Matrix Rank Explained with Meaning and Applications

To define the rank of a matrix, we should have prior knowledge of sub-matrices and minors of a matrix. Let A be a given matrix. Matrix obtained by deleting some rows and some columns of matrix A is known as the sub-matrix of A. A matrix (Plural is matrices) is a rectangular array of numbers, symbols, or expressions, which are arranged in the form of rows and columns. Minor of the matrix is the determinant of the square matrix that is obtained by deleting one row and one column from some larger square matrix.

For the students of Class 12 studying subjects like Maths, Matrix is an important concept as it forms the base for more detailed concepts later, the topic is one of the most crucial from the perspective of the JEE main exam, and thus, it is imperative for the students to wisely learn the same.

The rank of a Matrix Definition

The rank of the matrix refers to the number of linearly independent rows or columns in the matrix. ρ(A) is used to denote the rank of matrix A.  A matrix is said to be of rank zero when all of its elements become zero. The rank of the matrix is the dimension of the vector space obtained by its columns. The rank of a matrix cannot exceed more than the number of its rows or columns. The rank of the null matrix is zero.

Key Points:

• Rank can never exceed the minimum of the number of rows and columns in the matrix.

• A zero matrix has a rank of 0, as all its rows/columns are linearly dependent (filled with zeros).

• An identity matrix has a rank equal to its dimension, as all its rows/columns are linearly independent.

Rank of Matrix on the basis of Linear Independent Vectors

The rank of a matrix on the basis of linearly independent vectors refers to the number of linearly independent vectors that can be formed from its columns or rows. In other words, it tells us the dimensionality of the vector space spanned by these vectors.

Matrix A with row vectors R1, R2, ..., Rm, or column vectors C1, C2, ..., Cn, then:

Row vectors: $A=\begin{bmatrix} R1\\ R2\\ .\\ .\\ .\\Rm \end{bmatrix}$

Column vectors: $A=\begin{bmatrix}C1 & C2 & . & . & . & Cn \end{bmatrix}$

Then, R1, R2, …., Rm or C1, C2, …, Cn are linearly independent if,

A1 R1 + A2 R2 + …..+ Am Rm = 0 ⇔ A1 = A2 = … = Am = 0 

k1 C1 + k2 C2 + … + kn Cn = 0 ⇔ k1 = k2 = … = kn = 0

Where, A1, A2, …, Am and k1, k2, …, kn are scalars.

Determining a Matrix's Rank through Minor of Matrix

The rank of a matrix is determined by the highest order of a non-zero minor within it. If a matrix has a rank 'r', it means that at least one minor in the matrix is of order 'r', and any minors of order greater than 'r' are all zero.

Determining Matrix Rank through Echelon Form

To find the rank of a matrix using its echelon form, count the number of non-zero rows in the reduced echelon form of the matrix. An echelon form is achieved when:

i.) Any zero rows are placed below non-zero rows.

ii.) The number of zeros in front of a row increases with the row number.

The non-zero rows in an echelon matrix represent its linearly independent row vectors.

The rank of a matrix A is usually denoted as r(A) or ⍴(A).

Nullity of a Matrix

The nullity of a matrix is defined as the number of vectors present in the null space of a given matrix. In other words, it can be defined as the dimension of the null space of matrix A called the nullity of A. Rank + Nullity is the number of all columns in matrix A.

Properties of the Rank of the Matrix: 

• Rank linear algebra refers to finding column rank or row rank collectively known as the rank of the matrix. 

• Zero matrices have no non-zero row. Hence it has an independent row (or column). So, the rank of the zero matrices is zero.

• When the rank equals the smallest dimension it is called the full rank matrix.

How to Find the Rank of the Matrix?

Let A = (a_ij)\[_{m\times n}\],  be a matrix. A positive integer r is said to be the rank of matrix A if

• Matrix A has at least one r-rowed minor which is different from zero 

• Every (r + 1) row minor of matrix A is zero.

