the matrix.

Example: Find the rank of a matrix \begin{bmatrix}1 & 2 & 3\\4 & 5 & 6 \\ 7 & 8 & 7\end{bmatrix} using the minor method.

Given A = \begin{bmatrix}1 & 2 & 3\\4 & 5 & 6 \\ 7 & 8 & 7\end{bmatrix}

• Step 1: Calculate the determinant of A

det(A) = 1 (35 - 48) - 2 (28 - 42) + 3 (32 - 35)
det(A) = -13 + 28 - 9 = 6

• As det(A) ≠ 0,
  ρ(A) = order of A = 3

2. Elementary Operations

Elementary operations are reversible transformations that preserve rank. They come in three types

• Row/Column Swap (R_i ↔ R_j​)
• Row/Column Scaling (R_i → kR_i, k≠0)
• Row/Column Replacement (R_i ​→ R_i ​+ kR_j​)

We use these elementary row operations to find Rank by converting the matrix into either:

(a) Row/Column Echelon Form

The following steps are followed to calculate the rank of a matrix using the Echelon form:

• Locate the leftmost non-zero column (this is the pivot column) of the matrix.
• If needed, swap rows to bring a non-zero entry to the top of this column.
• Use row replacement to make all entries below the pivot zero.
• Move to the next row and next column (diagonally down-right).
• Repeat these steps until no more pivots can be formed.
• Count the non-zero rows → that’s the rank.

Example 1: Find the rank of a matrix \begin{bmatrix}1 & 2 & 3\\4 & 5 & 6 \\ 7 & 8 & 9\end{bmatrix}using the Elementary Operations.

Given A = \begin{bmatrix}1 & 2 & 3\\4 & 5 & 6 \\ 7 & 8 & 9\end{bmatrix}

• Convert A to echelon form

Apply R₂ = R₂ - 4R₁
Apply R₃ = R₃ - 7R₁

A = \begin{bmatrix}1 & 2 & 3\\0 & -3 & -6 \\ 0 & -6 & -12\end{bmatrix}

Apply R₃ = R₃ - 2R₂

A = \begin{bmatrix}1 & 2 & 3\\0 & -3 & -6 \\ 0 & 0 & 0\end{bmatrix}

As matrix A is now in lower triangular form, it is in Echelon Form.

• Number of non-zero rows in A = 2.
  Thus ρ(A) = 2

Example 2: Find the rank of a matrix \begin{bmatrix}1 & 2 & 3\\2 & 4 & 6 \\ 1 & 1 & 1\end{bmatrix}using the Elementary Operations.

Given \begin{bmatrix}1 & 2 & 3\\2 & 4 & 6 \\ 1 & 1 & 1\end{bmatrix}

• Convert A to echelon form

Apply R₂ = R₂ - 2R₁
Apply R₃ = R₃ - R₁

A = \begin{bmatrix}1 & 2 & 3\\0 & 0 & 0 \\ 0 & -1 & -2\end{bmatrix}

Swap R₃ and R₂

A = \begin{bmatrix}1 & 2 & 3\\0 & -1 & -2 \\ 0 & 0 & 0\end{bmatrix}

As matrix A is in Echelon Form.

• Number of non-zero rows in A = 2.
  Thus ρ(A) = 2

Example 3: Find the rank of a matrix \begin{bmatrix}2 & 4 & -2\\4 & 9 & -3 \\ -2 & -3 & 7\end{bmatrix}using the Elementary Operations.

Given A =\begin{bmatrix}2 & 4 & -2\\4 & 9 & -3 \\ -2 & -3 & 7\end{bmatrix}

• Convert A to echelon form

Apply R₂ = R₂ - 2R₁
Apply R₃ = R₃ - R₁

A = \begin{bmatrix}2 & 4 & -2\\0 & 1 & 1 \\ 0 & 1 & 5\end{bmatrix}

R₃ = R₃ - R₂

A = \begin{bmatrix}2 & 4 & -2\\0 & 1 & 1 \\ 0 & 0 & 4\end{bmatrix}

As matrix A is in Echelon Form.

