Question

If A and B are orthogonal matrices then prove that AB and BA are also orthogonal​

Answer 1

$\green{\large\underline{\sf{Solution-}}}$

Given that

A is Orthogonal matrix.

$\bf :\longmapsto\: {AA}^{T} = I - - - (1)$

B is Orthogonal matrix.

$\bf :\longmapsto\: {BB}^{T} = I - - - (2)$

1. We have to prove that AB is Orthogonal.

So, it is sufficient to prove that

$\: \: \: \: \: \: \: \: \: \: {\red{\boxed{ \quad \sf \: (AB) {(AB)}^{T} = I \quad}}}$

Consider,

$\rm :\longmapsto\:(AB) {(AB)}^{T}$

$\: \: \rm = \: \: AB( {B}^{T} {A}^{T})$

$\: \: \rm = \: \: A(B{B}^{T}) {A}^{T}$

$\: \: \rm = \: \: A(I) {A}^{T}$

$\: \: \rm = \: \: A {A}^{T}$

$\: \: \rm = \: \:I$

$\bf\implies \:AB( {AB}^{T}) = I$

$\: \: \: \: \: \: \: \: \: \: \: \: \: \boxed{ \sf \: \bf\implies \:AB \: is \: orthogonal}$

2. Now, we have to prove that BA is Orthogonal.

So, it is sufficient to show that

$\: \: \: \: \: \: \: \: \: \: {\red{\boxed{ \quad \sf \: (BA) {(BA)}^{T} = I \quad}}}$

Consider,

$\rm :\longmapsto\:BA( {BA)}^{T}$

$\: \: \rm = \: \: BA( {A}^{T} {B}^{T})$

$\: \: \rm = \: \: B(A{A}^{T}){B}^{T}$

$\: \: \rm = \: \: B(I){B}^{T}$

$\: \: \rm = \: \: B{B}^{T}$

$\: \: \rm = \: \: I$

$\bf\implies \:BA( {BA}^{T}) = I$

$\: \: \: \: \: \: \: \: \: \: \: \: \: \boxed{ \sf \: \bf\implies \:BA \: is \: orthogonal}$

Additional Information :-

$\boxed{ \sf \: {(A + B)}^{T} = {A}^{T} + {B}^{T}}$

$\boxed{ \sf \: {(A - B)}^{T} = {A}^{T} - {B}^{T}}$

$\boxed{ \sf \: ( {xA)}^{T} = x {A}^{T} \: where \: x \: is \: real \: number}$

$\boxed{ \sf \: {(AB)}^{T} = {B}^{T} {A}^{T} }$

$\boxed{ \sf \: {( {A}^{T}) }^{T} = A}$

$\boxed{ \sf \: A \: is \: symmetric \implies \: {A}^{T} = A}$

$\boxed{ \sf \: A \: is \: skew - symmetric \implies \: {A}^{T} = - A}$

$\boxed{ \sf \: A \: is \: idempotent \implies \: {A}^{2} = A}$

$\boxed{ \sf \: A \: is \: involutary \implies \: {A}^{2} = I}$