Rank of a Matrix and Some Special Matrices

Table of Contents

• Introduction to Matrices
• Types of Matrices
• Matrix Operations
• Adjoint and Inverse of a Matrix
• Rank of a Matrix and Special Matrices
• Solving Linear Equations using Matrix

How to Determine the Rank of a Matrix?

If \(A={{\left( {{a}_{ij}} \right)}_{m\,\,\times n}}\) is a matrix, and B is its sub-matrix of order r, then\(\left |\beta \right |\) the determinant is called r-rowed minor of A.

Definition:

Let \(A={{\left( {{a}_{ij}} \right)}_{m\,\,\times n}}\) be a matrix. A positive integer r is said to be a rank of A if

• A possesses at least one r-rowed minor which is different from zero; and
• Every (r + 1) rowed minor of A is zero.

From (ii), it automatically follows that all minors of higher order are zeros. We denote rank of A by \(\rho \left( A \right)\)

Note:

The rank of a matrix does not change when the following elementary row operations are applied to the matrix:

(a) Two rows are interchanged \(\left( {{R}_{i}}\leftrightarrow {{R}_{j}} \right);\)

(b) A row is multiplied by a non-zero constant, \(\left( {{R}_{i}}\to k{{R}_{i}},\,with\,k\ne 0 \right);\)

(c) A constant multiple of another row is added to a given row \(\left( {{R}_{i}}\to {{R}_{i}},+k{{R}_{j}} \right)\) where \(i\ne j.\)

Note: The arrow → means “replaced by”.

Note that the application of these elementary row operations does not change a singular matrix to a non-singular matrix nor does a non-singular matrix change to a singular matrix. Therefore, the order of the largest non-singular square sub-matrix is not affected by the application of any of the elementary row operations. Thus, the rank of a matrix does not change by the application of any of the elementary row operations. A matrix obtained from a given matrix by applying any of the elementary row operations is said to be equivalent to it. If A and B are two equivalent matrices, we write A ~ B. Note that if A ~ B, then ρ(A) = ρ(B)

By using the elementary row operations, we shall try to transform the given matrix in the following

Form\(\begin{pmatrix} 1 &* &* &* \\ 0& 1& * &* \\ 0& 0 & 1& *\\ .& . & . & .\\ .& . & . &. \\ .& .& .& .\\ 0& 0 &0.. &* \end{pmatrix}\).

Where * stands for zero or non-zero element. That is, we shall try to make a_ii as 1 and all the elements below a_ij as zero.

Illustration: For what values of x does the matrix \(\left[ \begin{matrix} 3+x & 5 & 2 \\ 1 & 7+x & 6 \\ 2 & 5 & 3+x \\ \end{matrix} \right]\) have the rank 2?

Solution:

The given matrix has only one 3rd-order minor. In order that the rank arrive at 2, we must bring about its determinant to zero. Hence, by applying the invariance method we can obtain values of x.

Now, using \({{R}_{1}}\to {{R}_{1}}-{{R}_{3}}\) \(\left| \begin{matrix} 3+x & 5 & 2 \\ 1 & 7+x & 6 \\ 2 & 5 & 3+x \\ \end{matrix} \right|=\left| \begin{matrix} 1+x & 0 & -1-x \\ 1 & 7+x & 6 \\ 2 & 5 & 3+x \\ \end{matrix} \right|;\;\;using\;{{C}_{3}}\to {{C}_{3}}+{{C}_{1}}=\left| \begin{matrix} 1+x & 0 & 0 \\ 1 & 7+x & 7 \\ 2 & 5 & 5+x \\ \end{matrix} \right|\) \(=\left( 1+x \right)\left| \begin{matrix} 7+x & 7 \\ 5 & 5+x \\ \end{matrix} \right|=\left( 1+x \right)\left[ \left( 7+x(5+x \right)-35 \right]=\left( 1+x \right)\left( {{x}^{2}}+12x \right)=x\left( 1+x \right)\left( x+12 \right)\)

(i) holds for \(x=0,\,\,-1,\,\,-12\)

When x = 0 the matrix \(=\left[ \begin{matrix} 3 & 5 & 2 \\ 1 & 7 & 6 \\ 2 & 5 & 3 \\ \end{matrix} \right]\)

Clearly, a minor \(\left[ \begin{matrix} 3 & 5 \\ 1 & 7 \\ \end{matrix} \right]\ne 0,\)

So, the rank = 2

The matrix has the rank 2 if x=0,-1,-12.

Positive Integral Powers of a Square Matrix

The positive integral powers of a matrix A are defined only when A is a square matrix.

Also then, \({{A}^{2}}=A.A;{{A}^{3}}=A.A.A={{A}^{2}}A.\) Alsofor any positive integers m, n:

Matrix Polynomial

If \(f\left( x \right)={{a}_{0}}{{x}^{n}}+{{a}_{1}}{{x}^{n}}+{{a}_{2}}{{x}^{n-2}}+…….+{{a}_{n}}{{x}^{0}},\) then we define a matrix polynomial a, b \(f\left( A \right)={{a}_{0}}{{A}^{n}}+{{a}_{1}}{{A}^{n-1}}+{{a}_{2}}{{A}^{n-2}}+……+{{a}_{n}}{{I}_{n}}\) where A is the given square matrix. If f(A) is a null matrix, then A is called the zero or root of the matrix polynomial f(A).

Illustration: Let \(A=\left[ \begin{matrix} 2 & 0 & 1 \\ 2 & 1 & 3 \\ -1 & -1 & 0 \\ \end{matrix} \right]\) and \(f (x)={x}^{2}-5x+6{I}_{3}\). Find f (A).

