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No. Let $Q$ be the quiver $1\to 2$ over a field $k$, and take the representation of a quiverAn endomorphism is a pair of scalar maps $(a,b)$ satisfying $b=a$, so its endomorphism ring is $k$. Every nonzero endomorphism is therefore an isomorphism, making this representation a brick module. It nevertheless has the proper nonzero subrepresentation $0\to k$, so it is not an irreducible module.

Equivalently, this is a nonsimple module over the path algebra $kQ$ whose endomorphism ring is a division ring.

For a connected source, first integrate on its universal covering Lie group. The resulting map descends to the original group exactly when it kills the discrete kernel of the covering map. For the circle group, a real-linear map $t\mapsto ct$ integrates to a circle endomorphism exactly when $c\in \mathbb{Z}$, since $2\pi \mathbb{Z}$ must map into itself. This is a global period condition invisible to the zero bracket of the circle's Lie algebra.

For $M={⨁}_j(F_q[t]/(f^j){)}^{m_j}$ with $f$ irreducible of degree d and partition $\lambda =(j^{m_j})$, its automorphism group has order $q^{d{\sum }_i({\lambda }_i^{\prime }{)}^2}{\prod }_j{\prod }_{r=1}^{m_j}(1-q^{-dr})$. The endomorphism algebra has dimension $d{\sum }_{a,b}\min (a,b)m_a{m}_b=d{\sum }_i({\lambda }_i^{\prime }{)}^2$; its semisimple quotient is ${\prod }_j{M}_{m_j}(F_{q^d})$, and units are precisely lifts of units in that quotient.

An algebra representation is a unital algebra homomorphism into the endomorphisms of a vector space. It is equivalent to making that vector space a left module over the algebra, with $av=\rho (a)v$.

For a smooth function $f$ and vector field $X$,on every tensor field, with $X\otimes df$ interpreted as the endomorphism $Y\mapsto Y(f)X$. Both sides vanish on functions; on vector fields this follows from $[fX,Y]=f[X,Y]-Y(f)X$. Both are contraction-compatible tensor derivations, so agreement on functions and vector fields proves equality on all tensor types. In particular ${\mathcal{L}}_{fX}\omega =f{\mathcal{L}}_X\omega +\omega (X)df$.

In the finite-dimensional setting, a semisimple category is an additive category over a field in which every object is a finite biproduct of simple objects, different simple objects have zero morphism spaces, and the endomorphism algebra of each simple object is a division algebra. Such categories have finite matrix blocks and all their idempotents split. The semisimplicity of the reduced Temperley-Lieb category gives an example whose simple objects are the surviving Jones-Wenzl colors.

The Jones-Wenzl idempotent recursion decomposes $f_j\otimes 1$ into the orthogonal $f_{j+1}$ and $f_{j-1}$ channels. The Jones-Wenzl partial trace supplies the nonzero scalar normalizing the cap-cup channel. At the highest surviving color the $f_{r-1}$ channel is zero. Induction therefore decomposes every tensor power into the surviving colors. A planar matching between different projected colors has a turnback on the larger side and vanishes; between equal colors only the identity matching remains. Their endomorphism spaces are consequently one-dimensional, and the resulting finite matrix blocks split. Taking direct sums and splitting these idempotents produces a semisimple category, supplying the complete resolutions needed in fusion of Jones-Wenzl colors and the Kirby color handle-slide identity.

Let $T$ be a singular endomorphism of a finite-dimensional vector space. Choose $0\ne u\in \ker T$ and a nonzero linear functional $\lambda$ vanishing on $\mathrm{im}T$, possible because that image of a linear map is proper. Then $S(x)=\lambda (x)u$ is a nonzero rank-one endomorphism with $TS=ST=0$.

If $C$ is a commutative algebra of finite dimension over a field $K$, every prime ideal is maximal and $|\mathrm{Spec}C|\le {\dim }_{K}C$. Indeed, the quotient by a prime is a finite-dimensional integral domain. Multiplication by any nonzero element is an injective linear endomorphism, hence surjective, so the quotient is a field. Distinct maximal ideals are pairwise comaximal. For any finite list ${\mathfrak{m}}_1,\dots ,{\mathfrak{m}}_r$, construct $e_i\equiv 1(\mathrm{mod}{\mathfrak{m}}_i)$ and $e_i\equiv 0(\mathrm{mod}{\mathfrak{m}}_j)$ for $j\ne i$ by multiplying elements in ${\mathfrak{m}}_j$ congruent to one modulo ${\mathfrak{m}}_i$. These give a surjection $C\to {\prod }_{i}C/{\mathfrak{m}}_i$. Each factor has positive $K$-dimension, so $r\le {\dim }_{K}C$, excluding an infinite list.

A matrix with the same number of rows and columns represents an endomorphism of a finite-dimensional vector space after choosing a basis. Square matrices support trace, determinant, powers and the matrix exponential.

For a surjective ring endomorphism $f$ of a Noetherian ring, the ascending kernel chain $\ker f\subseteq \ker f^2\subseteq \cdots$ stabilizes. If $\ker f^m=\ker f^{m+1}$ and $f(a)=0$, choose $b$ with $f^m(b)=a$. Then $b\in \ker f^{m+1}=\ker f^m$, so $a=0$. Thus the endomorphism is a ring automorphism.

For an abelian variety $A$ and a prime number $\ell$ different from the characteristic, the inverse limit of its geometric ${\ell }^n$-torsion, with transition multiplication by $\ell$, is its Tate module. For an elliptic curve, it is free of rank two over the p-adic integers and satisfies $T_{\ell }(E)/{\ell }^n{T}_{\ell }(E)\simeq E[{\ell }^n]$. The absolute Galois group acts continuously and commutes with all base-defined endomorphisms.

For a nonzero finitely generated indecomposable module $P$ that is a projective module over a right Artinian ring $A$, its top $P/PJ(A)$ is a simple module. The Jacobson radical $J(A)$ is a nilpotent ideal, so the top is nonzero; the quotient is a semisimple module over the semisimple ring $A/J(A)$. If it decomposed, a nontrivial idempotent of its endomorphism ring would lift by projectivity to an endomorphism of $P$. The Hopkins-Levitzki theorem gives finite composition length, and the Fitting lemma makes the lift either invertible or nilpotent. Its induced map on the top would have the same property, impossible for that nontrivial idempotent.

If $\sigma$ has ${\alpha }_r$ cycles of length $r$, contraction of matrix entries around its cycles gives $\mathrm{tr}(\sigma {\xi }^{\otimes n})={\prod }_r\mathrm{tr}({\xi }^r{)}^{{\alpha }_r}$. The formula holds for every endomorphism, not only diagonalizable ones. The Schur–Weyl duality decomposition equates it with a sum of products of Specht and Schur characters.