Question

Define a relation $R$ over a class of $n\times n$ real matrices $A$ and $B$ as $“ARB$ iff there exists a non-singular matrix $P$ such that $PA{P}^{–1}=B”.$ Then which of the following is true ?

• A. $R$ is reflexive, symmetric but not transitive
• B. $R$ is symmetric, transitive but not reflexive
• C. $R$ is an equivalence relation
• D. $R$ is reflexive, transitive but not symmetric

Answer 1

The correct option is C $R$ is an equivalence relation

For reflexive,
$B=PB{P}^{-1}$
which is true for $P=I$
$(B,B)\in R$
$\therefore R$ is reflexive.

For symmetry,
let $(A,B)\in R$ for matrix $P$
$\Rightarrow B=PA{P}^{-1}$
$\Rightarrow P^{-1}B=P^{-1}PA{P}^{-1}$
$\Rightarrow P^{-1}BP=IA{P}^{-1}P=IAI$
$\Rightarrow P^{-1}BP=A$
or $A=P^{-1}BP$
$\therefore (B,A)\in R$ for matrix $P^{-1}$
$\therefore R$ is symmetric.

For transitivity,
let $(A,B)\in R$ for matrix $P$ and $(B,C)\in R$ for matrix $Q$
$\Rightarrow B=PA{P}^{-1}$ and $C=QB{Q}^{-1}$
$\Rightarrow C=Q\left(PA{P}^{-1}\right)Q^{-1}$
$\Rightarrow C=\left(QP\right)A\left(P^{-1}{Q}^{-1}\right)$
$\Rightarrow C=\left(QP\right)A(QP{)}^{-1}$
$\therefore (A,C)\in R$ for matrix $QP$
$\therefore R$ is transitive.

So, $R$ is an equivalence relation.