Can somebody please explain me the difference between Linear transformations which epimorphism, isomorphism, endomorphism or automorphism. I would appreciate if somebbody can explain the idea with examples or guide to some good source to clear the concept.


For any algebraic structure, a homomorphism preserves the structure, and some types of homomorphisms are:

  • Epimorphism: a homomorphism that is surjective (AKA onto)
  • Monomorphism: a homomorphism that is injective (AKA one-to-one, 1-1, or univalent)
  • Isomorphism: a homomorphism that is bijective (AKA 1-1 and onto); isomorphic objects are equivalent, but perhaps defined in different ways
  • Endomorphism: a homomorphism from an object to itself
  • Automorphism: a bijective endomorphism (an isomorphism from an object onto itself, essentially just a re-labeling of elements)

So a linear transformation $A\colon\mathbb{R}^{n}\to\mathbb{R}^{m}$ is a homomorphism since it preserves the vector space structure (vector addition, scalar addition and multiplication, scalar multiplication of vectors), e.g. $A(av+w)=aA(v)+Aw$. It is an epimorphism if its image is $\mathbb{R}^{m}$, a monomorphism if it has zero kernel, an endomorphism if $n=m$, and an automorphism (as well as an isomorphism) if all of these are true.

The below figure might be helpful. More details here.

Types of mappings between sets

  • $\begingroup$ I'm new to math.stackexchange, can someone explain the downvote? I found that I could not embed the image without more reputation, but if I could it would have been self-explanatory I think. $\endgroup$ – Adam Marsh Feb 1 '18 at 3:47
  • $\begingroup$ Added more details to perhaps address the downvote. Also, I cannot comment on the above answer, but as @al-jebr mentions, an endomorphism need not be injective. $\endgroup$ – Adam Marsh Feb 1 '18 at 19:47

In linear algebra, an epimorphism between vector spaces is a surjective linear application $A:V_{1}\to V_{2}$, that is $\text{Im}(A)=V_{2}$. For example, $B:\mathbb{R}^{2}\to\mathbb{R}:(x,y)\mapsto x+y$.

In linear algebra, an endomorphism from a vector space to itself. For example, $B:\mathbb{R}\to\mathbb{R}:x\mapsto 2x$. Edited (30/08/2018): removed 'injective' and corrected the definition.

In linear algebra, an isomorphism between vector spaces is a both surjective and injective linear application $A:V_{1}\to V_{2}$, that is $\text{Ker}(A)=\{0_{V_{1}}\}$ and $\text{Im}(A)=V_{2}$. An automorphism is an isomorphism between a vector space and itself. For example, $B:\mathbb{R}\to\mathbb{R}:x\mapsto x$.

In a more general setting, a morphism $\phi$ between two groups $(G,\cdot)$ and $(H,\star)$ is an application $G\to H:g\mapsto \phi(g)$ such that, for all $g,g'\in G$, we have $\phi(g\cdot g')=\phi(g)\star\phi(g')$ and such that $\phi(e_{G})=e_{H}$ where $e_{I}$ is the identity element of $I=G,H$.

An epimorphism between such two groups is a surjective morphism $\phi:G\to H$, i.e. for any $h\in H$, there exists $g\in G$ such that $h=\phi(g)$.

An endomorphism between such two groups is an injective morphism $\phi:G\to H$, i.e. for all $g,g'\in G$ with $\phi(g)=\phi(g')$, it implies $g=g'$.

An isomorphism between two such groups is a both injective and surjective morphism. An automorphism is an isomorphism with $(H,\star)=(G,\cdot)$.

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    $\begingroup$ Dear Corzer thanks for the relpy, can you explain what does {0_v1} means $\endgroup$ – EngDR Nov 3 '15 at 19:13
  • $\begingroup$ I use the notation $0_{V_{1}}$ to denote the null vector from the vector space $V_{1}$. For example, if $V_{1}=\mathbb{R}^{2}$, $0_{V_{1}}=(0,0)$. If $V_{1}=\mathbb{R}^{5}$, $0_{V_{1}}=(0,0,0,0,0)$. $\endgroup$ – MoebiusCorzer Nov 3 '15 at 23:21
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    $\begingroup$ Endomorphisms are not necessarily injective. That is a monomorphism. Endomorphisms are maps from a mathematical object to itself. $\endgroup$ – Al Jebr Nov 16 '17 at 2:04

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