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Algebraic irreducibility excludes every proper nonzero invariant vector subspace, including those that are not closed. It is stronger than topological irreducibility, which excludes only closed invariant subspaces. For a nonzero irreducible action, every nonzero vector is cyclic in the algebraic sense: $A\xi =X$.

A finite-dimensional complex representation of a compact group is a direct sum of irreducible representations. Average a positive Hermitian inner product over normalized Haar measure. The orthogonal complement of an invariant subspace is then invariant, so induction on dimension gives the decomposition. The same argument works for finite-dimensional real representations with a positive real inner product.

An eigenline is a one-dimensional vector subspace spanned by an eigenvector. If $Av=\lambda v$ with $v\ne 0$, then $A(av)=\lambda av$ for every scalar $a$, so the line is an invariant subspace. It can be contained in an eigenspace of larger dimension; an eigenline need not be the whole eigenspace.

If a $k$-dimensional invariant subspace is mapped to the span of the first $k$ coordinate vectors by a similarity transformation, the transformed matrix has block formIts characteristic polynomial factors as $\det (\lambda I-B)\det (\lambda I-D)$, so its eigenvalues are the eigenvalues of the two diagonal blocks, counted with algebraic multiplicity.

For a linear group representation, the common fixed vectors of a subgroup form an invariant subspace for any equivariant map restricted to it. Its dimension is central to symmetry-based branching theorems. It need not be invariant under the entire group; conjugate subgroups have conjugate fixed-point subspaces.

A flag is a strictly nested chain of vector subspaces of a fixed vector space. A complete flag contains a subspace of every possible dimension. For an action by linear maps, an invariant flag means that each of its vector subspaces is an invariant subspace. A basis adapted to an invariant complete flag makes all these maps upper triangular, connecting complete flags to simultaneous triangularization of a Lie algebra representation.

The Floquet growth rate measures the exponential trend of a periodic linear differential equation over many coefficient periods. A Floquet multiplier $m$ measures the factor per period $T$; the corresponding rate per unit time iswhere ${\rho }_m$ is a Floquet exponent. Taking the real part removes the ambiguity in the complex logarithm.

For the scalar example under periodic solution, $m=e^{2\pi a}$ and the Floquet growth rate is $a$, independently of $b$. Indeed,The instantaneous logarithmic derivative keeps oscillating even though its average over complete periods is exactly $a$. Sampling at the same phase in each period removes the modulation altogether: $y(nT)=m^{n}y(0)$.

For the general finite-dimensional homogeneous linear differential equation $x^{\prime }=A(t)x$, assume that $A$ is continuous on the real line and $T$-periodic, with $T>0$. Normalize its fundamental matrix by $\Phi (0)=I$ and set $M=\Phi (T)$. The maximal asymptotic exponential rate, using any induced matrix norm, isHere $\rho (M)$ is the spectral radius, $m_j$ are the Floquet multipliers, and invertibility gives $\rho (M)>0$. This concerns growth of the solution operator; it does not say that every initial condition attains the maximum.^[ref]

To prove the formula, write $t=nT+s$ with $0\le s<T$. Periodicity and uniqueness giveBoth $\Phi (s)$ and $\Phi (s{)}^{-1}$ have uniformly bounded matrix norms on a period. Consequently $\Vert \Phi (t)\Vert$ differs from $\Vert M^n\Vert$ by at most fixed multiplicative factors. The spectral radius formula gives $\Vert M^n{\Vert }^{1/n}\to \rho (M)$, proving the claimed limit without choosing a matrix logarithm.

For a particular nonzero initial condition $v$, decompose $v$ into the generalized eigenspaces of $M$ over the complex numbers. The rate of $\Phi (t)v$ is the largest $T^{-1}\log |m_j|$ among components that are nonzero. Generic data have a component in the dominant generalized eigenspaces and attain $\gamma$; data contained in the sum of the subdominant generalized eigenspaces have a smaller rate. For example, with $A=\mathrm{diag}(0.2,-0.3)$, the maximal rate is $0.2$, while $v=(0,1)$ decays at rate $-0.3$. These invariant subspaces describe initial data at phase zero. Their transported versions $E(t)=\Phi (t)E(0)$ can vary with time and satisfy $E(t+T)=E(t)$.

