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Each endomorphism has three distinct eigenvalues in a three-dimensional complex vector space. Choose a nonzero eigenvector for each eigenvalue; these vectors are linearly independent and hence form an eigenbasis. Both operators are therefore diagonalizable with diagonal matrix $D=\mathrm{diag}(1,2,3)$.

Their characteristic polynomials areIf a polynomial $p$ annihilates either operator, applying it to each eigenvector gives $p(1)=p(2)=p(3)=0$. Thus its minimal polynomial must contain all three factors. Conversely, their product annihilates the diagonal matrix, soFinally, send an ordered eigenbasis of $S$ to the correspondingly ordered eigenbasis of $T$. The resulting invertible map $C$ satisfies $CS=TC$, and hence $T=CS{C}^{-1}$: the two endomorphisms are conjugate.

A connection on a vector bundle is a linear operator $\nabla :\Gamma (E)\to {\Omega }^1(M;E)$ satisfying $\nabla (fs)=df\otimes s+f\nabla s$. Choose coordinates $x^i$ and a local frame $e_1,\dots ,e_m$, and define the connection matrix by $\nabla e_b={\sum }_a{A}^a{}_b\otimes e_a$, with $A^a{}_b={\sum }_i{A}^a{}_{bi}d{x}^i$. For a section $s={\sum }_a{s}^a{e}_a$, the coefficient column satisfiesThe symbol $d$ on the right means the ordinary exterior derivative of the coefficient functions.

Extend uniquely to $E$-valued differential forms by the graded Leibniz ruleIn the same local trivialization, an $E$-valued form is a column $\sigma =({\sigma }^a)$ and the covariant exterior derivative isMoving the one-form $A$ past the scalar $r$-form produces exactly the $(-1{)}^r$ in the graded rule. This explains the sign rather than replacing the rule by an ungraded product formula.

The endomorphism bundle has the induced endomorphism bundle connection characterized on sections by ${\nabla }_X^{\mathrm{End}}(T)s={\nabla }_X(Ts)-T{\nabla }_{X}s$. Its coefficient formula is $dT+AT-TA$. On an endomorphism-valued $r$-form $T$, extend as a graded derivation of the endomorphism-valued exterior product, givingExplicitly its $(a,b)$ entry is $d{T}^a{}_b+{\sum }_c{A}^a{}_c\wedge T^c{}_b-(-1{)}^r{\sum }_c{T}^a{}_c\wedge A^c{}_b$. This version is characterized byThe endomorphism-valued version also satisfies $d_A(T\wedge S)=d_{A}T\wedge S+(-1{)}^{r}T\wedge d_{A}S$. Products here combine composition of endomorphisms with wedge products of coefficient forms; matrix factors must retain their order.

These local formulas are independent of the local frame. For $e^{\prime }=eg$, the change of frame of a vector-bundle connection gives $A^{\prime }=g^{-1}Ag+g^{-1}dg$, ${\sigma }^{\prime }=g^{-1}\sigma$ and $T^{\prime }=g^{-1}Tg$. Using $d(g^{-1})=-g^{-1}(dg)g^{-1}$ gives $d_{A^{\prime }}{\sigma }^{\prime }=g^{-1}{d}_A\sigma$ and $d_{A^{\prime }}{T}^{\prime }=g^{-1}(d_{A}T)g$. Thus both extended operators are globally defined.

Use the covariant exterior derivative on the endomorphism bundle and remember that $F$ has degree two. The Bianchi identity is $d_{A}F=dF+A\wedge F-F\wedge A=0$. The local formula from part (b) givesThe two lines cancel, establishingBecause this is an endomorphism-valued equation and the operators transform covariantly under a change of local frame, this local calculation proves the global identity.

For a line bundle over $k=\mathbb{R}$ or $\mathbb{C}$, every fibre endomorphism is multiplication by a scalar. Thus $\mathrm{End}E$ is canonically the trivial scalar bundle, even when $E$ itself is nontrivial. In a local frame $A$ is a scalar one-form, so $A\wedge A=0$ and $F=dA$ locally. Also $A\wedge F=F\wedge A$, since their degrees are one and two. The Bianchi identity therefore reduces to the ordinary closedness conditionThis does not assert that the local scalar one-forms $A$ combine into a global one-form.

