Define a relation $R$ over a class of $n imes n$ real matrices $A$ and $B$ as "$ARB$ iff there exists a non-singular matrix $P$ such that $PAP^{-1} = B$". Then which of the following is true?
(A)
$R$ is reflexive, transitive but not symmetric
(B)
$R$ is symmetric, transitive but not reflexive.
(C)
$R$ is reflexive, symmetric but not transitive
(D)
$R$ is an equivalence relation
MEDIUM
Correct Answer:D
Explanation
For reflexive relation, $\forall (A, A) \in R$ for matrix $P$. $\Rightarrow A = PAP^{-1}$ is true for $P = I$. So, $R$ is reflexive relation.
For symmetric relation, Let $(A, B) \in R$ for matrix $P$. $\Rightarrow A = PBP^{-1}$. After pre-multiply by $P^{-1}$ and post-multiply by $P$, we get $P^{-1}AP = B$. So, $(B, A) \in R$ for matrix $P^{-1}$. So, $R$ is a symmetric relation.
For transitive relation, Let $ARB$ and $BRC$. So, $A = PBP^{-1}$ and $B = PCP^{-1}$. Now, $A = P(PCP^{-1})P^{-1} \Rightarrow A = (P)^2C(P^{-1})^2 \Rightarrow A = (P)^2 \cdot C \cdot (P^2)^{-1}$. $\therefore (A, C) \in R$ for matrix $P^2$. $\therefore R$ is transitive relation. Hence, $R$ is an equivalence relation.