A negative Floquet growth rate allows temporary amplification. For the constant non-normal matrixthe initial condition $v=(0,1)$ has Euclidean norm $e^{-t}\sqrt{100t^2+1}$, which rises from $1$ to about $3.697$ at $t=1$ before eventually decaying. Its exponential rate is still $-1$: the Jordan block contributes a polynomial factor, whose logarithm divided by $t$ tends to zero. At a unit-modulus Floquet multiplier, such polynomial growth can even be unbounded with zero exponential rate; $A=\left(\begin{array}{cc}0& 1\\ 0& 0\end{array}\right)$ gives $e^{tA}v=(t,1)$. Thus zero Floquet growth rate alone proves neither boundedness nor a periodic solution.

For a finite-dimensional continuous complex representation $R$ of a compact group, normalize Haar measure to mass one and average any positive-definite Hermitian inner product:The resulting inner product is positive-definite and invariant. If $W$ is an invariant subspace, then $W^{\perp }$ is invariant as well, since $⟨R(g)v,w{⟩}_K=⟨v,R(g^{-1})w{⟩}_K$. Thus $V=W\oplus W^{\perp }$. Restriction to a compact real form, followed by integration of a Lie-algebra representation, proves the Weyl complete reducibility theorem.

Here a Hilbert space module means a Hilbert space with a unital star-representation of a Von Neumann algebra, rather than a Hilbert C-star module with an algebra-valued inner product. A closed invariant subspace is reducing because the algebra is closed under adjoints. Module equivalence means an intertwining unitary. The polar decomposition of a bounded operator converts an intertwiner into a unitary between its initial and final support subspaces.

The solvable Lie algebra $\mathbb{C}x\oplus \mathbb{C}y$, with $[x,y]=y$, has an infinite-dimensional Irreducible Lie algebra representation on $\mathbb{C}[t]$: $xf(t)=tf(t)$ and $yf(t)=f(t-1)$. Indeed $[x,y]f=yf$. An invariant subspace is an ideal $p(t)\mathbb{C}[t]$ because it is invariant under multiplication by $t$. Shift invariance gives $p(t)\mid p(t-1)$, hence $p(t-1)=p(t)$ and $p$ is constant. This shows why the Lie theorem needs finite-dimensional representations, even when the algebra itself is finite-dimensional.

An irreducible group representation is nonzero and has no nonzero proper invariant subspace. These are the pieces that cannot be decomposed further by taking a smaller symmetry-preserved vector space.

The real two-dimensional rotation through $120$ degrees illustrates the role of the coefficient field. A one-dimensional real invariant subspace would supply a real eigenvector, but the characteristic polynomial ${\lambda }^2+\lambda +1$ has no real root. The representation is therefore irreducible over the real numbers. Over the complex numbers, that polynomial has distinct roots of unity $e^{2\pi i/3}$ and $e^{-2\pi i/3}$, giving two invariant eigenlines. Irreducibility over the real numbers does not imply absolute irreducibility.

Every algebraically irreducible representation of a Banach algebra that is normed is continuous. The continuous orbit vectors form an invariant subspace. If all nonzero orbit maps were discontinuous, apply the countable annihilation sequence in an irreducible Banach module and the gliding-hump continuity principle to right multiplication on the algebra and evaluation of the operator-valued representation. Off-diagonal maps vanish, whereas diagonal maps are discontinuous, a contradiction. Once every orbit map is continuous, the Uniform boundedness principle on the Banach algebra gives the common operator-norm bound. Completing the original normed representation space supplies the Banach operator space needed in the argument.

