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Use the standard Hermitian inner product $⟨w,v⟩=w^{†}v$. Since $W$ is invariant under every $D_2(g)$, it is in particular invariant underLet $v\in W^{\perp }$ and $w\in W$. Thenbecause $D_1(g{)}^{†}w\in W$ and $v$ is orthogonal to $W$. Thus $D_1(g)v$ is orthogonal to every $w\in W$, provingTherefore $W^{\perp }$ is an invariant subspace for $D_1$. If $D_1$ is irreducible, $W^{\perp }$ is either $\{0\}$ or ${\mathbb{C}}^n$. In finite-dimensional inner product space, $(W^{\perp }{)}^{\perp }=W$, so these alternatives force $W={\mathbb{C}}^n$ or $W=\{0\}$. HenceNo unitarity of $D_1$ was assumed; the inverse-adjoint representation is exactly what makes this orthogonal complement argument work.

A rigid rotation of a regular tetrahedron permutes its four vertices, giving an injective group homomorphism from its rotation group to $S_4$: fixing every vertex forces the rotation to be the identity. Its seven nontrivial rotation axes are four lines through a vertex and the opposite face centre, and three lines through midpoints of opposite edges.

Each vertex-face axis admits rotations by $\pm 2\pi /3$, giving eight three-cycles on the vertices. Each opposite-edge axis admits a half-turn, giving three double transpositions. Along with the identity there are twelve rotations. All their induced permutations are even, so the injective image is a twelve-element subgroup of $A_4$, hence the whole group. ThereforeIts action on ${\mathbb{R}}^3$ supplies a three-dimensional representation. To prove irreducibility even after complexification, choose verticesThe group contains the half-turns $D_x=\mathrm{diag}(1,-1,-1)$ and their cyclic variants, as well as the coordinate cycle $(x,y,z)\mapsto (y,z,x)$. If a nonzero invariant subspace over the complex numbers contains a vector with nonzero $x$ component, applying $(I+D_x)/2$ shows it contains the $x$ axis. If another component is nonzero, use the corresponding projection instead. The coordinate cycle then supplies the other two axes. Every nonzero invariant subspace is therefore the full space, provingThis also gives an elementary proof without presupposing the character table.

The initial displacement lies in an invariant subspace of axial motions. Let the three opposite vertex pairs have positionsFor a vertex on the positive $x$ axis, the two springs towards the $y$ vertices contribute $-k(q_x+q_y)$ in the $x$ direction; the two towards the $z$ vertices contribute $-k(q_x+q_z)$. Transverse components cancel in pairs. Cyclically,The matrix is $I+11^{\mathsf{T}}$. It has eigenvalue $4$ on the breathing vector $(1,1,1)$, and eigenvalue $1$ on the orthogonal plane $q_x+q_y+q_z=0$. This identifies the axial normal modes of an octahedral spring framework without a large characteristic determinant.

Put ${\omega }_0=\sqrt{k/m}$ and decompose the initial vector asSince all initial velocities are zero, only cosine time factors occur. The complete motion to first order in the displacements iswith the vertex positions given by the first displayed formula. No translation or rotation is excited: the initial displacements are orthogonal to those zero modes and their initial velocities vanish. The two checks ${ϵ}_1={ϵ}_2$ and ${ϵ}_1+2{ϵ}_2=0$ respectively isolate pure breathing and pure traceless axial oscillation. Nonlinear corrections beyond the small-displacement approximation have been omitted.

The plane spanned by $u_1,v_1$ is an invariant subspace for $M$, sinceIts orthogonal complement is also invariant: for $w$ orthogonal to both, $u_1^{T}Mw=-(M{u}_1{)}^{T}w=0$ and $v_1^{T}Mw=-(M{v}_1{)}^{T}w=0$. The restriction of the real symmetric matrix $M^2$ to this nonempty complement therefore has a unit eigenvector $u_2$, by the spectral theorem for real symmetric matrices.

If its eigenvalue is $-{\mu }_2^2<0$, the argument in part (b) shows that $v_2=M{u}_2/{\mu }_2$ is a unit eigenvector orthogonal to $u_2$. It also lies in the same complement; explicitly,Repeating the construction decomposes every negative eigenspace into orthogonal pairs. The zero eigenspace has dimension $n-2r$ for some number $r$ of pairs, and therefore has the same parity as $n$. This is the paired eigenspaces of the square of a real skew-symmetric matrix property.

For $n=2m$, there can be $m$ distinct negative eigenvalues and no zero, or at most $m-1$ negative ones plus zero, because a nonempty zero eigenspace has dimension at least two. For $n=2m+1$, there are at most $m$ distinct negative ones plus the mandatory zero. HenceThe bound is attained by skew two-by-two blocks with distinct positive parameters, adding one zero direction when $n$ is odd.

