# How do I show this

Given invertible matrices $A,B$ and $P$ such that $A = PB$, then we say that $A$ is left equivalent to $B$. Show that left equivalence is indeed an equivalence relation.

• This isn't true as it is formulated. Correct would be: given $A,B$, if there exists $P$ such that $A=PB$... May 25, 2014 at 14:33
• You need to show that the relation is reflexive, symmetric and transitive. Which of these is a problem? May 25, 2014 at 14:35

Reflexivity is trivial: $A=IA$
Symmetry: If $A=PB$ and $P$ is invertible, then $B=P^{-1}A$
Transitivity: If $A=PB$ and $B=QC$ and $P,Q$ are invertible, then $PQ$ is also invertible and we have $A=PB=P(QC)=(PQ)C$
• Reflexive: for all invertible $A$, take $P\stackrel{\rm def}{=}I$.
• Symmetric: for $A,B$ invertible such that $A$ is left equivalent to $B$, let $P$ be an invertible matrix proving it: $A=PB$. Then $B=P^{-1}A$, and $B$ is left equivalent to $A$.
• Transitive: let $A,B,C$ be 3 invertible matrices such that $A$ is l.e. to $B$ and $B$ l.e. to $C$; and let $P,Q$ be two invertible matrices proving it: $A=PB$, $B=QC$. Then $R\stackrel{\rm def}{=}PQ$ is invertible, and $A=RC$ -- so $A$ is l.e. to $C$.