# Terminology and Properties of Matrices

Go back to  'Determinants and Matrices'

Before we proceed, lets get acquanted with some terminology.

 $$(1) \quad Zero\; matrix\qquad:$$ All elements of the matrix are zero. $$(2) \quad\; Row \;matrix\qquad:$$ Also called a row vector, such a matrix has only row. Eg. $$[1\quad-1\quad2\qquad4]$$ $$(3) \quad Column \; matrix\qquad:$$ Also called a column vector, such a matrix has only one column. Eg.\quad\left[ \begin{align} \; \ \ 1 \\ \; \ \ 3 \\ \; -1 \\ \end{align} \right] $$(4) \quad Diagonal \;matrix\qquad:$$ All non-diagonal elements are zero. $$\text{Eg.}\left[ \ \begin{matrix} 1 & 0 & 0 \\ 0 & -1 & 0 \\ 0 & 0 & 2 \\ \end{matrix}\ \right]$$ $$(5) \quad Upper / lower \;triangular\; matrix\qquad:$$ All elements below / above the diagonal are zero. \begin{align} & \text{Eg.}\,\,\,\left[ \ \begin{matrix} 1 & 2 & -1 \\ 0 & 3 & \ \ 4 \\ 0 & 0 & \ \ 1 \\ \end{matrix}\ \right]\left[ \ \begin{matrix} 1 & 0 & 0 \\ -2 & 4 & \ \ 0 \\ -1 & 7 & \ \ 1 \\\end{matrix}\ \right] \\& \qquad\qquad U-T\ \qquad\qquad L-T \\ \end{align} $$(6) \quad Singular\; matrix \qquad:$$ A matrix where determinant is zero. $$(7) \quad Transpose \qquad:$$ For a matrix A, the transpose of A, which we represent as AT, is obtained by interchanging the rows and columns of A. Example, $$A=\left[ \ \begin{matrix} 2 & 4 \\ 1 & 1 \\ 3 & 7 \\\end{matrix}\ \right]\quad\Rightarrow\qquad {{A}^{T}}=\left[ \ \begin{matrix} 2 & 1 & 3 \\ 4 & 1 & 7 \\\end{matrix}\ \right]$$ Note that $${{\left( {{A}^{T}} \right)}^{T}}=A$$

Example - 24

For two matrices A and B whose product is defined, prove that $${{\left( AB \right)}^{T}}={{B}^{T}}{{A}^{T}}.$$

Solution: Let A be of order $$m\times n$$ and B of order $$n\times p.$$ Then AB is of order $$m\times p.$$ Now $${{A}^{T}}$$ is of order $$n\times m\ \text{and}\ {{B}^{T}}$$ is of order $$p\times n.$$ So $${{B}^{T}}{{A}^{T}}$$ is of order $$p\times m$$ which is the same as the order of $${{\left( AB \right)}^{T}}.$$

Let us understand this using a concrete example:

\begin{align}A&=\left[ \begin{matrix} a & b \\ c & d \\\end{matrix} \right]B=\left[ \ \begin{matrix} p & q & r \\ x & y & z \\\end{matrix}\ \right]\\\\\Rightarrow \left( AB \right)&=\left[ \ \begin{matrix} ap+bx & aq+by & ar+bz \\ cp+dx & cq+dy & cr+dz \\\end{matrix}\ \right]\\\\\Rightarrow {{\left( AB \right)}^{T}}&=\left[ \ \begin{matrix} ap+bx & cp+dx \\ \begin{gathered} aq+by \\ ar+bz \\ \end{gathered} & \begin{gathered} cq+dy \\ cr+dz \\ \end{gathered} \\\end{matrix}\ \right]\end{align}

Also,

${{B}^{T}}{{A}^{T}}=\left[ \begin{matrix} p & x \\ q & y \\ r & z \\\end{matrix} \right]\ \ \left[ \begin{matrix} a & c \\ b & d \\\end{matrix} \right]\ \ =\left[ \begin{matrix} ap+bx & cp+dx \\ \begin{gathered} aq+by \\ ar+bz \\ \end{gathered} & \begin{gathered} cq+dy \\ cr+dz \\ \end{gathered} \\\end{matrix} \right]$

Focus on a particular term, say $$ap+bx,$$ and note how it is generated at the position in both $${{\left( AB \right)}^{T}}\ \text{and}\ {{B}^{T}}{{A}^{T}}.$$

Now let us generalize this.