Let A = (a_ij)\[_{m\times n}\] is a matrix and B  is its sub-matrix of order r, then ∣β∣ the determinant is called an r-rowed minor of A.

To Calculate Rank of Matrix There are Two Methods:

•  Minor method 

•  Echelon form

Steps to Find the Rank of the Matrix by Minor Method:

(i) If a matrix contains at least one non zero element, then ρ (A) ≥ 1

(ii) The rank of the identity matrix I_n is n.

(iii) If the rank of matrix A is r, then there exists at least one minor of order r which does not vanish.  Every minor of matrix A of order (r + 1) and higher-order (if any) vanishes.

(iv) If A is a matrix of m × n , then

        ρ(A) ≤ min {m, n}  

(v) A square matrix A of order n has to inverse if and only if ρ(A)  =  n.

Steps to Find the Rank of the Matrix by Echelon Form:

(i) The first element of every non-zero row should be 1.

(ii) The row in which every element is zero, then that row should be below the non-zero rows.

(iii) Total number of zeroes in the next non-zero row should be more than the number of zeroes in the previous non-zero row.

By elementary operations, we can easily bring the given matrix to the echelon form.

Note: The rank of a matrix does not change if we perform the following elementary row operations are applied to the matrix:

(a) Two rows are interchanged (R_i ↔ R_j)

(b) A row is multiplied by a non-zero constant, (Ri ↔ kR_i) where k ≠ 0

(c) A constant multiple of another row is added to a given row (Rᵢ ⟶ Rᵢ + kRⱼ), where i ≠ j.

Step-by-Step Methods to Find the Rank of Matrix?

Let's learn how to find the rank of a matrix using an example. Consider matrix A of size 4 × 4.

$A=\begin{bmatrix} 2 & 3 & 0 & 1 \\1 & 0 & 1 & 2 \\-1 & 1 & 1 & -2 \\1 & 5 & 3 & -1\end{bmatrix} $

Method 1:

First, perform elementary row operations on matrix A.

$A=\begin{bmatrix} 2 & 3 & 0 & 1 \\1 & 0 & 1 & 2 \\-1 & 1 & 1 & -2 \\1 & 5 & 3 & -1\end{bmatrix} $

Step 1: Applying elementary row operations, Exchange row 1 with row 2: R1 ↔ R2.

$A \sim\begin{bmatrix} 1 & 0 & 1 & 2 \\2 & 3 & 0 & 1 \\-1 & 1 & 1 & -2 \\1 & 5 & 3 & -1\end{bmatrix} $

Modify rows 2, 3, and 4 accordingly: R2 → R2 + (-2)R1, R3 → R3 + R1, R4 → R4 + (-1)R1.

$A \sim\begin{bmatrix} 1 & 0 & 1 & 2 \\0 & 3 & -2 & -3 \\0 & 1 & 2 & 0 \\0 & 5 & 2 & -3\end{bmatrix} $

Now, exchange row 2 with row 3: R2 ↔ R3.

$A \sim\begin{bmatrix} 1 & 0 & 1 & 2 \\0 & 1 & 2 & 0 \\ 0 & 3 & -2 & -3 \\0 & 5 & 2 & -3\end{bmatrix} $

Adjust rows 3 and 4: R3 → R3 + (-3)R2, R4 → R4 + (-5)R2.

$A \sim\begin{bmatrix} 1 & 0 & 1 & 2 \\0 & 1 & 2 & 0 \\0 & 0 & -8 & -3 \\0 & 0 & -8 & -3\end{bmatrix} $

Finally, Apply R3 → (-⅛)R3 and R4 → R4 + (8)R3.

$A \sim\begin{bmatrix} 1 & 0 & 1 & 2 \\0 & 1 & 2 & 0 \\0 & 0 & 1 & 3 / 8 \\0 & 0 & 0 & 0\end{bmatrix} $

$\mathrm{B} \text { (let) }$

The resulting matrix is denoted as B.