Choose a real unit eigenvector $e_1$ of the real symmetric matrix $Q$, with eigenvalue ${\lambda }_1$. Its orthogonal complement is an invariant subspace: for any $v$ with $e_1^{T}v=0$,The restriction of $Q$ to this $(n-1)$-dimensional space is again real and symmetric. Applying the same argument inductively produces an orthonormal basis $e_1,\dots ,e_n$ of real eigenvectors. The base case $n=1$ is immediate. This proves the spectral theorem for real symmetric matrices without assuming that distinct eigenvalues are available.

One can also check that eigenvectors with different eigenvalues are automatically orthogonal: ${\lambda }_i{e}_i^T{e}_j=(Q{e}_i{)}^T{e}_j=e_i^{T}Q{e}_j={\lambda }_j{e}_i^T{e}_j$. Within a repeated eigenspace, the Gram-Schmidt process preserves the eigenvector property. Thus repeated eigenvalues cause no obstruction, and $\boxed{Q=OD{O}^T\text{ for an orthogonal }O\text{ and real diagonal }D}$.

From part (a), $[N,a_j]=-a_j$ and $[N,a_j^{\ast }]=a_j^{\ast }$. The commutator product rule givesConsequentlyEach term in $L$ transfers one excitation between the two modes, preserving the total number operator eigenvalue. This is an instance of the oscillator bilinear commutator; every number operator eigenspace is an invariant subspace of $L$.

Write $u={\psi }_{1,0}=a_1^{\ast }\psi$ and $v={\psi }_{0,1}=a_2^{\ast }\psi$. The canonical commutation relations for the creation and annihilation operators giveFor example, $a_1{a}_1^{\ast }\psi =(I+a_1^{\ast }{a}_1)\psi =\psi$. The expression for $L$ from part (b), with distinct modes reordered, then yieldsThus $\mathrm{span}\{u,v\}$ is an invariant subspace, and the matrix of $L$ in the ordered orthonormal basis $(u,v)$ isIts characteristic polynomial is ${\ell }^2-{\hslash }^2$, giving the eigenvalues and normalized eigenstatesIndeed, $L(u\pm iv)=\pm \hslash (u\pm iv)$ and $\Vert u\pm iv{\Vert }^2=2$ by the assumed orthonormality. Both eigenstates remain in the $N=1$ eigenspace.

A linear operator $T:V\to V$ on a vector space over a field $k$ makes $V$ a module over the polynomial ring $k[t]$: let $t$ act as $T$, let constants act as the original scalars, and define $p(t)\cdot v=p(T)v$. The identity $(pq)(T)=p(T)q(T)$ for polynomials $p,q$ verifies the multiplication axiom. The submodules are precisely the invariant subspaces of $T$. This construction is described in Keith Conrad's Introductory Notes on Modules, section 6.

For a computed example, take $k=\mathbb{Q}$, $V=k^2$, andThen $T{e}_2=e_1$, $T{e}_1=0$, and $T^2=0$. ConsequentlyEvery $(a,b)$ equals $(b+at)\cdot e_2$. The module homomorphism $k[t]\to V$, $p\mapsto p(T)e_2$, is therefore surjective, and its kernel is the ideal $(t^2)$. It induces a module isomorphism $V\cong k[t]/(t^2)$. This is a cyclic module and a torsion module over $k[t]$, since $t^{2}V=0$. Over $k$ it is a free module with basis $(e_1,e_2)$ and has no nonzero torsion elements of a module. Over $\mathbb{Q}$ its nonzero elements also have infinite additive order of a group element. Changing scalars changes the questions one asks about the same addition law.

Suppose $p(T)q(T)=0$ for a linear operator $T$, and Bézout's identity for polynomials gives $ap+bq=1$. The kernel decomposition for coprime polynomials is $V=\ker p(T)\oplus \ker q(T)$. The displayed operators are the corresponding projection operators: $P$ is the identity on $\ker p(T)$ and zero on $\ker q(T)$, while $Q$ has the opposite action. Thus $P+Q=I$, $P^2=P$, $Q^2=Q$ and $PQ=QP=0$. These actions also prove independence from the choice of Bézout coefficients. Every summand is an invariant subspace of $T$, since the projections are polynomials in $T$ and commute with it.

For a linear map $T:V\to V$ on a finite-dimensional vector space over a field $F$, factor its minimal polynomial as $m_T={\prod }_i{p}_i^{a_i}$ with distinct monic polynomials $p_i$ that are irreducible polynomials. ThenEach $V_i$ is an invariant subspace, since $T$ commutes with every polynomial in $T$. These are canonical vector subspaces, determined by $T$, without requiring any eigenvalues in $F$.

For a proof, put $m_i=p_i^{a_i}$ and $M_i=m_T/m_i$. The coprime polynomials $m_i,M_i$ admit a Bézout's identity for polynomials $b_i{M}_i+c_i{m}_i=1$. Set $E_i=b_i(T)M_i(T)$. Its image is contained in $V_i$ because $m_i(T)E_i=0$. The polynomial ${\sum }_i{b}_i{M}_i-1$ is divisible by every $m_i$, hence by their product, so ${\sum }_i{E}_i=I$. On $V_i$, $E_i=I$; on $V_j$ with $j\ne i$, $E_i=0$ because $m_j$ divides $M_i$. These facts prove the direct sum and identify $E_i$ as its projection operators. Finally, if the minimal polynomial of any restriction had a smaller exponent, $m_T/p_i$ would annihilate every summand and hence all of $V$, a contradiction.

