Rank of a Matrix

Rank of a Matrix

To define the rank of a matrix, we have to know about sub-matrices and minors of a matrix.

Let A be a given matrix. A matrix obtained by deleting some rows and some columns of A is called a sub-matrix of A. A matrix is a sub-matrix of itself because it is obtained by leaving zero number of rows and zero number of columns.

Recall that the determinant of a square sub-matrix of a matrix is called a minor of the matrix.

Definition 1.6

The rank of a matrix A is defined as the order of a highest order non-vanishing minor of the matrix A. It is denoted by the symbol ρ (A). The rank of a zero matrix is defined to be 0.

Note

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

iv.                     If A is an m × n matrix, then ρ (A) ≤ min{m, n} = minimum of m, n.

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

Example 1.15

Find the rank of each of the following matrices:

Solution

(i) Let A =. Then A is a matrix of order 3× 3. So ρ(A) ≤ min {3, 3}  = 3. The highest  order of minors of A is 3 . There is only one third order minor of A .

 It is  =  3 (6− 6) − 2 (6−6) +  5 (3 − 3) = 0. So, ρ(A) < 3.

 Next consider the second-order minors of A .

We find that the second order minor  = 3 − 2 = 1 ≠ 0 . So ρ(A) = 2 .

(ii) Let A = . Then A is a matrix of order 3×4 . So ρ(A) ≤ min {3, 4}  = 3.

The highest order of minors of A is 3 . We search for a non-zero third-order minor of A . But

we find that all of them vanish. In fact, we have

 So, ρ(A) < 3. Next, we search for a non-zero second-order minor of A .

We find that  = -4+9 =5 ≠ 0 . So, ρ(A) = 2 .

Remark

Finding the rank of a matrix by searching a highest order non-vanishing minor is quite tedious when the order of the matrix is quite large. There is another easy method for finding the rank of a matrix even if the order of the matrix is quite high. This method is by computing 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 a₁₁ , a₂₂ , a₃₃ ,L. of the matrix) are zeros. So, checking whether a minor is zero or not, is quite simple.

Example 1.16

Find the rank of the following matrices which are in row-echelon form :

Solution

(i) Let A = . Then A is a matrix of order 3 3 × and ρ(A) ≤ 3

 The third order minor |A| =   = (2) (3)( 1) = 6 ≠ 0 . So, ρ(A) = 3 .

Note that there are three non-zero rows.

(ii) Let A = . Then A is a matrix of order 3× 3 and ρ(A) ≤ 3.

The only third order minor is |A| =  = (-2) (5) (0) = 0 . So ρ(A) ≤ 2 .

There are several second order minors. We find that there is a second order minor, for  example,  = (-2)(5) = -10 ≠ 0 . So, ρ(A) = 2.

Note that there are two non-zero rows. The third row is a zero row.

(iii) Let A = . Then A is a matrix of order 4 × 3 and ρ(A) ≤ 3.

The last two rows are zero rows. There are several second order minors. We find that there  is a second order minor, for example,  = (6) (2) = 12 ≠ 0 . So, ρ(A) = 2.

Note that there are two non-zero rows. The third and fourth rows are zero rows.

We observe from the above example that the rank of a matrix in row echelon form is equal  to the number of non-zero rows in it. We state this observation as a theorem without proof.

Theorem 1.11

The rank of a matrix in row echelon form is the number of non-zero rows in it.

The rank of a matrix which is not in a row-echelon form, can be found by applying the following result which is stated without proof.

Theorem 1.12

The rank of a non-zero matrix is equal to the number of non-zero rows in a row-echelon form of the matrix.

Example 1.17

Find the rank of the matrix  by reducing it to a row-echelon form.

Solution

Let A = . Applying elementary row operations, we get

The last equivalent matrix is in row-echelon form. It has two non-zero rows. So, ρ (A)= 2.

Example 1.18

Find the rank of the matrix  by reducing it to a row-echelon form.

Solution

Let A be the matrix. Performing elementary row operations, we get

The last equivalent matrix is in row-echelon form. It has three non-zero rows. So, ρ(A) = 3 .

Elementary row operations on a matrix can be performed by pre-multiplying the given matrix by a special class of matrices called elementary matrices.

Definition 1.7

An elementary matrix is defined as a matrix which is obtained from an identity matrix by applying only one elementary transformation.

Remark

If we are dealing with matrices with three rows, then all elementary matrices are square matrices of order 3 which are obtained by carrying out only one elementary row operations on the unit matrix I₃. Every elementary row operation that is carried out on a given matrix A can be obtained by pre-multiplying A with elementary matrix. Similarly, every elementary column operation that is carried out on a given matrix A can be obtained by post-multiplying Awith an elementary matrix. In the present chapter, we use elementary row operations only.

For instance, let us consider the matrix A = 

Suppose that we do the transformation R₂  → R₂ + λR₃ on A, where λ ≠ 0 is a constant. Then, we get

The matrix   is an elementary matrix, since we have 

Pre-multiplying A by   , we get

From (1) and (2), we get 

So, the effect of applying the elementary transformation R₂ → R₂ + λR₃ on A is the same as that of pre-multiplying the matrix A with the elementary matrix 

Similarly, we can show that

(i) the effect of applying the elementary transformation R₂↔ R₃ on A is the same as that of  pre-multiplying the matrix A with the elementary matrix 

(ii) the effect of applying the elementary transformation R₂ → R₂λ on A is the same as that of  pre-multiplying the matrix A with the elementary matrix 

We state the following result without proof.

Theorem 1.13

Every non-singular matrix can be transformed to an identity matrix, by a sequence of elementary row operations.

As an illustration of the above theorem, let us consider the matrix A = 

Then, |A| =  12+ 3 = 15 ≠  0. So, A is non-singular. Let us transform A into I₂ by a sequence of  elementary row operations. First, we search for a row operation to make a₁₁ of A as 1. The elementary row operation needed for this is R₁ → (1/2) R₁. The corresponding elementary matrix is 

Next, let us make all elements below a₁₁ of E₁A as 0. There is only one element a₂₁.

The elementary row operation needed for this is R₂  → R₂  + (−3) R₁ .

The corresponding elementary matrix is E₂ =

Next, let us make a₂₂ of E₂(E₂A) as 1. The elementary row operation needed for this is  

The corresponding elementary matrix is E₃ = 

Then, we get E₃(E₂(E₁A)) = 

Finally, let us find an elementary row operation to make a₁₂ of E₃(E₂(E₁A)) as 0. The elementary  row operation needed for this is R₁ → R₁ + (1/2) R₂. The corresponding elementary matrix is 

We write the above sequence of elementary transformations in the following manner:

Example 1.19

Show that the matrix  is non-singular and reduce it to the identity matrix by elementary row transformations.

Solution

Let A = .Then, |A| = 3 (0+2 ) – 1(2+5) + 4(4-0) = 6-7+16 ≠ 0. So, A is non-singular. Keeping the identity matrix as our goal, we perform the row operations sequentially on A as follows: