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When the displayed equality holds, the two nonzero eigenvalues of the three-state cyclic conversion model coincide at $\lambda =-(a+b+c)/2$. The positive cyclic rates force each eigenspace to have dimension at most one: if its $A$ coordinate vanished, the first eigenvector equation would force $C=0$, and then the third would force $B=0$. Thus the repeated eigenvalue has only one eigenvector direction and the matrix is not diagonalizable. A length-two Jordan chain produces both $e^{\lambda t}$ and $t{e}^{\lambda t}$ transients. The rates $(a,b,c)=(1,4,1)$ give an explicit example with $\lambda =-3$.

A generalized eigenfunction solves the eigenvalue equation for a linear operator in an enlarged space, often a space of distributions, although it may fail to belong to the underlying Hilbert space. A plane wave $e^{ikx}$ on the whole real line is a generalized momentum eigenstate, with momentum $\hslash k$, and is not square-integrable. This usage differs from a generalized eigenvector in a Jordan chain, where a higher power of $L-\lambda$ vanishes rather than $L-\lambda$ itself.

A Jordan basis is a basis consisting of Jordan chains of generalized eigenvectors for a linear operator. Its matrix is in Jordan normal form. Over the complex numbers every finite-dimensional linear operator has such a basis; a real matrix with nonreal eigenvalues requires complexification for this form.

Using ${\chi }_M(t)=\det (tI-M)$, expansion or triangular reordering givesThe eigenspace for $1$ is spanned by $e_1$, so its algebraic multiplicity two requires a size-two Jordan block. A convenient Jordan chain isFor eigenvalue $2$, the eigenspace has basis $w_1=e_1+e_3$, $w_2=e_4$, hence both corresponding blocks have size one. Therefore a Jordan normal form and the minimal polynomial areThe ordered basis $(v_1,v_2,w_1,w_2)$ supplies the change-of-basis matrixIts determinant is one, and the chain equations give $MP=PJ$, proving $P^{-1}MP=J$. The square at $t=1$ in the minimal polynomial is essential: $M$ is not diagonalizable.

A Jordan normal form of a complex $n\times n$ matrix $A$ is a similarity transformation $P^{-1}AP=J$ in which $J$ is a direct sum of Jordan blocksFor each eigenvalue, the multiset of block sizes is uniquely determined by $A$; only the order of the blocks is arbitrary. The change-of-basis matrix $P$ is generally not unique.

For the given matrix, expansion of the characteristic polynomial givesThe upper-left $2\times 2$ minor of $A-I$ is $-5$, so its rank is $2$ and the eigenspace for $1$ has dimension $1$. Thus the two occurrences of the eigenvalue $1$ belong to a single size-$2$ Jordan block. More explicitly, the Jordan chain and the remaining eigenvector can be chosen asConsequently, with $P=(v_1{v}_2{v}_3)$, for which $\det P=1$, the answer is

(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

An idempotent $e$ gives a projective module $Ae$, since $A=Ae\oplus A(1-e)$. It is primitive when it cannot be written as a sum of two nonzero orthogonal idempotents. This is equivalent to indecomposability of $Ae$: a projection onto a direct summand is an idempotent of ${\mathrm{End}}_A(Ae)\cong (eAe{)}^{\mathrm{op}}$, and conversely such an idempotent splits $Ae$. Starting with $1$, split any nonprimitive summand repeatedly. The process terminates because each splitting increases the number of nonzero vector-space summands in a finite-dimensional algebra. Thuswith orthogonal primitive idempotents and indecomposable projective modules. Each summand is generated by its idempotent.

Let $D={\mathrm{Hom}}_k(-,k)$. A Frobenius algebra is a finite-dimensional algebra with $A\cong D(A)$ as left modules over $A$, where $(a\varphi )(x)=\varphi (xa)$. Equivalently there is a linear functional $\lambda :A\to k$ for which $(x,y)\mapsto \lambda (xy)$ is nondegenerate. A symmetric algebra satisfies the stronger bimodule isomorphism $A\cong D(A)$, with $(a\varphi b)(x)=\varphi (bxa)$; equivalently the functional can be chosen with $\lambda (xy)=\lambda (yx)$. This is the associative-algebra meaning of symmetric, rather than the symmetric algebra of a vector space.

For a finite-dimensional module,The socle is its largest semisimple submodule, and the radical of a module gives its largest semisimple quotient. With $J=J(A)$ one has $\mathrm{Rad}M=JM$ and $\mathrm{Soc}M=\{m:Jm=0\}$: $A/J$ is semisimple, and $J$ annihilates every simple module.

Now let $P=Ae$ for a primitive idempotent, allowing the field $k$ to be nonsplit. Symmetry identifies $Ae$ with $D(eA)$ through the pairingIt is nondegenerate on these corners. If $v=ve\ne 0$, choose $a$ with $\lambda (av)\ne 0$; then $\lambda (eav)=\lambda (ave)=\lambda (av)\ne 0$. The other variable is treated similarly. The pairing respects the left $A$-action, soThe right-module quotient $eA/eAJ$ is simple. A primitive idempotent remains primitive modulo the nilpotent radical: idempotents lift through that radical, so a decomposition in the quotient would lift to one in $eAe$. In the semisimple algebra $A/J$, the left simple $Ae/JAe$ and the dual of the right simple $eA/eAJ$ belong to the same matrix block. There is exactly one left simple isomorphism class in that block, even over a nonalgebraically closed field. Hence the head-socle identity for a symmetric algebra isThis is an isomorphism of left modules, and need not be canonical.