Linear operator theory studies linear maps, their spectra, invariant subspaces, and normal forms such as Jordan normal form. In a finite-dimensional vector space, eigenvalues identify parameters for which a shifted linear operator cannot be inverted. In an infinite-dimensional vector space, inversion can fail even without an eigenvector. The spectrum of a bounded operator captures this larger obstruction and connects linear algebra with the approximation questions of functional analysis.

For a compact Lie group $G$, the complex representation ring of a compact group $R(G)$ is the Grothendieck group of finite-dimensional continuous complex group representations, with the relations $[M\oplus N]=[M]+[N]$. Multiplication is $[M][N]=[M\otimes N]$, and the unit is the trivial one-dimensional representation. The ring of class functions iswith pointwise addition and multiplication. The character of $M$ is ${\chi }_M(g)=\mathrm{tr}{\rho }_M(g)$. The trace identities for direct sums and tensor products makea well-defined ring homomorphism $\chi :R(G)\to c\ell (G)$.

Normalize Haar measure by ${\int }_{G}dg=1$. Averaging a positive definite Hermitian inner product giveswhich is positive definite and $G$-invariant. This proves unitarization of a compact-group representation. In a unitary representation the orthogonal complement of an invariant subspace is invariant. Induction on dimension therefore gives complete reducibility of compact-group representations. Hence every element of $R(G)$ is a finite integral linear combination of irreducible classes.

We establish the needed character orthogonality for compact groups carefully. For irreducible unitary representations $V,W$, let $G$ act on ${\mathrm{Hom}}_{\mathbb{C}}(W,V)$ byIts average $P={\int }_G(g\cdot )dg$ is a projection onto ${\mathrm{Hom}}_G(W,V)$: averaging makes every image invariant, and it fixes every intertwiner. The trace of this action is ${\chi }_V(g)\overline{{\chi }_W(g)}$, since the inverse of a unitary matrix has conjugate trace. Taking the trace of the projection givesFor completeness, Schur's lemma follows here from invariance of the kernel and image of an intertwiner. A nonzero intertwiner between irreducibles is an isomorphism. An endomorphism of an irreducible complex representation has an eigenvalue $\lambda$; the noninvertible intertwiner $A-\lambda I$ must vanish. Thus the last dimension is one for isomorphic irreducibles and zero for inequivalent ones.

If a virtual character ${\sum }_V{n}_V{\chi }_V$ vanishes, integrating it against $\overline{{\chi }_W}$ gives $n_W=0$ for every irreducible $W$. The irreducible multiplicities therefore distinguish the classes, andThis also proves that two finite-dimensional representations with the same character are isomorphic.

For SU(2), the irreducibles are $V_k={\mathrm{Sym}}^k({\mathbb{C}}^2)$, $k\ge 0$, of dimension $k+1$. One can see this classification through the usual ${\mathfrak{sl}}_2$ operators $H,E,F$: a highest-weight vector $v$ has $Hv=kv$, $Ev=0$, and the relations $[H,E]=2E$, $[H,F]=-2F$, $[E,F]=H$ give, by induction,The torus weights are integers. If $F^{r}v\ne 0$ but $F^{r+1}v=0$, the identity at $j=r+1$ forces $k=r\ge 0$. The vectors $v,Fv,\dots ,F^{k}v$ span an invariant irreducible module, which in an irreducible representation is the whole space. This is the symmetric-power module. Conversely its successive monomial weight vectors are linked by $E,F$ with nonzero coefficients, proving its irreducibility. Thus the classification of finite-dimensional representations of SU2 gives every irreducible, rather than just a list of examples.

