Many linear maps compare one [vector space](/page/Vector%20Space) with another, but some of the richest questions begin when the source and target are the same. Then a map can be iterated, represented by a square matrix, tested for fixed points and eigenvectors, and inserted into polynomials. This is why finite-dimensional linear algebra treats maps $T: V \to V$ as objects in their own right rather than merely as arrows between spaces. For a finite-dimensional vector space $V$ over a field $F$, these self-maps form the endomorphism algebra $\operatorname{End}_F(V)$. It packages projections, nilpotent operators, changes of basis, kernels, images, and matrix similarity into one setting. The same idea also appears more broadly in [Function](/page/Function), [Linear Map](/page/Linear%20Map), [Group Homomorphism](/page/Group%20Homomorphism), and Ring Homomorphism, but the linear finite-dimensional case is the guiding model on this page. ## Definition In undergraduate linear algebra, the object being studied is usually a finite-dimensional vector space, and the structure to preserve is linear structure. Requiring the codomain to be the same vector space is what makes powers $T^m$, eigenvectors, characteristic polynomials, and matrix similarity available. [definition: Endomorphism] Let $V$ be a finite-dimensional vector space over a field $F$. An endomorphism of $V$ is a linear map \begin{align*} T: V &\to V. \end{align*} The set of all endomorphisms of $V$ is denoted $\operatorname{End}_F(V)$. [/definition] ## Variants The finite-dimensional hypothesis is not needed to define a linear self-map, but it is the setting where every endomorphism can be represented by a square matrix after choosing a basis. This distinction motivates the following definition, which keeps the self-map condition while dropping any unnecessary finite-dimensional assumption. [definition: Linear Endomorphism] Let $V$ be a vector space over a field $F$. A linear endomorphism of $V$ is a linear map \begin{align*} T: V &\to V. \end{align*} The set of all linear endomorphisms of $V$ is denoted $\operatorname{End}_F(V)$ or $\mathcal{L}(V,V)$. [/definition] The word "endomorphism" also has a categorical form, useful when the same pattern is repeated for groups, rings, modules, and topological spaces. This broader version is needed to state the common pattern once: the ambient category tells us which structure the self-map must preserve. [definition: Categorical Endomorphism] Let $\mathcal{C}$ be a category and let $X$ be an object of $\mathcal{C}$. A categorical endomorphism of $X$ is a morphism \begin{align*} f: X &\to X \end{align*} in $\mathcal{C}$. The set of endomorphisms of $X$ is denoted $\operatorname{End}_{\mathcal{C}}(X)$. [/definition] The linear condition means that for all $v,w \in V$ and all $a \in F$, \begin{align*} T(v+w) &= T(v)+T(w), & T(av) &= aT(v). \end{align*} Because the codomain is again $V$, iterates such as $T^2=T\circ T$ and $T^3=T\circ T\circ T$ can be formed. The later algebraic structure theorem and polynomial-closure theorem state the formal closure results that justify building new endomorphisms from these iterates. For groups, an arbitrary self-map need not respect the multiplication, so iterating it can destroy the algebraic information one wants to study. The useful self-maps are those compatible with the group law, because their powers, kernels, quotients, and induced actions remain tied to the same group structure. [definition: Group Endomorphism] Let $G$ be a group. A group endomorphism of $G$ is a group homomorphism \begin{align*} \varphi: G &\to G. \end{align*} [/definition] Rings require a separate self-map notion because they carry two interacting operations. Preserving addition alone is not enough: an additive self-map can fail to respect products, so its iterates may no longer reflect the multiplication that makes the ring a ring. The obstruction is that many natural additive operators, such as multiplication by a fixed element, are self-maps of the underlying [abelian group](/page/Abelian%20Group) without being compatible with multiplication in the ring. To study symmetries, dynamics, or repeated operations internal to a ring, one needs a self-map whose powers still preserve both ring operations. The next definition isolates exactly those self-maps: the map must be a homomorphism of the ring back to itself. [definition: