# Multiplying by invertible matrices maintains similarity

I know that given any two similar matrices $A,B\in M^F_{n\times n}$ and any two invertible matrices $P,Q$ of the same order then $P^{-1}AP$ is similar to $Q^{-1}BQ$.

However I don't completely understand why. Could someone explain it?

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We use the fact that if $A_1$ is similar to $A_2$ and $A_2$ to $A_3$ then $A_1$ is similar to $A_3$. To see that, write $A_2=P^{-1}A_1P$ and $A_3=Q^{-1}A_2Q$. Then $$A_3=Q^{-1}P^{-1}A_1PQ=(PQ)^{-1}A_1PQ$$ and $PQ$ is invertible.

Now, to solve the problem, note that $P^{-1}AP$ is similar to $A$ and $Q^{-1}BQ$ is similar to $B$.

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So this is a result of similarity being an equivalence relation? Specifically from it's transitivity? –  Robert S. Barnes Jun 24 '12 at 16:17
Exactly. We just need transitivity. –  Davide Giraudo Jun 24 '12 at 16:26

if $A \sim B$ then there exists an invertible $R$ such that $A=R^{-1}BR$ hence given any invertible matrices $P$ and $Q$ we have that $$P^{-1}AP=(P^{-1}R^{-1}Q)Q^{-1}BQ(Q^{-1}RP)$$ which implies that $P^{-1}AP\sim Q^{-1}BQ$

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