For the modular examples put $r=(123)$, $s=(12)$, so $r^3=s^2=1$ and $srs=r^{-1}$. The coefficient-of-identity functional on the group algebra is symmetric and nondegenerate, so the identity just proved applies.

In characteristic two, the central idempotents are $b_0=1+r+r^2$ and $b_1=1-b_0$. The first block has basis $b_0,b_{0}s$ and is $k{C}_2\cong k[u]/(u^2)$. For the second, choose a primitive cube root $\omega$ and useThe two $r$-eigenlines are interchanged by $s$, proving irreducibility. Polynomial interpolation in $r$ gives diagonal matrix units, and multiplying by $s$ gives the off-diagonal units. Thus the four-dimensional block maps onto $M_2(k)$ and is isomorphic to it:There are two simple modules, the trivial one-dimensional module $T$ and the two-dimensional module $V$; the sign module equals $T$. Their projective covers are $P_T=k[u]/(u^2)$, with layers $T,T$, and $P_V=V$. The regular module is $P_T\oplus V^{\oplus 2}$, and the Cartan matrix of a group algebra is $\mathrm{diag}(2,1)$. Every module is a sum of $T,P_T,V$: the first block is classified by Jordan chains for a nilpotent linear map of lengths one and two, while the matrix block is semisimple. Thus there are exactly three indecomposable isomorphism classes. This describes the 2-modular blocks of S3.

In characteristic three put $u=r-1$. The ideal $I=uA$ satisfies $I^3=0$, and $A/I\cong k{C}_2\cong k\oplus k$, so $J(A)=I$. The only simples are the trivial and sign modules $S_{+},S_{-}$. Put $e_{\pm }=(1\pm s)/2$. The projectives $P_{\pm }=A{e}_{\pm }$ have dimension three. A convenient radical generator is $v=r-r^{-1}$:The basis $e_{\pm },v{e}_{\pm },v^2{e}_{\pm }$ gives the radical layersThus $A=P_{+}\oplus P_{-}$ and its Cartan matrix of a group algebra isThe elements $e_{-}v{e}_{+}$ and $e_{+}v{e}_{-}$ give one arrow in each direction between the two simples. All paths of length three vanish, and the six paths of lengths zero, one and two are linearly independent. Hence $A$ is the path algebra of that two-vertex oriented cycle modulo the ideal of paths of length three. Its connected quiver also proves there is a single block of S3 in characteristic three.

Any module is a $\mathbb{Z}/2$-graded vector space given by the $s$-eigenspaces, with $v$ interchanging the grades and $v^3=0$. Constructing Jordan chains with homogeneous bases in the kernel and image filtrations and homogeneous lifts decomposes it into graded chains. Each chain has length $1,2$ or $3$ and is determined by its top sign. It is a uniserial module and hence indecomposable. Consequently there are exactly six indecomposables, the length-$\ell$ quotients $P_{\pm }/J^{\ell }{P}_{\pm }$, $\ell =1,2,3$.

Suppose $\mathbf{v}=\alpha {\mathbf{e}}_1+\beta {\mathbf{e}}_2$. Applying $M-{\lambda }_{2}I$ would give ${\mathbf{e}}_2=\alpha ({\lambda }_1-{\lambda }_2){\mathbf{e}}_1$, contradicting linear independence of the two eigenvectors. Therefore the generalized eigenvector $\mathbf{v}$ lies outside their span and $({\mathbf{e}}_1,{\mathbf{e}}_2,\mathbf{v})$ is a basis. Write $\mathbf{x}=a{\mathbf{e}}_1+b{\mathbf{e}}_2+d\mathbf{v}$. Since $M\mathbf{v}={\lambda }_2\mathbf{v}+{\mathbf{e}}_2$, the linear system of ordinary differential equations givesSolving these equations gives $a=A_1{e}^{{\lambda }_{1}t}$, $d=A_3{e}^{{\lambda }_{2}t}$ and $b=(A_2+A_{3}t)e^{{\lambda }_{2}t}$. ThusThe length-two Jordan chain solution for the Jordan chain $({\mathbf{e}}_2,\mathbf{v})$ produces the $t{e}^{{\lambda }_{2}t}$ mode. The three arbitrary constants span all initial data, so this is the full general solution.

Three nonnegative amounts convert in the directed cycle $A\to B\to C\to A$ at positive first-order rates $a,b,c$. The linear system of ordinary differential equations conserves $A+B+C=N$. Its characteristic polynomial is $\lambda [{\lambda }^2+(a+b+c)\lambda +ab+bc+ca]$. The two nonzero eigenvalues have negative real part, so every trajectory converges to the unique steady state with that total:This is also a column-convention continuous-time Markov chain after normalizing by $N>0$: every component loses to one successor and receives from one predecessor. A complex-conjugate pair produces damped oscillations about equilibrium even though the amounts remain nonnegative. A repeated nonzero eigenvalue is defective and requires a Jordan chain.