Matrix Rank

5.5. Matrix Rank#

The rank of a matrix is the number of linearly independent rows or columns.

  • The rank of a matrix \(A\) is denoted as \(\text{rank}(A)\) or \(r(A)\).

  • The rank of an \(m \times n\) matrix is always less than or equal to both \(m\) and \(n\): \(\text{rank}(A) \leq \min(m, n)\)

  • The rank of a zero matrix (all elements are zero) is 0.

  • A square \(n \times n\) matrix has full rank (rank \(n\)) if and only if it is invertible (non-singular), its determinant is non-zero, and all its rows and columns are linearly independent. The following are equivalent:

    1. Full Rank
      \begin{equation} \text{rank}(A) = n \end{equation}

    2. Invertible (Non-Singular)
      \begin{equation} A^{-1} \text{ exists} \end{equation}

    3. Non-Zero Determinant
      \begin{equation} \det(A) \neq 0 \end{equation}

    4. Linearly Independent Rows and Columns

      • Rows of \(A\) are linearly independent

      • Columns of \(A\) are linearly independent

  • The row rank (maximum number of linearly independent rows) is always equal to the column rank (maximum number of linearly independent columns).

  • Elementary row (or column) operations do not change the rank of a matrix.

5.5.1. Gaussian Elimination to Find the Rank#

  • Transform the matrix into its row echelon form (REF) or reduced row echelon form (RREF) using elementary row operations.

  • The rank of the matrix is the number of non-zero rows in its REF or RREF. A non-zero row is a row that contains at least one non-zero element.

Example

(5.62)#\[\begin{equation} A = \begin{bmatrix} 1 & 2 & 3 \\ 2 & 5 & 7 \\ 3 & 7 & 10 \end{bmatrix} \end{equation} \]

Step 1: Eliminate elements below the first pivot (the element in the first row and first column).

  • \(R_2 = R_2 - 2R_1\): Subtract 2 times the first row from the second row.

    (5.63)#\[\begin{equation} \begin{bmatrix} 1 & 2 & 3 \\ 2 - 2(1) & 5 - 2(2) & 7 - 2(3) \\ 3 & 7 & 10 \end{bmatrix} = \begin{bmatrix} 1 & 2 & 3 \\ 0 & 1 & 1 \\ 3 & 7 & 10 \end{bmatrix} \end{equation} \]

  • \(R_3 = R_3 - 3R_1\): Subtract 3 times the first row from the third row.

    (5.64)#\[\begin{equation} \begin{bmatrix} 1 & 2 & 3 \\ 0 & 1 & 1 \\ 3 - 3(1) & 7 - 3(2) & 10 - 3(3) \end{bmatrix} = \begin{bmatrix} 1 & 2 & 3 \\ 0 & 1 & 1 \\ 0 & 1 & 1 \end{bmatrix} \end{equation} \]

Step 2: Eliminate elements below the second pivot (the element in the second row and second column).

  • \(R_3 = R_3 - 1R_2\): Subtract the second row from the third row.

    (5.65)#\[\begin{equation} \begin{bmatrix} 1 & 2 & 3 \\ 0 & 1 & 1 \\ 0 - 0 & 1 - 1 & 1 - 1 \end{bmatrix} = \begin{bmatrix} 1 & 2 & 3 \\ 0 & 1 & 1 \\ 0 & 0 & 0 \end{bmatrix} \end{equation} \]

Step 3: Identify the number of non-zero rows.

The matrix is now in row echelon form. The non-zero rows are:

  • \([1 \quad 2 \quad 3]\)

  • \([0 \quad 1 \quad 1]\)

There are two non-zero rows; therefore, the rank is 2.

Here is the example in NumPy.

import numpy as np
import sympy as sym 

# NumPy rank 
A = np.array([[1,2,3],[2,5,7], [3, 7,10]]) 
np.linalg.matrix_rank(A) 

np.int64(2)

Here is the example in SymPy. SymPy can get the matrix in reduced row echelon form with the command Matrix.rref() and the rank with the command Matrix.rank().

A = sym.Matrix([[1,2,3],[2,5,7], [3, 7,10]])
A
\[\begin{split}\displaystyle \left[\begin{matrix}1 & 2 & 3\\2 & 5 & 7\\3 & 7 & 10\end{matrix}\right]\end{split}\]
# SymPy reduced row echelon form
A.rref()

(Matrix([
 [1, 0, 1],
 [0, 1, 1],
 [0, 0, 0]]),
 (0, 1))

# Sympy rank 
A.rank()

2

# A few more NumPy Examples 
A = np.array([[1,0,1],[0,0,1],[0, 0, 2]]) 
np.linalg.matrix_rank(A) 

np.int64(2)

A = np.array([[10, 8, 9], [-3, 4, 0],[2, 60, -100]])
np.linalg.matrix_rank(A) 

np.int64(3)
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