• Number of non-zero rows in A = 3.
  Thus ρ(A) = 3

(b) Normal Form

A matrix is said to be in normal form if it can be reduced to the form \begin{bmatrix} I_r & 0\\ 0 & 0\\ \end{bmatrix} . Here, I_r represents the identity matrix of order r.

If a matrix can be converted to its normal form, then the rank of the matrix is said to be r.

Example: Find the rank of a matrix \bold{\begin{bmatrix}1 & 2 & 1 & 2\\1 & 3 & 2 & 2 \\ 2 & 4 & 3 & 4 \\3 & 7 & 4 & 6\end{bmatrix}}using the normal form method.

Solution:

Given A = \begin{bmatrix}1 & 2 & 1 & 2\\1 & 3 & 2 & 2 \\ 2 & 4 & 3 & 4 \\3 & 7 & 4 & 6\end{bmatrix}

Apply R₂ = R₂ - R₁ , R₃ = R₃ - 2R₁ and R₄ = R₄ - 3R₁

A = \begin{bmatrix}1 & 2 & 1 & 2\\0 & 1 & 1 & 0 \\ 0 & 0 & 1 & 0 \\0 & 1 & 1 & 0\end{bmatrix}

Apply R₁ = R₁ - 2R₂ and R₄ = R₄ - R₂

A = \begin{bmatrix}1 & 0 & -1 & 2\\0 & 1 & 1 & 0 \\ 0 & 0 & 1 & 0 \\0 & 0 & 0 & 0\end{bmatrix}

Apply R₁ = R₁ + R₃ and R₂ = R₂ - R₃

A = \begin{bmatrix}1 & 0 & 0 & 2\\0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\0 & 0 & 0 & 0\end{bmatrix}

Apply C₄ → C₄ - 2C₁

A = \begin{bmatrix}1 & 0 & 0 & 0\\0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\0 & 0 & 0 & 0\end{bmatrix}

Thus A can be written as \begin{bmatrix} I_3 & 0\\ 0 & 0\\ \end{bmatrix} .

Thus, ρ(A) = 3

Properties of Rank of a Matrix

The properties of the rank of a matrix are as follows:

• Rank ≤ min(m, n) for an m×n matrix
• The rank of a matrix is equal to the number of non-zero rows if it is in Echelon Form.
• Rank(A) = Rank(A^T), Rank of a matrix and it's transpose are equal.
• The rank of the identity matrix is equal to the order of the identity matrix.
• [Rank of matrix] < [Order of matrix], if it is a singular matrix.
• [Rank of matrix] < [minimum {m, n}], if it is a rectangular matrix of order m x n.
• The rank of a zero matrix or a null matrix is zero.

Solved Examples

Example 1: Find the rank of a matrix \begin{bmatrix}-1 & -2 & -3\\-4 & -5 & -6 \\ -7 & -8 & -7\end{bmatrix}using the minor method.

Solution:

Given A = \begin{bmatrix}-1 & -2 & -3\\-4 & -5 & -6 \\ -7 & -8 & -7\end{bmatrix}

Step 1: Calculate the determinant of A
det(A) = -1 (35 - 48) + 2 (28 - 42) - 3 (32 - 35)
det(A) = 13 - 28 - 9 = -24
As det(A) ≠ 0, ρ(A) = order of A = 3

Example 2: Find the rank of a matrix \bold{\begin{bmatrix}2 & 4 & 6\\8 & 10 & 12 \\ 14 & 16 & 0\end{bmatrix}}using the minor method.

Solution:

Given A = \begin{bmatrix}2 & 4 & 6\\8 & 10 & 12 \\ 14 & 16 & 0\end{bmatrix}

Step 1: Calculate the determinant of A
det(A) = 2(0-192) - 4(0-168) + 6(128-140)
det(A) = -384 + 672 - 72 = 216
As det(A) ≠ 0, ρ(A) = order of A = 3

Example 3: Find the rank of a matrix \bold{\begin{bmatrix}-1 & -2 & -3\\-4 & -5 & -6 \\ -7 & -8 & -9\end{bmatrix}}using the Echelon form method.

Solution:

Given A = \begin{bmatrix}-1 & -2 & -3\\-4 & -5 & -6 \\ -7 & -8 & -9\end{bmatrix}

Step 1: Convert A to echelon form
Apply R₂ = R₂ - 4R₁
Apply R₃ = R₃ - 7R₁

A = \begin{bmatrix}-1 & -2 & -3\\0 & 3 & 6 \\ 0 & 6 & 12\end{bmatrix}

Apply R₃ = R₃ - 2R₂

A = \begin{bmatrix}-1 & -2 & -3\\0 & 3 & 6 \\ 0 & 0 & 0\end{bmatrix}

As matrix A is now in lower triangular form, it is in Echelon Form.

Step 2: Number of non-zero rows in A = 2.
Thus ρ(A) = 2

Example 4: Find the rank of a matrix \bold{\begin{bmatrix}2 & 4 & 6\\8 & 10 & 12 \\ 14 & 16 & 18\end{bmatrix}}using the Echelon form method.

Solution:

Given A = \begin{bmatrix}2 & 4 & 6\\8 & 10 & 12 \\ 14 & 16 & 18\end{bmatrix}

Step 1: Convert A to echelon form
Apply R₂ = R₂ - 4R₁
Apply R₃ = R₃ - 7R₁

A = \begin{bmatrix}2 & 4 & 6\\0 & -6 & -12 \\ 0 & -12 & -24\end{bmatrix}

Apply R₃ = R₃ - 2R₂

A = \begin{bmatrix}2 & 4 & 6\\0 & -6 & -12 \\ 0 & 0 & 0\end{bmatrix}

As matrix A is now in lower triangular form, it is in Echelon Form.

Step 2: Number of non-zero rows in A = 2.
Thus ρ(A) = 2

Example 5: Find the rank of a matrix \bold{\begin{bmatrix}2 & 4 & 2 & 4\\2 & 6 & 4 & 4 \\ 4 & 8 & 6 & 8 \\6 & 14 & 8 & 12\end{bmatrix}}using the normal form method.

Solution:

Given A = \begin{bmatrix}2 & 4 & 2 & 4\\2 & 6 & 4 & 4 \\ 4 & 8 & 6 & 8 \\6 & 14 & 8 & 12\end{bmatrix}

Apply R₂ = R₂ - R₁ , R₃ = R₃ - 2R₁ and R₄ = R₄ - 3R₁

A = \begin{bmatrix}2 & 4 & 2 & 4\\0 & 2 & 2 & 0 \\ 0 & 0 & 2 & 0 \\0 & 2 & 2 & 0\end{bmatrix}

Apply R₁ = R₁ - 2R₂ and R4 = R₄ - R₂

A = \begin{bmatrix}2 & 0 & -2 & 4\\0 & 2 & 2 & 0 \\ 0 & 0 & 2 & 0 \\0 & 0 & 0 & 0\end{bmatrix}

Apply R₁ = R₁ + R₃ and R₂ = R₂ - R₃

A = \begin{bmatrix}2 & 0 & 0 & 4\\0 & 2 & 0 & 0 \\ 0 & 0 & 2 & 0 \\0 & 0 & 0 & 0\end{bmatrix}

Apply C₄ → C₄ - 2C₁

A = \begin{bmatrix}2 & 0 & 0 & 0\\0 & 2 & 0 & 0 \\ 0 & 0 & 2 & 0 \\0 & 0 & 0 & 0\end{bmatrix}

Apply R₁ = R₁/2, R₂ = R₂/2, R₃ = R₃/2

A = \begin{bmatrix}1 & 0 & 0 & 0\\0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\0 & 0 & 0 & 0\end{bmatrix}

Thus A can be written as \begin{bmatrix} I_3 & 0\\ 0 & 0\\ \end{bmatrix}
Thus, ρ(A) = 3

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