Solution:

By using methods of multiplication and addition of matrices we will obtain the required result. Here \(f\left( A \right)={{A}^{2}}-5A+5{{I}_{3}}\) \(={{\left[ \begin{matrix} 2 & 0 & 1 \\ 2 & 1 & 3 \\ -1 & -1 & 0 \\ \end{matrix} \right]}^{2}}-5\left[ \begin{matrix} 2 & 0 & 1 \\ 2 & 1 & 3 \\ -1 & -1 & 0 \\ \end{matrix} \right]+6\left[ \begin{matrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \\ \end{matrix} \right]\) \(=\left[ \begin{matrix} 2 & 0 & 1 \\ 2 & 1 & 3 \\ -1 & -1 & 0 \\ \end{matrix} \right]\times \left[ \begin{matrix} 2 & 0 & 1 \\ 2 & 1 & 3 \\ -1 & -1 & 0 \\ \end{matrix} \right]-\left[ \begin{matrix} 10 & 0 & 5 \\ 10 & 5 & 15 \\ -5 & -5 & 0 \\ \end{matrix} \right]+\left[ \begin{matrix} 6 & 0 & 0 \\ 0 & 6 & 0 \\ 0 & 0 & 6 \\ \end{matrix} \right]\) \(=\left[ \begin{matrix} 2\times 2+0\times 2+1\left( -1 \right) & 2\times 0+0\times 1+1\left( -1 \right) & 2\times 1+0\times 3+1\times 0 \\ 2\times 2+1\times 2+3\left( -1 \right) & 2\times 0+1\times 1+3\left( -1 \right) & 2\times 1+1\times 3+3\times 0 \\ \left( -1 \right)2+\left( -1 \right)2+0\left( -1 \right) & \left( -1 \right)0+\left( -1 \right)1+0\left( -1 \right) & \left( -1 \right)1+\left( -1 \right)3+0\times 0 \\ \end{matrix} \right]+\left[ \begin{matrix} 6-10 & 0-0 & 0-5 \\ 0-10 & 6-5 & 0-15 \\ 0-\left( -5 \right) & 0-\left( -15 \right) & 6-0 \\ \end{matrix} \right]\) \(=\left[ \begin{matrix} 3 & -1 & 2 \\ 3 & -2 & 5 \\ -4 & -1 & -4 \\ \end{matrix} \right]+\left[ \begin{matrix} -4 & 0 & -5 \\ -10 & 1 & -15 \\ 5 & 5 & 6 \\ \end{matrix} \right]=\left[ \begin{matrix} 3-4 & -1+0 & 2-5 \\ 3-10 & -2+1 & 5-15 \\ -4+5 & -1+5 & -4+6 \\ \end{matrix} \right]=\left[ \begin{matrix} -1 & -1 & -3 \\ -7 & -1 & -10 \\ 1 & 4 & 2 \\ \end{matrix} \right]\)

Illustration: Let \(A =\left[ \begin{matrix} a & b \\ c & d \\ \end{matrix} \right]\) be such that

then \({A}^{3}=0, \;\;but\;\;{{A}^{3}}\ne 0\)

Solution:

(a) As \({{A}^{3}}=0, we \;get \left| {{A}^{3}} \right|=0; \left| {{A}^{3}} \right|=0\Rightarrow \left| A \right|=0\Rightarrow ad-bc=0\)

In this problem, \({A}^{3}=0\) means | A | also is equal to 0. Therefore, by calculating \({A}^{2}\\) we can obtain the result.

Also, \({{A}^{3}}=\left( \begin{matrix} {{a}^{2}}+bc & \left( a+d \right)b \\ \left( a+d \right)c & bc+{{d}^{2}} \\ \end{matrix} \right)=\left( \begin{matrix} {{a}^{2}}+ad & \left( a+d \right)b \\ \left( a+d \right)c & ad+{{d}^{2}} \\ \end{matrix} \right)=\left( a+d \right)A\) \(If \;a+d=0,\; we get \;{{A}^{2}}=0.\;But,\; if \;a+d\ne 0,\; then \;{{A}^{3}}={{A}^{2}}A=\left( a+d \right){{A}^{2}}\Rightarrow 0=\left( a+d \right){{A}^{2}}\Rightarrow {{A}^{2}}=0\)

Illustration: If \(A=\left[ \begin{matrix} 1 & 0 \\ 1 & 1 \\ \end{matrix} \right] \;and\; I=\left[ \begin{matrix} 1 & 0 \\ 0 & 1 \\ \end{matrix} \right]\)

then which one of the following holds for all \(n\ge 1,\) by the principle of mathematical induction.

\((c) {{A}^{n}}=nA-\left( n-1 \right)I \;\;\;\;\;\;\;\;\;\;\;\;\;\; (d) {{A}^{n}}={{2}^{n-1}}A-\left( n-1 \right)I\)Solution:

By substituting n = 2 we can determine the correct answer.

For \(n=2,\;RHS \;of \;\;\;(b) =2A+I\ne {{A}^{2}}\) So possible answer is (c) or (d)

In fact \({{A}^{n}}=\left[ \begin{matrix} 1 & 0 \\ n & 1 \\ \end{matrix} \right]\; which \;equals \;nA-\left( n-1 \right)I;\)

Alternatively. Write A = I + B Where \(B=\left[ \begin{matrix} 0 & 0 \\ 1 & 0 \\ \end{matrix} \right]\) \(As \;{{B}^{2}}=0, we\; get\; {{B}^{r}}=0\,\,\forall r\ge 2\)

By the binomial theorem, \({{A}^{n}}=I+nB=I+n\left( A-I \right)=n\left( n-1 \right)I\)