Ring Endomorphism] Let $R$ be a ring. A ring endomorphism of $R$ is a ring homomorphism \begin{align*} \psi: R &\to R. \end{align*} [/definition] The convention about whether ring homomorphisms preserve the multiplicative identity is inherited from the ambient category of rings being used. This is another instance of the same warning: an endomorphism is always a self-map preserving the structure that has been specified. ## Algebraic Structure on Endomorphisms Composition is the first operation to examine on endomorphisms. The theorem below records the formal closure and identity properties: composing two endomorphisms of the same object gives another endomorphism, and the identity morphism is the neutral element for this operation. [quotetheorem:9551] For vector spaces there is more structure because linear maps can be added and scaled. To use $\operatorname{End}_F(V)$ as more than a set of self-maps, we need the theorem below: it records the algebra structure coming from linear combination and composition. [quotetheorem:9552] Finite-dimensional endomorphisms are the coordinate-free version of square matrices. If $\dim_F V=n<\infty$ and an ordered basis $\mathcal B$ of $V$ is chosen, every $T \in \operatorname{End}_F(V)$ is represented by an $n \times n$ matrix $[T]_{\mathcal B}$ with entries in $F$. The theorem below is needed to make this identification precise, including its compatibility with addition, scalar multiplication, and composition. [quotetheorem:382] Once individual endomorphisms have matrix representatives, the remaining issue is whether the whole algebra of endomorphisms is faithfully captured by matrix algebra. The coordinate description is useful only if sums, scalar multiples, products, and identities correspond exactly to the same operations on square matrices. [quotetheorem:383] The matrix model is useful only after separating intrinsic features of an endomorphism from features introduced by a chosen basis. This raises the next structural question: if the same endomorphism is written in two different bases, what precise relation connects the two resulting matrices? [quotetheorem:400] This basis-change theorem is the reason many questions about an endomorphism become similarity-invariant questions about square matrices. The examples below use concrete self-maps to show how the abstract definitions appear in computations. ## Examples The simplest endomorphisms are often the most revealing. They show the difference between a self-map that respects structure and a random function whose iterates may not interact with the structure of the object. [example: Scalar Multiplication Endomorphism] Let $V$ be a vector space over a field $F$, and fix $a\in F$. Define $T_a:V\to V$ by \begin{align*} T_a(v)=av. \end{align*} For every $v\in V$, scalar multiplication in $V$ gives $av\in V$, so $T_a$ is a well-defined self-map of $V$. We show that this self-map is linear. Let $v,w\in V$. By distributivity of scalar multiplication over vector addition, \begin{align*} a(v+w)=av+aw. \end{align*} Using the definition of $T_a$ at $v+w$, at $v$, and at $w$, \begin{align*} T_a(v+w)=a(v+w)=av+aw=T_a(v)+T_a(w). \end{align*} Thus $T_a$ preserves addition. Now let $c\in F$ and $v\in V$. By the definition of $T_a$, \begin{align*} T_a(cv)=a(cv). \end{align*} Associativity of scalar multiplication gives \begin{align*} a(cv)=(ac)v. \end{align*} Since $F$ is a field, multiplication in $F$ is commutative, so \begin{align*} ac=ca. \end{align*} Substituting $ac=ca$ into $(ac)v$ gives \begin{align*} (ac)v=(ca)v. \end{align*} Associativity of scalar multiplication gives \begin{align*} (ca)v=c(av). \end{align*} Using $T_a(v)=av$, \begin{align*} c(av)=cT_a(v). \end{align*} Combining the displayed equalities, \begin{align*} T_a(cv)=a(cv)=(ac)v=(ca)v=c(av)=cT_a(v). \end{align*} Thus $T_a$ preserves scalar multiplication. Therefore $T_a$ is a linear map from $V$ to itself, so $T_a\in\operatorname{End}_F(V)$. Each scalar $a\in F$ determines the endomorphism that multiplies every vector of $V$ uniformly by $a$. [/example] Scalar multiplication endomorphisms are more than isolated examples; they provide the canonical way for field elements to act inside $\operatorname{End}_F(V)$. This raises a classification problem inside the endomorphism algebra: if an endomorphism commutes with the structure in the same uniform way as scalar multiplication, one must know