Consider the term in $${{\left( AB \right)}^{T}}$$ at the position $$\left( i,j \right)$$ say t. In AB, this same term is at the position $$\left( j.\ i \right).$$ So, t is generated from $${{R}_{j}}\ \text{in}\ A\ \text{and}\ {{C}_{i}}\ \text{in}\ B$$:

$t=\sum\limits_{k=1}^{n}{{{a}_{jk}}{{b}_{ki}}}$

Let us now consider the term at the position $$\left( i,j \right)\ \ \text{in}\ \ {{B}^{T}}{{A}^{T}}.$$ This will be generated from $${{R}_{i}}\ \text{in}\ {{B}^{T}}\ {{C}_{j}}\ \text{in}\ {{A}^{T}},$$ $${{C}_{i}}\ \text{in}\ B\ \text{and}\ {{R}_{j}}\ \text{in}\ A,$$ which will generate t again. Thus,

${{\left( AB \right)}^{T}}={{B}^{T}}{{A}^{T}}$

Example -25

A square matrix A is said to be symmetric if $$A={{A}^{T}},$$ and skew-symmetric if $$A=-{{A}^{T}}.$$ Show that every square matrix can be expressed as the sum of a symmetric and a skew-symmetric matrix.

Solution: If A be the given matrix, then A can be written as

$A=\underbrace{\frac{1}{2}\left( A+{{A}^{T}} \right)}_{\text{Symmetric}}\ \ +\ \ \underbrace{\frac{1}{2}\left( A-{{A}^{T}} \right)}_{\text{Skew - Symmetric}}$

$$A+{{A}^{T}}$$ is symmetric because $${{\left( A+{{A}^{T}} \right)}^{T}}={{A}^{T}}+{{\left( {{A}^{T}} \right)}^{T}}={{A}^{T}}+A.$$

$$A-{{A}^{T}}$$ skew symmetric because $${{\left( A-{{A}^{T}} \right)}^{T}}={{A}^{T}}-A=-\left( A-{{A}^{T}} \right)$$

For example, let

$A=\left[ \ \begin{matrix} 1 & -1 & \ \ 0 \\ 3 & \ 2 & \ \ 6 \\ 2 & \ 4 & -3 \\\end{matrix}\ \right]$

Then,

$X=\frac{1}{2}\left( A+{{A}^{T}} \right)=\left[ \ \begin{matrix} 1 & 1 & \ \ 1 \\ 1 & \ 2 & \ \ 5 \\ 1 & \ 5 & -3 \\\end{matrix}\ \right]Y=\frac{1}{2}\left( A-{{A}^{T}} \right)=\left[ \ \begin{matrix} 0 & -2 & \ -1 \\ 2 & \ 0 & \ \ \ \ 1 \\ 1 & \ -1 & \ \ \ \ 0 \\\end{matrix}\ \right]$

Note that $$X+Y=A$$

Example -26

Prove that

(a) $$\widetilde{AB}=\widetilde{B}\widetilde{A}.$$              (b) $${{\left( {{A}^{T}} \right)}^{-1}}={{\left( {{A}^{-1}} \right)}^{T}}.$$

Solution: (a) The solution will involve evaluating $$AB\left( \widetilde{B}\widetilde{A} \right)\ \text{and}\ \widetilde{B}\widetilde{A}\left( AB \right):$$

\begin{align}AB\left( \widetilde{B}\widetilde{A} \right)\ &=\ A\left( B\widetilde{B} \right)\widetilde{A}\\&=A\left( \left| B \right|I \right)\widetilde{A}\\&=\left| B \right|\left( A\widetilde{A} \right)\\&=\left| B \right|\left| A \right|=\left| AB \right|=AB\left( \widetilde{A}B \right)\end{align}

Similarly,

$\widetilde{B}\widetilde{A}\left( AB \right)=\widetilde{AB}\left( AB \right)$

This must mean that $$\widetilde{AB}=\widetilde{B}\widetilde{A}$$

(b) Since

\begin{align} &\qquad \qquad A{{A}^{-1}}={{A}^{-1}}A=I, \\ & \qquad \qquad{{\left( A{{A}^{-1}} \right)}^{T}}={{\left( {{A}^{-1}}A \right)}^{T}}=I \\ & \Rightarrow \qquad {{\left( {{A}^{-1}} \right)}^{T}}{{A}^{T}}={{A}^{T}}{{\left( {{A}^{-1}} \right)}^{T}}=I \\ & \Rightarrow \qquad {{\left( {{A}^{T}} \right)}^{-1}}={{\left( {{A}^{-1}} \right)}^{T}} \\ \end{align}

## TRY YOURSELF

A square matrix A is said to be orthogonal if  $${{A}^{T}}A=I.$$ For example,

$A=\frac{1}{3}\left[ \ \begin{matrix} -1 & 2 & -2 \\ -2 & 1 & \ 2 \\ \ 2 & 2 & \ 1 \\\end{matrix}\ \right]\ \ \text{and}\ {{A}^{T}}=\frac{1}{3}\left[ \ \begin{matrix} -1 & -2 & 2 \\ \ \ 2 & 1 & \ 2 \\ -2 & 2 & \ 1 \\\end{matrix}\ \right]\ .$

Verify that $${{A}^{T}}A=I.$$

Q1.      Show that the transpose of an orthogonal matrix is also orthogonal

Q2.     Show that every orthogonal matrix is non-singular

Q3.     If A is orthogonal, show that $$\left| A \right|=\pm 1$$

Q4.     Show that the product of two orthogonal matrices is also orthogonal

Q5.     Show that the inverse of an orthogonal matrix is also orthogonal.