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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 continuous 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.

For $C_n=⟨g:g^n=e⟩$, the complex irreducible representations are precisely the $n$ one-dimensional maps $g\mapsto e^{2\pi ij/n}$, with $0\le j<n$. Their characters have the displayed values, forming the discrete Fourier character table. Indeed a representing matrix $A$ satisfies $A^n=I$. Applying the fundamental theorem of algebra to its characteristic polynomial gives an eigenvalue and an eigenvector over the complex numbers. Its line is an invariant subspace under every power of $A$; irreducibility therefore forces the entire space to be that line. Its eigenvalue is an $n$th root of unity. Different roots give inequivalent one-dimensional representations, and there are exactly $n$ possibilities.

All conjugacy classes are singletons because the group is Abelian. The number of the displayed irreducibles equals the group order, and their squared dimensions sum to $n$, in agreement with the sum of squares of irreducible representation dimensions. Over the real numbers, conjugate nonreal characters instead combine into the planes described by the irreducible real representations of a finite cyclic group.

A stationary iterative method for a linear system converges to its specified fixed point for every initial vector if and only if its iteration matrix satisfies $\rho (H)<1$, where the spectral radius is the largest modulus of an eigenvalue.^[ref]

For a diagonalizable matrix, write $H=V\Lambda V^{-1}$. Then $H^k=V{\Lambda }^k{V}^{-1}$, so each eigenvector coefficient tends to zero exactly when its eigenvalue has modulus less than one. For a general matrix, the Jordan normal form gives the same result: a Jordan block $J=\lambda I+K$ with a nilpotent matrix $K$ of size $m$ satisfiesIf $0<|\lambda |<1$, every coefficient tends to zero because exponential decay beats any fixed polynomial in $k$; if $\lambda =0$, the block vanishes once $k\ge m$. Conversely, an eigenvector with $|\lambda |\ge 1$ provides an initial error that cannot tend to zero. For a real matrix with complex eigenvalues, the same obstruction appears in the real invariant subspace spanned by the real and imaginary parts of an eigenvector.

The words "every initial vector" matter. For example, $H=\mathrm{diag}(3/2,1/2)$ fails the criterion, but $e^{(0)}=(0,1{)}^T$ still tends to zero. Here a generic initial error has a nonzero first coordinate and diverges; the exceptional errors lie on the second coordinate axis. With $H=\mathrm{diag}(-1,1/2)$, generic errors oscillate rather than diverge in norm, while the same exceptional axis still converges. Both are valid matrix splittings with $M=I$ and $A=I-H$. In general the convergent errors form the direct sum of the generalized eigenspaces for $|\lambda |<1$; if $\rho (H)\ge 1$, this is a proper linear subspace.

A small spectral radius describes eventual convergence, but does not imply that the Euclidean norm of the error decreases on every step for a non-normal matrix. An operator norm bound $\Vert H\Vert <1$ does give the stepwise estimate $\Vert e^{(k)}\Vert \le \Vert H{\Vert }^k\Vert e^{(0)}\Vert$. For a real symmetric matrix, the matrix 2-norm equals its spectral radius, making this particularly direct.

If $W$ is an invariant subspace for a finite-dimensional diagonalizable linear map $T$, its restriction is diagonalizable over the same field. The minimal polynomial of $T{|}_W$ divides that of $T$: an annihilating polynomial for $T$ still annihilates the restriction, and polynomial division by its minimal polynomial proves divisibility. A divisor of a product of distinct linear polynomials remains a product of distinct linear polynomials, proving the assertion. This is the key step in simultaneous diagonalization of commuting linear maps.

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.

The permutation matrices of the three displayed double transpositions, together with $I_4$, form a faithful representation of the Klein four-group. They satisfy $P^2=Q^2=I_4$ and $PQ=QP$. Every nonidentity element fixes no basis vector, so this is its four-dimensional regular representation.

The constant vector $(1,1,1,1{)}^T$ spans a nonzero proper invariant subspace, proving reducibility. More completely, the columns ofform a common orthonormal basis of eigenvectors. Direct multiplication gives $S^{T}PS=\mathrm{diag}(1,1,-1,-1)$ and $S^{T}QS=\mathrm{diag}(1,-1,1,-1)$. Thus the representation splits into the four distinct real one-dimensional characters, each once. The basis change is a normalized Hadamard matrix.

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.

For a linear map $T:V\to V$ and an invariant subspace $U$, the rule $\overline{T}(v+U)=T(v)+U$ defines a linear map on the quotient vector space $V/U$. Indeed, two representatives differ by an element of $U$, so their images also differ by an element of $U$. Conversely, this rule is well-defined only if $T(U)\subseteq U$, by comparing every $u\in U$ with zero.

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.

For a linear map $T:V\to V$, an invariant subspace $U$ has an annihilator of a vector subspace $U^0$ invariant under the dual map $T^{\prime }$. Indeed, for $\phi \in U^0$ and $u\in U$,For a finite-dimensional vector space, the converse also holds. If $T^{\prime }(U^0)\subseteq U^0$, every $\phi \in U^0$ vanishes on $Tu$ for each $u\in U$. The dual basis description of $U^0$ shows that the common zero set of its linear functionals is exactly $U$, so $Tu\in U$. ThusThe dimension identity $\dim U^0=\dim V-\dim U$ exchanges a one-dimensional invariant subspace with an invariant hyperplane in the dual space.

For a quarter-turn rotation $T$ of ${\mathbb{R}}^3$ about a unit axis vector $n$, the only real invariant subspaces are $\{0\}$, the axis $\mathbb{R}n$, its perpendicular plane $n^{\perp }$, and all of ${\mathbb{R}}^3$.

Indeed $T^2$ fixes the axis and negates the perpendicular plane. Hence $P=(I+T^2)/2$ is the linear projection onto the axis, while $I-P$ is the linear projection onto the perpendicular plane. An invariant subspace $L$ is preserved by these polynomials in $T$, soThe first intersection is either zero or the axis. If the second contains a nonzero $v$, it also contains $Tv$. These are perpendicular nonzero vectors, so they span the perpendicular plane. This proves the classification.

Over the complex numbers the situation changes: the rotation matrix has eigenvalues $1,i,-i$, giving three one-dimensional eigenspaces. The absence of real eigenvalues on the perpendicular plane explains why it has no real one-dimensional invariant subspace.

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.