whether it is forced to be multiplication by a single scalar. The answer also has to account for the degenerate case $V=0$, where there are no nonzero vectors on which to test uniqueness. [quotetheorem:9553] Scalar endomorphisms show how field elements act uniformly on every vector. The next example turns from uniform scaling to a map that keeps one summand of a [direct sum](/page/Direct%20Sum) and discards the other. [example: Projection Endomorphism] Let $V=U\oplus W$ be a direct sum of vector spaces over $F$. Thus every $v\in V$ has a unique decomposition $v=u+w$ with $u\in U$ and $w\in W$, and $U\cap W=\{0\}$. Define $P:V\to V$ by \begin{align*} P(u+w)=u. \end{align*} Because $u\in U$ and $U\subset V$, the value $P(u+w)=u$ lies in $V$, so the formula has the correct codomain. We first check that the formula does not depend on the chosen expression of the vector. Suppose \begin{align*} u+w=u'+w', \end{align*} where $u,u'\in U$ and $w,w'\in W$. Adding $-u'$ to both sides gives \begin{align*} u-u'+w=w'. \end{align*} Adding $-w$ to both sides gives \begin{align*} u-u'=w'-w. \end{align*} Since $U$ is a subspace, $u-u'\in U$. Since $W$ is a subspace, $w'-w\in W$. Therefore the common vector $u-u'=w'-w$ belongs to $U\cap W$. But $V=U\oplus W$, so $U\cap W=\{0\}$, and hence \begin{align*} u-u'=0. \end{align*} Adding $u'$ to both sides gives \begin{align*} u=u'. \end{align*} Thus two decompositions of the same vector have the same $U$-component, so $P$ is well-defined. Now let $x,y\in V$ and $c\in F$. Write \begin{align*} x=u_1+w_1 \end{align*} with $u_1\in U$ and $w_1\in W$, and write \begin{align*} y=u_2+w_2 \end{align*} with $u_2\in U$ and $w_2\in W$. Since $U$ and $W$ are subspaces, \begin{align*} u_1+u_2\in U \end{align*} and \begin{align*} w_1+w_2\in W. \end{align*} Using associativity and commutativity of vector addition, \begin{align*} x+y=(u_1+w_1)+(u_2+w_2)=(u_1+u_2)+(w_1+w_2). \end{align*} Applying the definition of $P$ to this direct-sum decomposition, \begin{align*} P(x+y)=P((u_1+u_2)+(w_1+w_2))=u_1+u_2. \end{align*} Also, \begin{align*} P(x)=P(u_1+w_1)=u_1 \end{align*} and \begin{align*} P(y)=P(u_2+w_2)=u_2. \end{align*} Adding these two equalities gives \begin{align*} P(x)+P(y)=u_1+u_2. \end{align*} Therefore \begin{align*} P(x+y)=P(x)+P(y). \end{align*} For scalar multiplication, since $U$ and $W$ are subspaces, \begin{align*} cu_1\in U \end{align*} and \begin{align*} cw_1\in W. \end{align*} By distributivity of scalar multiplication over vector addition, \begin{align*} cx=c(u_1+w_1)=cu_1+cw_1. \end{align*} Applying $P$ to this direct-sum decomposition gives \begin{align*} P(cx)=P(cu_1+cw_1)=cu_1. \end{align*} Since $P(x)=u_1$, multiplying by $c$ gives \begin{align*} cP(x)=cu_1. \end{align*} Thus \begin{align*} P(cx)=cP(x). \end{align*} So $P$ preserves addition and scalar multiplication. Since its domain and codomain are both $V$, the map $P$ is an endomorphism of $V$. Finally, let $v=u+w$ with $u\in U$ and $w\in W$. Since $0\in W$, the vector $u$ has direct-sum decomposition \begin{align*} u=u+0. \end{align*} By the definition of $P$, \begin{align*} P(v)=P(u+w)=u. \end{align*} Applying $P$ again, \begin{align*} P^2(v)=P(P(v)). \end{align*} Substituting $P(v)=u$ gives \begin{align*} P(P(v))=P(u). \end{align*} Using the decomposition $u=u+0$, \begin{align*} P(u)=P(u+0)=u. \end{align*} Therefore \begin{align*} P^2(v)=u. \end{align*} Since $P(v)=u$, this gives \begin{align*} P^2(v)=P(v). \end{align*} This holds for every $v\in V$, so $P^2=P$. The map $P$ keeps exactly the $U$-component and discards the $W$-component; it is an idempotent endomorphism. [/example] The computation $P^2=P$ is not just a feature of this example; it is the algebraic signature of a projection. The key question is whether every idempotent endomorphism actually comes from splitting the space into the part it keeps and the part it kills. For finite-dimensional vector spaces, the answer can be stated without introducing module language. If $P:V\to V$ is linear and $P^2=P$, then every vector $v\in V$ splits as \begin{align*} v=P(v)+(v-P(v)). \end{align*} The first summand lies in $\operatorname{im}(P)$, the image of $P$, and the second lies in $\ker(P)$, the set of vectors sent to $0$. Moreover, these two parts do not overlap except at $0$, because a vector that is both kept by $P$ and killed by $P$ must equal $0$. Thus an idempotent linear endomorphism