On the maximal torus $T=\{\mathrm{diag}(z,z^{-1}):|z|=1\}$, their characters areWith $x={\chi }_1=z+z^{-1}$, multiplication gives ${\chi }_{k+1}=x{\chi }_k-{\chi }_{k-1}$, starting with ${\chi }_0=1$. Hence each ${\chi }_k$ is a monic integral polynomial of degree $k$ in $x$. For example ${\chi }_2=x^2-1$ and ${\chi }_3=x^3-2x$. They form a basis over $\mathbb{Z}$, soThe character map realizes this ring as the finite integral polynomials in $z+z^{-1}$, interpreted as continuous class functions. Its image is not the whole infinite-dimensional ring of continuous class functions; injectivity is the assertion being proved.

Take $V={\mathbb{R}}^3$, $\alpha =\mathrm{diag}(0,1,0)$ and $W=\mathrm{span}(e_1,e_2)$. Then $W$ is a proper invariant subspace, and the restriction has matrix $\mathrm{diag}(0,1)$. In each case $X(X-1)$ annihilates the operator, while any annihilator must vanish at both zero and one. ConsequentlyThis same example works over any scalar field.

A representation of a Banach algebra is a complex-linear multiplicative map $\pi :A\to {\mathrm{End}}_{\mathbb{C}}(X)$. A nonzero irreducible action of a unital algebra automatically has $\pi (1)=I$: the image of the idempotent $\pi (1)$ is a nonzero invariant subspace and hence is all of $X$. An algebraically irreducible representation of a Banach algebra is a nonzero action if it has no nonzero proper invariant submodule, whether closed or not. A normed representation of a Banach algebra has $\pi (a)$ bounded for each $a$ on a preassigned normed vector space $X$; continuity of $\pi$ as an operator-norm-valued map is a further conclusion. This must not be confused with topological irreducibility, which excludes only closed invariant subspaces. Unitizing the algebra extends a nonunital action by $\pi (a+\lambda 1)=\pi (a)+\lambda I$, so it is enough to discuss the unital case.

For $\xi \ne 0$, irreducibility gives $A\xi =X$. The annihilator $L_{\xi }=\{a:\pi (a)\xi =0\}$ is a maximal left ideal: an intermediate left ideal gives an intermediate submodule of $X$. It is closed by the Neumann-series argument for maximal left ideals. Consequently $A/L_{\xi }$ transports a complete quotient norm to $X$, and left multiplication is contractive in this norm. Every algebraically irreducible representation of a Banach algebra can therefore be realized continuously on a Banach space. This alone does not prove continuity for its preassigned norm.

The kernel of $\pi$ is the intersection of the closed left ideals $L_{\xi }$ and is a closed primitive ideal. The Jacobson radical is the intersection of all primitive ideals, equivalently the intersection of maximal left ideals. Thus these quotient representations detect semisimplicity. Equivalent algebra representations are related by a bijective module homomorphism.

The algebra of module endomorphisms is exactly $\mathbb{C}I$. Indeed, if a module endomorphism $D$ sends a fixed cyclic vector $\xi$ to $a\xi$, then $D(b\xi )=ba\xi$. It is induced by bounded right multiplication by $a$ on $A/L_{\xi }$, so is bounded for the quotient norm on an irreducible Banach module. Every nonzero module endomorphism is injective and surjective by irreducibility, and its inverse is bounded by the same argument. This commutant of an operator algebra is a closed division subalgebra of the bounded operators on this Banach module; the Gelfand-Mazur theorem makes it $\mathbb{C}I$.

The Jacobson density theorem now says that for finitely many linearly independent ${\xi }_1,\dots ,{\xi }_n$ and arbitrary ${\eta }_1,\dots ,{\eta }_n$, one element of $A$ sends all ${\xi }_j$ to ${\eta }_j$. Here is the induction behind it. The one-vector assertion is $A{\xi }_1=X$. Assuming the first $n-1$ coordinates can be prescribed, set $L=\{a:a{\xi }_j=0(j<n)\}$. Its image $L{\xi }_n$ is an invariant subspace. If it is $X$, it adjusts the last image without changing the first ones. If it is zero, the rule $(a{\xi }_1,\dots ,a{\xi }_{n-1})\mapsto a{\xi }_n$ is a well-defined module map $X^{n-1}\to X$. Each coordinate map is scalar by the module endomorphism result, forcing ${\xi }_n$ to be a linear combination of the earlier vectors, a contradiction. This proves the assertion.

We can now give Johnson's continuity argument with its crucial construction. If $X$ is finite dimensional, the closed kernel has finite codimension and $A/\ker \pi$ is finite dimensional, so $\pi$ is continuous. Suppose instead that $X$ is infinite dimensional. LetThis is an invariant submodule, because the orbit map of $\pi (b)\xi$ is the orbit map of $\xi$ composed with the bounded map $a\mapsto ab$. Thus either $Y=X$ or $Y=0$.

To rule out $Y=0$, choose linearly independent unit vectors ${\xi }_0,{\xi }_1,\dots$. There are elements $a_n\in A$ such that, with $c_n=a_n\cdots a_1$,For completeness, construct them in the canonical quotient-norm Banach module. At stage $n$ the current tail vectors are independent. In the closed left annihilator of its first vector, density permits arbitrary images of any finite list of the remaining vectors. The condition that finitely many such images are independent is open in this Banach subspace and dense: perturb any element by $t$ times an element prescribing independent images; a suitable determinant is a nonzero polynomial in $t$, with only finitely many exceptional values. The Baire category theorem intersects these open dense conditions for all initial finite tail lists. The resulting $a_n$ kills the first current vector and retains independence of the entire remaining tail. This is the countable annihilation sequence in an irreducible Banach module.

The gliding-hump continuity principle supplies the other ingredient: if $E,F$ are Banach spaces, $S:E\to F$ is linear, $T_j:E\to E$ and $U_n:F\to F_n$ are bounded, and $U_{n}S{T}_1\cdots T_m$ is continuous whenever $m>n$, then $U_{n}S{T}_1\cdots T_n$ is continuous eventually. The product form of this principle is recorded in Proposition 1.3 of Thomas's paper. Here is the contradiction mechanism. Rescale so $\Vert T_j\Vert ,\Vert U_n\Vert \le 1$ and put $P_m=T_1\cdots T_m$. If infinitely many diagonal maps are discontinuous, choose increasing $n_j$ and successively tiny $x_j$ soAlso choose $\Vert x_j\Vert \le 2^{-j}/(1+{\max }_{i<j}\Vert U_{n_i}S{P}_{n_i+1}\Vert )$. The convergent sum $x={\sum }_j{P}_{n_j}{x}_j$ has, after its $j$th term, a tail factoring through $P_{n_j+1}$. Applying that continuous composite bounds the tail contribution by one. The display forces $\Vert U_{n_j}Sx\Vert >j$, contradicting $\Vert U_{n_j}Sx\Vert \le \Vert Sx\Vert$. Only finite decompositions and the continuous tail composites are used; no unjustified interchange of a discontinuous map with an infinite sum occurs.

Apply this principle with $E=A$, $F=\mathcal{B}(\overline{X})$, where $\overline{X}$ is the completion in the original norm, and $S(a)$ the bounded extension of $\pi (a)$. Set $T_j(a)=a{a}_j$ and $U_n(R)=R{\xi }_n$. ThenFor $m>n$ this is zero, hence continuous. For $m=n$ it is the orbit map of a nonzero vector, hence discontinuous if $Y=0$. This contradicts the principle. Therefore $Y=X$.

Finally, apply the Uniform boundedness principle on the Banach space $A$ to the continuous maps $a\mapsto \pi (a)\xi$ indexed by $\Vert \xi \Vert \le 1$. They are pointwise bounded because each individual $\pi (a)$ is bounded in the original norm. It givesThis proves Johnson's continuity theorem for irreducible normed representations, including when the original normed vector space is incomplete. The quotient norm, countable density construction, and gliding hump explain the substantive steps rather than assuming the desired continuity in advance.

For $x\in \ker (\alpha -\lambda I)$, commutation givesThus each eigenspace of $\alpha$ is an invariant subspace for $\beta$.

