Proof that if $A$ is similar to $B$, then $B$ is similar to $A$

Question

$A$ is similar to $B$ if there is an invertible matrix $S$ such that $B = S^{-1}AS$.

Prove that if $A$ is similar to $B$, then $B$ is similar to $A$.

So if $A$ is similar to $B$ then $B = S^{-1}AS$ for some invertible matrix $S$. Then we have

$$SB=SS^{-1}AS$$

$$SB = AS$$

$$SBS^{-1}=ASS^{-1}$$

$$A=SBS^{-1}$$

Does this mean that $B$ is similar to $A$? By definition, $B$ is similar to $A$ if there is an invertible matrix such that $A = S^{-1}BS$, but this is not what the final result shows.

Answer 1

Observe that $$ A=SBS^{-1}=(S^{-1})^{-1}BS^{-1}=U^{-1}BU, $$ where $U=S^{-1}$.

In fact, similarity is an equivalence relation between square matrices of the same order.