is exactly a projection onto its image along its kernel. The condition $P^2=P$ is essential. An arbitrary endomorphism may have an image and a kernel, but it need not preserve the image pointwise or split the ambient space as image plus kernel. This is why projections are so useful: they turn an operator equation into a decomposition of the underlying vector space. Linear endomorphisms are only one instance of this general self-map idea. The next example turns to groups, where the same word means a homomorphism from a group to itself. [example: Doubling Endomorphism of the Integers] The map \begin{align*} \varphi: \mathbb{Z} &\to \mathbb{Z} \end{align*} defined by $\varphi(n)=2n$ is a self-map of $\mathbb{Z}$ because $2\in\mathbb{Z}$, $n\in\mathbb{Z}$, and $\mathbb{Z}$ is closed under multiplication, so $2n\in\mathbb{Z}$. We show that $\varphi$ is a group homomorphism for the additive group $(\mathbb{Z},+)$. Let $m,n\in\mathbb{Z}$. By the definition of $\varphi$, \begin{align*} \varphi(m+n)=2(m+n). \end{align*} By distributivity of multiplication over addition in $\mathbb{Z}$, \begin{align*} 2(m+n)=2m+2n. \end{align*} Again by the definition of $\varphi$, \begin{align*} \varphi(m)=2m. \end{align*} Also, \begin{align*} \varphi(n)=2n. \end{align*} Substituting $2m=\varphi(m)$ and $2n=\varphi(n)$ into $2m+2n$ gives \begin{align*} 2m+2n=\varphi(m)+\varphi(n). \end{align*} Therefore \begin{align*} \varphi(m+n)=\varphi(m)+\varphi(n). \end{align*} Thus $\varphi$ preserves the addition operation. Since its domain and codomain are both $\mathbb{Z}$, the map $\varphi$ is a group endomorphism of $(\mathbb{Z},+)$. The same map is not an automorphism. If an endomorphism $\alpha:G\to G$ is an automorphism, then there is a homomorphism $\beta:G\to G$ such that $\alpha\circ\beta=\operatorname{id}_G$. For every $y\in G$, taking $x=\beta(y)$ gives \begin{align*} \alpha(x)=\alpha(\beta(y)). \end{align*} By the definition of composition, \begin{align*} \alpha(\beta(y))=(\alpha\circ\beta)(y). \end{align*} Since $\alpha\circ\beta=\operatorname{id}_G$, \begin{align*} (\alpha\circ\beta)(y)=\operatorname{id}_G(y). \end{align*} By the definition of the identity map, \begin{align*} \operatorname{id}_G(y)=y. \end{align*} Hence every automorphism is surjective. It remains to show that $\varphi$ is not surjective. We prove that $1\notin\operatorname{im}(\varphi)$. Suppose, for contradiction, that some $n\in\mathbb{Z}$ satisfies $\varphi(n)=1$. By the definition of $\varphi$, \begin{align*} 2n=1. \end{align*} If $n\geq 1$, then multiplying the inequality by the positive integer $2$ gives $2n\geq 2$, so $2n\neq 1$. If $n=0$, then \begin{align*} 2n=2\cdot 0=0, \end{align*} so again $2n\neq 1$. If $n\leq -1$, then multiplying the inequality by the positive integer $2$ gives $2n\leq -2$, so $2n\neq 1$. These three cases cover every integer $n$, so no integer satisfies $2n=1$. Therefore $1\notin\operatorname{im}(\varphi)$, so $\varphi$ is not surjective and hence is not an automorphism. This example separates endomorphisms from invertible structure-preserving maps. [/example] The failure of invertibility matters. Endomorphisms may collapse information, have nonzero kernels, or fail to be surjective. Automorphisms are precisely the endomorphisms that preserve all information in both directions. ## Related Concepts Endomorphisms need not preserve all information. A projection may discard a summand, a nilpotent operator may eventually send every vector to zero, and a non-surjective group endomorphism may miss elements. Automorphisms are the special endomorphisms for which no information is lost: they are invertible self-maps that preserve the same structure in both directions. An automorphism is an invertible endomorphism. In a category, this means an endomorphism $f: X \to X$ has a two-sided inverse $g: X \to X$ satisfying $g\circ f=\operatorname{id}_X$ and $f\circ g=\operatorname{id}_X$. The formal definition isolates the invertible part of the endomorphism monoid. [definition: Automorphism] Let $\mathcal{C}$ be a category and let $X$ be an object of $\mathcal{C}$. An automorphism of $X$ is an endomorphism $f: X \to X$ that has an inverse morphism $g: X \to X$ satisfying $g\circ f=\operatorname{id}_X$ and $f\circ g=\operatorname{id}_X$. [/definition] To understand a linear endomorphism, one often looks for smaller subspaces on which the same map can be studied