For two connections on a vector bundle $\nabla ,{\nabla }^{\prime }$, the difference is $C^{\infty }(M)$-linear in the section: the two $df\otimes s$ terms cancel. It is consequently a global endomorphism-valued one-form $a$. In rank one, $a$ is a global ordinary scalar one-form. Locally $A^{\prime }=A+a$, and the scalar cross terms and $a\wedge a$ vanish, givingThis is a global equality, so the two closed differential forms differ by an exact differential form. It proves the connection-independent curvature class of a line bundle:

Choose mutually inverse maps $i:D^D\to D$ and $j:D\to D^D$. This is an extensional reflexive object in the Cartesian closed category. Define its application morphismFor endomorphisms $f,g:D\to D$, application in the underlying set $\mathrm{End}(D)$ is $f\cdot g=\mathrm{app}\circ ⟨f,g⟩$.

It is useful to define term interpretation uniformly over any context object $A$. An environment assigns each free variable $x$ a morphism $\rho (x):A\to D$. Variables are interpreted by their assigned maps, and application is interpreted by $\mathrm{app}$ and pairing. For lambda abstraction, lift the old environment along ${\pi }_A:A\times D\to A$, interpret the new variable by ${\pi }_D$, and let $h:A\times D\to D$ interpret the body. Then setTaking $A=D$ gives interpretations in the requested set of endomorphisms. Other context objects are needed in the recursive definition to handle bound variables correctly.

The beta reduction rule follows from $ji={\mathrm{id}}_{D^D}$ and the evaluation identity for currying. Applying the interpreted abstraction to a map $g:A\to D$ givesInduction on terms identifies this with interpretation of the capture-avoiding substitution of the argument for $x$.

For eta conversion, suppose $x$ is not free in $M$ and write $f:A\to D$ for the interpretation of $M$. The body $Mx$ on $A\times D$ is the uncurried form of $jf$. HenceThus beta-eta-equivalent terms have equal interpretations in $\mathrm{End}(D)$. The full isomorphism, rather than just a retract of the function object, provides the inverse law used in the eta step.

Write products of endomorphisms as composition, with the rightmost map applied first. Because $1$ is terminal, the only hom-sets that can contain multiple arrows are $\mathrm{Hom}(1,D)$ and $\mathrm{End}(D)$. If the latter contained only the identity, any point $p:1\to D$ would have $p!={\mathrm{id}}_D$, where $!:D\to 1$. Then any other point $q$ would satisfy $q=p!q=p$. Thus the category would be a preorder. The hypothesis therefore implies $|\mathrm{End}(D)|>1$.

If the product object $D\times D$ were $1$, its universal property applied to arrows from $D$ would give $\mathrm{End}(D{)}^2\cong \mathrm{Hom}(D,1)$, a singleton, which is impossible. Since there are only two objects, $D\times D=D$. Similarly, if $D^D=1$, the exponential adjunction at $1$ would giveagain impossible. Hence $D\times D=D^D=D$ as chosen objects.

Let $M=\mathrm{End}(D)$, let $\pi ,{\pi }^{\prime }$ be the product projections, and let $\epsilon :D^D\times D\to D$ be evaluation, all now endomorphisms. Pairing is $⟨-,-⟩$, and for $x:D\times D\to D$ write $x^{\ast }$ for its currying. The product identities and the two inverse exponential identities giveThis is the structure of a cartesian closed monoid.

For the converse, start only from the monoid and these identities. The first three make pairing a bijection from $M^2$ to $M$, inverse to $z\mapsto (\pi z,{\pi }^{\prime }z)$. In particularas follows by applying both projections and using uniqueness. Define $U(y)=\epsilon ⟨y\pi ,{\pi }^{\prime }⟩$. The last two identities say that $U$ and $(-{)}^{\ast }$ are inverse bijections. These facts also derive, rather than assume, the naturality lawIndeed $U(x^{\ast }z)=\epsilon ⟨x^{\ast }z\pi ,{\pi }^{\prime }⟩=U(x^{\ast })⟨z\pi ,{\pi }^{\prime }⟩=x⟨z\pi ,{\pi }^{\prime }⟩$; applying the inverse bijection proves the law.

Put $e=({\pi }^{\prime }{)}^{\ast }$. The law applied to $x={\pi }^{\prime }$ gives, for every $z\in M$,Taking $z=e$ proves $e^2=e$. This stronger absorption identity is what makes the split object terminal.

Assume first that $M$ is nontrivial; the degenerate case is handled below. Construct the required category using the two idempotents $d=1_M$ and $e$, as a two-object part of the Karoubi envelope. Call their objects $D$ and $1$. A morphism from the object with idempotent $a$ to the object with idempotent $b$ is an element $f\in M$ satisfying $bfa=f$. Composition is monoid multiplication, and the identity on each object is its idempotent. These laws are well-defined and associative. Its hom-sets are concretelyFor example, $ef=e$ for every $f$, so the condition for $D\to 1$ forces $f=e$. Thus $1$ is terminal, and the original monoid appears unchanged at $D$.