Therefore, B is the row reduced echelon form of A. So, ⍴(A), the rank of A, is 3, as there are 3 non-zero rows in B.

Method 2:

For finding non-singular minors of the given matrix A.

$A=\begin{bmatrix} 2 & 3 & 0 & 1 \\ 1 & 0 & 1 & 2 \\ -1 & 1 & 1 & -2 \\ 1 & 5 & 3 & -1 \end{bmatrix} $

When we check its determinant (|A|), it turns out to be zero. This means that A is a singular matrix.

However, if we focus on a specific part of A, which we call a minor,

$\begin{aligned}& \left|\begin{array}{ccc}2 & 3 & 0 \\1 & 0 & 1 \\-1 & 1 & 1\end{array}\right| \\& =-8 \neq 0\end{aligned}$

We find that it's not zero; in fact, it's -8.

This tells us that there is a non-singular part in A.

The highest level or order of this non-singular part is 3.

So, the conclusion is that the rank of A (denoted by ⍴(A)) is 3.

Solved Problems:

Ques: Find the Rank of a Matrix Using the Echelon Form.

\[\begin{bmatrix}1 &2  &3 \\ 2 & 3 & 4\\ 3 & 5 & 7\end{bmatrix}\]

Sol: First we will convert the given matrix into Echelon form and then find a number of non zero rows.

The order of A is 3 × 3. Hence ρ(A) ≤ 3

A = \[\begin{bmatrix}1 &2  &3 \\ 2 & 3 & 4\\ 3 & 5 & 7\end{bmatrix}\]

Convert R₂ ⟶ R₂ - 2R₁ and R₃ ⟶ R₃ - 3R₁

~ \[\begin{bmatrix}1 &2  &3 \\ 0 & -1 & -2\\ 0 & -1 & -2\end{bmatrix}\]

Again R₃ ⟶ R₃ - R₂

~ \[\begin{bmatrix}1 &2  &3 \\ 0 & -1 & -2\\ 0 & 0 & 0\end{bmatrix}\]

Now, the above matrix is in echelon form. In this number non zero rows is 2. Hence rank of matrix 2.

Properties of Rank of Matrix

Some important properties to remember when it comes to matrices are:

1. If you have a matrix A and its rank is less than its order, then A is a singular matrix.

2. The rank of a Null Matrix (a matrix filled with zeros) is always zero.

3. For an Identity Matrix I, its rank is the same as its order.

4. The rank of a matrix with dimensions m × n is the minimum of m and n.

5. If you take the transpose (flipping rows and columns) or the tranjugate (complex conjugate transpose) of a matrix A, the rank remains the same.

6. Two matrices that are row equivalent have the same rank.

7. When you add two matrices A and B, the rank of the result is less than or equal to the sum of their individual ranks. (⍴(A + B) ≤ ⍴(A) + ⍴(B))

8. When you subtract matrix B from A, the rank of the result is greater than or equal to the difference of their individual ranks. (⍴(A – B) ≥ ⍴(A) – ⍴(B))

9. For matrices A and B that can be multiplied, the rank of their product AB is less than or equal to the minimum of their individual ranks. (⍴(AB) ≤ min{⍴(A), ⍴(B)})

10. If you have a square matrix A of order m, then the nullity of A is m minus its rank (m – ⍴(A)).

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Conclusion: 

From the above discussion, we can conclude that if we have to find the rank of a matrix by searching the highest order non-vanishing minor is quite tedious when the order of the matrix is quite large. There is another easy method to find the rank of a matrix even if the order of the matrix is quite high. This method is used to find the rank of an equivalent row-echelon form of the matrix. If a matrix is in row-echelon form, then all entries below the leading diagonal (it is the line joining the positions of the diagonal elements like a₁₁, a₂₂, a₃₃ of the matrix) are zeros. So, to check whether a minor is zero or not is quite simple.