without leaving the subspace. This leads to invariant subspaces. [definition: Invariant Subspace] Let $V$ be a vector space over a field $F$, and let $T \in \operatorname{End}_F(V)$. A subspace $U \subset V$ is invariant under $T$ if $T(U) \subset U$. [/definition] Invariant subspaces make it possible to decompose an endomorphism into smaller pieces. Eigenvectors, kernels, images, generalized eigenspaces, and cyclic subspaces are all ways of finding structure inside a self-map. This motivates the theorem below: the kernel and image are the first invariant subspaces attached to any linear endomorphism. [quotetheorem:9554] The theorem is a first indication that an endomorphism carries its own internal geometry. Its kernel records what is collapsed to zero, and its range records the part of the vector space that the map reaches. ## What Endomorphisms Measure Endomorphisms measure internal dynamics rather than relationships between two different objects. Iterating $T: V \to V$ produces the sequence \begin{align*} v,\quad T(v),\quad T^2(v),\quad T^3(v),\ldots \end{align*} inside the same vector space. This makes questions about stability, nilpotence, periodicity, eigenvalues, and fixed points meaningful. Because $T(v)$ and $v$ live in the same vector space, one can ask whether a vector stays fixed or merely changes by a scalar factor. The following definitions make those two recurring comparisons explicit. [definition: Fixed Point of an Endomorphism] Let $V$ be a vector space over a field $F$, and let $T\in \operatorname{End}_F(V)$. A fixed point of $T$ is a vector $v\in V$ such that \begin{align*} T(v) &= v. \end{align*} [/definition] Fixed points are the most rigid comparison between a vector and its image. The next definition weakens that comparison in the direction most useful for linear dynamics: instead of asking that the vector itself be unchanged, it asks when the one-dimensional subspace it spans is preserved and the action along that line is given by a scalar. [definition: Eigenvector of an Endomorphism] Let $V$ be a vector space over a field $F$, and let $T\in \operatorname{End}_F(V)$. An eigenvector of $T$ is a nonzero vector $v\in V$ for which there is a scalar $\lambda\in F$ such that \begin{align*} T(v) &= \lambda v. \end{align*} [/definition] Fixed points and eigenvectors both use the self-map condition to compare $T(v)$ directly with $v$. This prepares the polynomial construction below, where repeated composition turns one endomorphism into many related endomorphisms. [quotetheorem:9555] This construction is the common language for minimal polynomials, characteristic polynomials, and, when the relevant polynomial splits, Jordan form. The point is that a polynomial in $T$ packages repeated composition and scalar combination into one operator. This lets questions about a linear map be converted into questions about polynomial equations it satisfies. For example, if $p(T)=0$, then the polynomial $p$ records an algebraic constraint on the dynamics of repeatedly applying $T$. This perspective also explains the limits of the construction. The polynomial $p(T)$ is still built from one endomorphism $T$, so all its powers commute with each other; it does not describe arbitrary collections of noncommuting operators. Its strength is precisely in single-operator linear algebra: it gives a bridge from endomorphisms to invariant subspaces, eigenvalues, diagonalization criteria, and canonical forms. ## Beyond and Connections Endomorphisms appear throughout algebra as structure-preserving self-maps. In linear algebra, projections and polynomial expressions in an operator lead toward invariant subspaces, eigenspaces, minimal polynomials, characteristic polynomials, diagonalization, and Jordan form. In group theory and ring theory, endomorphisms organize symmetries internal to a single object; invertible endomorphisms are automorphisms. The categorical viewpoint abstracts this further: an endomorphism is simply a morphism whose source and target are the same object. That viewpoint explains why the same word applies to vector spaces, groups, modules, rings, and many other settings, while the local meaning still depends on which structure the maps are required to preserve. ## References Axler, *Linear Algebra Done Right* (2015). Dummit and Foote, *Abstract Algebra* (2004). Mac Lane, *Categories for the Working Mathematician* (1971).