Question

Define a relation R over a class of n × n real matrices A and B as "ARB iff there exists a non-singular matrix P such that P A P⁻¹ = B". Then which of the following is true ?

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

Hint:

Matrices $A$ and $B$ related this way are called {Similar Matrices}. They share the same eigenvalues, determinant, and trace.

Answer 1

The Correct Option is D

Step 1: Reflexive: Let $P=I$. Then $I A I^{-1} = A$, so $ARA$.
Step 2: Symmetric: If $ARB$, then $B = PAP^{-1}$. Premultiply by $P^{-1}$ and postmultiply by $P$: $P^{-1}BP = A$. Let $Q = P^{-1}$. Then $QBQ^{-1} = A$, so $BRA$.
Step 3: Transitive: If $ARB$ and $BRC$, then $B=PAP^{-1}$ and $C=QBQ^{-1}$. $C = Q(PAP^{-1})Q^{-1} = (QP)A(P^{-1}Q^{-1}) = (QP)A(QP)^{-1}$. Let $M = QP$. Then $C = MAM^{-1}$, so $ARC$.
Step 4: Since it is reflexive, symmetric, and transitive, it is an equivalence relation.