# Rank of a Matrix and Some Special Matrices

A rectangular array of m x n numbers in the form of m rows and n columns, is called a matrix of order m by n, written as m x n matrix. This article gives an idea of the rank of a matrix and some special matrices.

## 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 aii as 1 and all the elements below aij 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.

$\left| \begin{matrix} 3+x & 5 & 2 \\ 1 & 7+x & 6 \\ 2 & 5 & 3+x \\ \end{matrix} \right|=0$

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

$When \; x=-1, the\; matrix \;=\left[ \begin{matrix} 2 & 5 & 2 \\ 1 & 6 & 6 \\ 2 & 5 & 2 \\ \end{matrix} \right] \;Clearly,\; a\; minor\; \left[ \begin{matrix} 2 & 5 \\ 1 & 6 \\ \end{matrix} \right]\ne 0, So\; the\; rank = 2$ $When \;x=-12,\; the\; matrix\; =\left[ \begin{matrix} -9 & 5 & 2 \\ 1 & -5 & 6 \\ 2 & 5 & -9 \\ \end{matrix} \right] Clearly,\; a\; minor \left[ \begin{matrix} -9 & 5 \\ 1 & -5 \\ \end{matrix} \right]\ne 0,So\; the\; rank = 2$

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:

$(a) {{A}^{m}}{{A}^{n}}={{A}^{m+n}} \;\;\;\;\; (b) {{\left( {{A}^{m}} \right)}^{n}}={{A}^{mn}}={{\left( {{A}^{n}} \right)}^{m}}\;\;\;\;\; (c){{I}^{n}}=I,{{I}^{m}}=I\;\;\;\;\; (d) {{A}^{0}}={{I}_{n}}$

### 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 A2 we can obtain the result.

$(a) {{A}^{2}}=0 \;\;\;\; (b) {{A}^{2}}=A \;\;\;\; (c){{A}^{2}}=I-A$ (d) None of these

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.

$(a) {{A}^{n}}=nA+\left( n-1 \right)I \;\;\;\;\;\;\;\;\;\;\; (b) {{A}^{n}}={{2}^{n-1}}A+\left( n-1 \right)I$

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

Solution:

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

$For n=2,{{A}^{2}}=\left[ \begin{matrix} 1 & 0 \\ 0 & 1 \\ \end{matrix} \right] \;For \;n = 2, \;RHS \;of\; (a) =2A+I=3\left[ \begin{matrix} 3 & 0 \\ 2 & 3 \\ \end{matrix} \right]\ne {{A}^{2}}$

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, ${{A}^{n}}=I+nB=I+n\left( A-I \right)=n\left( n-1 \right)I$