If $\alpha ={\beta }^2$, the two endomorphisms commute. The $n$ distinct real eigenvalues of $\alpha$ have one-dimensional real eigenspaces, since their direct sum is the $n$-dimensional vector space. Take a nonzero real eigenvector $x_i$. Its eigenspace is invariant under $\beta$, so $\beta x_i=c_i{x}_i$ for a real scalar $c_i$. Squaring givesThe simple-spectrum hypothesis is essential: the real $90$-degree rotation matrix squares to $-I$ in dimension two, whose negative eigenvalue has a two-dimensional eigenspace.

Work over $\mathbb{C}$ with finite-dimensional Lie algebras and Lie algebra representation spaces. The adjoint form of Engel theorem isHere a nilpotent Lie algebra has terminating lower central series, ${\gamma }_1\mathfrak{g}=\mathfrak{g}$, ${\gamma }_{j+1}\mathfrak{g}=[\mathfrak{g},{\gamma }_j\mathfrak{g}]$. We prove the stronger linear statement: if $V\ne 0$, $L\subseteq \mathfrak{gl}(V)$ and every member of $L$ is a nilpotent endomorphism, then there is a nonzero $v\in V$ killed by all of $L$, and $L$ can be simultaneously represented by strictly upper triangular matrices.

For the proof of Engel theorem by induction and normalizers, first establish a useful nilpotence fact. If $x^m=0$, left and right multiplication by $x$ on $\mathrm{End}(V)$ commute. Expanding their difference givesFor $N=2m-1$, every summand contains a power of $x$ at least $m$, so $(\mathrm{ad}x{)}^{2m-1}=0$. Restrictions and induced quotient maps remain nilpotent.

Now prove the common-kernel statement, the Engel lemma, by induction on $\dim L$. The zero algebra is immediate. Choose a maximal proper Lie subalgebra $M$. Its action on $L/M$ by commutators consists of nilpotent maps by the preceding fact. Applying the lower-dimensional induction hypothesis to its image gives a nonzero coset $y+M$ with $[M,y]\subseteq M$. HenceThe normalizer of a Lie subalgebra is a Lie subalgebra by the Jacobi identity. Maximality implies $N_L(M)=L$, so $M$ is an ideal of a Lie algebra. Moreover $L/M$ has no proper nonzero Lie subalgebra, and every one-dimensional subspace is a Lie subalgebra. Thus $\dim L/M=1$.

Induction also provides a nonzero common annihilatorIt is stable under $L$: for $m\in M$, $x\in L$, $v\in W$,because $[m,x]\in M$. Choose $x$ representing a basis of $L/M$. The restriction of the nilpotent endomorphism $x$ to the nonzero space $W$ has nonzero kernel. A vector in this kernel is killed by both $M$ and $x$, hence by all of $L$. This finishes the induction. When an action is not faithful, induction is applied to its image, whose dimension is no larger than $\dim M$; no faithfulness assumption is hidden.

Starting with the common-kernel line, apply the same statement to each successive quotient Lie algebra representation space. It constructs a complete flag of invariant subspacesAn adapted basis represents every member of $L$ by a strictly upper triangular matrix. Every product of $d$ such matrices is zero, so every iterated commutator with $d$ entries is zero as well.

Finally, if every $\mathrm{ad}x$ on $\mathfrak{g}$ is nilpotent, apply the linear statement to the Adjoint representation. With $d=\dim \mathfrak{g}$, every product of $d$ adjoint operators vanishes, so ${\gamma }_{d+1}\mathfrak{g}=0$. Conversely, if ${\gamma }_{c+1}\mathfrak{g}=0$, then $(\mathrm{ad}x{)}^{c}y=0$ for all $x,y$, proving the other implication. This proves Engel's theorem and its simultaneous-triangularization form. The hypothesis concerns every element of the algebra, not merely a chosen set of generators.