Set $D\times D=D$ with projections $\pi ,{\pi }^{\prime }$ and the given pairing. The product property works for arrows from either object: for global arrows $x,y$ satisfying $xe=x$, $ye=y$, pairing also satisfies $⟨x,y⟩e=⟨x,y⟩$. The original projection identities give existence and uniqueness. Products with $1$ are the other object, with its identity and the unique terminal arrow as projections; $1\times 1=1$.

For the exponential set $D^D=D$ with evaluation $\epsilon$. For a parameter object $D$, the given $(-{)}^{\ast }$ and $U$ already establish the universal property. We must additionally verify it for parameter object $1$. An arrow $f:1\times D=D\to D$ is an arbitrary element of $M$. Define its transpose byNaturality gives $\kappa (f)z=\kappa (f)$ for every $z$, because $f{\pi }^{\prime }⟨z\pi ,{\pi }^{\prime }⟩=f{\pi }^{\prime }$. In particular $\kappa (f)e=\kappa (f)$, so it really is an arrow $1\to D$.

Its uncurrying is $\epsilon ⟨\kappa (f),1_M⟩=f$. To see this, start from $U(\kappa (f))=f{\pi }^{\prime }$ and compose with $⟨e,1_M⟩$. Pairing naturality, ${\pi }^{\prime }⟨e,1_M⟩=1_M$, and the absorption property of $\kappa (f)$ give precisely the claimed equality. Conversely, if $g:1\to D$ satisfies $ge=g$, then $gz=gez=ge=g$ for every $z$. ThereforeThus transpose and uncurrying are inverse for the terminal parameter object too. This verifies the full exponential property, rather than only its endomorphism instance.

The remaining exponentials are $D^1=D$, $1^D=1$, and $1^1=1$, by the unit product law and terminality. Hence the constructed two-object category is cartesian closed and $\mathrm{End}(D)=M$.

For nontrivial $M$, the constructed $D$ is nonterminal because it has more than one endomorphism, and $e\ne 1_M$ by the absorption identity. This is the case corresponding to the forward non-preorder hypothesis.

If $M$ has one element, all the algebraic identities still hold, but splitting $e=1_M$ alone would make the original object terminal. To satisfy the literal converse's requirement of a nonterminal object, handle this case separately: take the two-object poset category $D<1$. Products are meets, $D^D=1$, $1^D=1$, $D^1=D$, and $1^1=1$, which verify the exponential adjunction on this two-element order. It is cartesian closed, $D$ is nonterminal, and its one-element endomorphism monoid is $M$. It is a preorder, so it does not recover the forward non-preorder setting or the self-exponential equality. Thus the converse's stated endomorphism realization works also in the degenerate case, while the stronger non-preorder reconstruction uses nontrivial $M$.

Let $\mathfrak{q}$ be a prime ideal of $C$. Its quotient ring $D=C/\mathfrak{q}$ is a nonzero integral domain and a finite-dimensional $K$-vector space. For $0\ne a\in D$, multiplication by $a$ is an injective $K$-linear map $D\to D$, because $D$ has no zero divisors. An injective endomorphism of a finite-dimensional vector space is surjective. In particular $ab=1$ for some $b\in D$. Every nonzero element is therefore invertible, so $D$ is a field and $\mathfrak{q}$ is a maximal ideal.

To prove finiteness, select any $r$ distinct maximal ideals ${\mathfrak{m}}_1,\dots ,{\mathfrak{m}}_r$. Distinct maximal ideals are comaximal: ${\mathfrak{m}}_i+{\mathfrak{m}}_j$ strictly contains ${\mathfrak{m}}_i$, so must equal $C$. The Chinese remainder theorem gives a surjectionHere is the needed surjectivity explicitly. For each $j\ne i$, choose $a_{ij}\in {\mathfrak{m}}_j$ with $a_{ij}\equiv 1(\mathrm{mod}{\mathfrak{m}}_i)$, using comaximality. Then $e_i={\prod }_{j\ne i}{a}_{ij}$ is one modulo ${\mathfrak{m}}_i$ and zero modulo the other selected ideals. Given any residue tuple represented by $b_i$, the element ${\sum }_i{b}_i{e}_i$ maps to that tuple. Thus the displayed map really is surjective.

Each nonzero field quotient has $K$-dimension at least one, soThere cannot be more than ${\dim }_{K}C$ maximal ideals, because otherwise one could select a finite list exceeding this bound. ConsequentlyThis proves the description of the spectrum of a finite-dimensional commutative algebra. If $C=0$, its spectrum is empty and the same conclusion holds.

