Why is "similarity" more specific than "equivalence"? At least regarding matrices, we have


*

*$A$ is similar to $B$ if $\exists S: B=S^{-1}AS$

*$A$ is equivalent to $B$ if $\exists P,Q: B=Q^{-1}AP$


I am confused about the usage of the terms "similar" and "equivalent". I would have thought that equivalence is more specific than similarity, i.e. equivalence is a special case (subset) of similarity. However, the above indicates the other way around.
How can this be explained? Is this specific to linear algebra, or is this a math-wide phenomenon?
 A: They are simply different concepts.
Similarity is a property related to endomorphisms on a vector space $V$. Two matrices $A, B$ are similar if they represent the same endomorphism $f: V \to V$ on picking bases $\mathcal{B}_A, \mathcal{B}_B$ of $V$ and taking the map with respect to the two bases respectively.
Equivalence is a property related to linear maps between pairs of spaces $V, W$. Two matrices (representing maps $f, g: V \to W$) are equivalent if we can pick an "input basis and an output basis" for each of $f, g$ such that $f, g$ have the same matrices with respect to those two pairs of bases respectively.
Both "similarity" and "equivalence" are equivalence relations, but they're on different spaces. Similarity is a sub-relation of equivalence.
In general, I'm not aware of other instances where "similarity" and "equivalence" relate to each other at all.
A: $A$ and $B$ are similar  if they represent the same endomorphism $f$ of a finite dimensional $K$-vector space $E$ in different bases. Thus similar matrices are equivalent, but  the converse is false:
They're similar if and only if they have the same Jordan normal form, and also if and only  if they have the same similarity invariants, which the invariant factors of $E$, seen as a $K[X]$-module  through $f$ (i. e. for any $v\in E$, $\;X\cdot v=f(v)$).
On another hand, $A$ and $B$ are equivalent if and only if $\;\operatorname{rank}A=\operatorname{rank}B$.