The ascending chain $\ker T\subseteq \ker T^2\subseteq \cdots$ must stabilize because $V$ is finite-dimensional. Once $\ker T^l=\ker T^{l+1}$, it stays stable: if $T^{l+2}v=0$, then $Tv\in \ker T^{l+1}=\ker T^l$, so $v\in \ker T^{l+1}=\ker T^l$; induction repeats the argument. Choose a positive such $l$, giving $\ker T^{2l}=\ker T^l$.

If $v\in \ker T^l\cap \mathrm{im}{T}^l$, write $v=T^{l}w$. Then $T^{2l}w=0$, so $w\in \ker T^l$ and $v=0$. By the rank-nullity theorem, the dimensions of those two subspaces sum to $\dim V$. HenceThis proves the finite-dimensional form of the Fitting lemma.

If $\det T=0$, the kernel contains a nonzero vector $u$, and $\mathrm{im}T$ is a proper vector subspace. Extend a basis of that image of a linear map to a basis of $V$ and choose a nonzero linear functional $\lambda$ which vanishes on the former basis. Define $S(v)=\lambda (v)u$. It is a nonzero endomorphism, since $\lambda$ takes a nonzero value somewhere, andThus a singular endomorphism has a nonzero two-sided annihilator, as required.

For an idempotent linear map $T$, the decomposition $v=(v-Tv)+Tv$ shows $V=\ker T\oplus \mathrm{im}T$: the first term is killed by $T$, the second is fixed, and a vector both fixed and killed is zero. An adapted basis makes its matrix $\mathrm{diag}(0,I_r)$, with $r=\mathrm{rank}T$. Consequently equal ranks give identical adapted matrices and therefore a change of basis relating $T_1,T_2$. Conversely, similar matrices have the same rank because pre- and postmultiplication by invertible matrices preserve rank. Hence $T_1$ and $T_2$ are similar exactly when their ranks agree, proving the similarity classification of idempotent linear maps.

(1) True. Over $\mathbb{C}$, $A$ has an eigenvalue $\lambda$ with nonzero eigenspace $E$. Commutativity makes $E$ invariant under $B$: $A(Bv)=B(Av)=\lambda Bv$. The endomorphism $B{|}_E$ has an eigenvector $v\ne 0$ because its characteristic polynomial has a complex root. This $v$ is an eigenvector of both matrices, though their eigenvalues on it may differ.

(2) False. Take $A=I$ and $B=2I$. They commute, but their only eigenvalues are respectively $1$ and $2$.

(3) True. Associativity gives $(BA{)}^{n+1}=B(AB{)}^{n}A=0$, so $T=BA$ is nilpotent. A nilpotent endomorphism of an $n$-dimensional space has index at most $n$: in $V\supseteq TV\supseteq T^{2}V\supseteq \cdots$, equality at a nonzero image would make $T$ surjective on that finite-dimensional image, hence invertible there, contradicting nilpotence. Thus dimensions strictly decrease until zero, in at most $n$ steps. Therefore $(BA{)}^n=0$.

(4) False. Take $T=0$ on a two-dimensional vector space and $\lambda =0$. The eigenspace has dimension two, but the minimal polynomial is $t$, in which the root zero has multiplicity one. Eigenspace dimension counts independent eigenvectors; the exponent in the minimal polynomial measures the longest corresponding Jordan chain, not that dimension.

(5) True. Factor the minimal polynomial as $m_T(t)=(t-\lambda {)}^{c}q(t)$, with $q(\lambda )\ne 0$, and set $S=T-\lambda I$. If $v\in W_{c+1}$, then $w=S^{c}v$ satisfies $Sw=0$, so $q(T)w=q(\lambda )w$. But $m_T(T)v=0$ also gives $q(T)w=0$. Thus $w=0$, so $v\in W_c$. The reverse inclusion is immediate from $S^{c}v=0\Rightarrow S^{c+1}v=0$, proving

If $\dim V=n$, then $\dim \mathrm{End}(V)=n^2$. The $n^2+1$ endomorphisms $I,T,\dots ,T^{n^2}$ are therefore linearly dependent, so some nonzero polynomial $p$ satisfies $p(T)=0$. The minimal polynomial is the monic polynomial of least degree annihilating $T$. Uniqueness follows because the difference of two such monic polynomials would be a nonzero annihilator of smaller degree.

A Jordan block $J_r(\lambda )$ has minimal polynomial $(x-\lambda {)}^r$, and the minimal polynomial of a direct sum is the least common multiple of those of its blocks. Thus an eight-dimensional example is