A nonzero representation is reducible when it has a nonzero proper invariant subspace; otherwise it is an irreducible representation.

The reflection representation is the action of the Weyl group on the real span of its root system, generated by its Weyl reflections. For an irreducible root system, it is an irreducible representation: if $U$ is an invariant subspace, each root either lies in $U^{\perp }$ or, by subtracting a reflected vector of $U$, lies in $U$. These alternatives partition the roots into orthogonal sets. Irreducibility and spanning force $U=0$ or the whole space.

Representation theory describes abstract symmetries by linear maps on vector spaces. A group representation assigns an invertible linear map to each group element, respecting multiplication. The payoff is that a symmetry problem becomes a problem about matrices, invariant subspaces, and eigenvalues. Decomposing the action can reveal combinations of variables that a symmetric system cannot mix.

Take the cyclic group $C_3=\{e,g,g^2\}$, with $g^3=e$. On three real coordinates, let $g$ send $(a,b,c)$ to $(c,a,b)$:This permutation representation does more than relabel a list: it acts linearly on every possible assignment of values to the three positions. The vector $(1,1,1)$ stays fixed. The vector subspace $W=\{(a,b,c):a+b+c=0\}$ is also preserved, since a permutation does not change the sum.

Every vector separates uniquely into these two types:For example, $(4,1,1)=2(1,1,1)+(2,-1,-1)$. The first term belongs to a trivial representation; the second is a contrast with zero coordinate sum. This is a direct sum decomposition ${\mathbb{R}}^3=\mathbb{R}(1,1,1)\oplus W$. The symmetry preserves the mean and rotates the contrasts.

The rotation is literal after choosing the orthonormal basis $u=(2,-1,-1)/\sqrt{6}$, $v=(0,1,-1)/\sqrt{2}$ of $W$. In this basis,It is a rotation through $120$ degrees, and three applications return to the starting point. There is no nonzero invariant real line, because this rotation has no real eigenvector. Thus the two-dimensional action is an irreducible representation over the real numbers. Over the complex numbers it instead splits into two eigenlines with eigenvalues $e^{2\pi i/3}$ and $e^{-2\pi i/3}$. The coefficient field is part of an irreducibility claim.

Why would this decomposition matter for a physical calculation? Consider three equal masses with scalar displacements $x_i$ and the symmetric quadratic coupling energyThe equations of motion are $m\ddot{x}=-kLx$, whereis a graph Laplacian and $J$ is the all-ones matrix. The constant vector has eigenvalue zero: translating all coordinates equally costs no coupling energy. On $W$, $J$ vanishes and $L=3I$, so both independent contrast directions have angular frequency $\sqrt{3k/m}$. A three-coordinate coupled problem reduces to a free mean and two identical harmonic oscillators.

For initial displacement $x(0)=(4,1,1)$ and initial velocity zero, the exact solution is $x(t)=2(1,1,1)+\cos (\sqrt{3k/m}t)(2,-1,-1)$. The fixed mean and oscillating contrast can be read directly from the decomposition. More generally, a linear operator commuting with the group action carries its fixed-vector subspace into itself; with an invariant inner product, it preserves the corresponding orthogonal complement too.

For a finite group, averaging over the action constructs an invariant inner product over the real numbers or complex numbers. The resulting invariant complements explain Maschke's theorem: a finite-dimensional group representation decomposes into irreducible representations. Characters compress the action into traces and help identify those pieces. The assumptions matter; in modular representation theory the coefficient field can have characteristic dividing the group order, and such decompositions can fail. Chapter 4 of Pavel Etingof and coauthors' notes develops these mechanisms. ^[ref].

For a bounded normal operator $T$ with spectral measure of a normal operator $P$, any projection $P(E)$ commutes with $T$ and $T^{\ast }$ by the Borel functional calculus for a normal operator. Its range is a closed reducing subspace. If the spectrum has two points, choose disjoint nonempty relative open sets around them. Full support of a faithful spectral measure makes both projections nonzero and their product zero, so either range is nonzero and proper, in particular an invariant subspace.

For the upper-left SU(3) subgroup of SU(4), the defining representation restricts as $4=3\oplus 1$. Write a traceless matrix in three-plus-one block form. Its upper-left traceless block is the SU(3) Adjoint representation, its upper-right column is $3$, its lower-left row is $\overline{3}$, and the remaining diagonal direction $\mathrm{diag}(1,1,1,-3)$ is fixed. These invariant subspaces have dimensions $8,3,3,1$ and exhaust the SU(4) Adjoint representation. This also proves the formula by restricting $4\otimes \overline{4}-1$.

A subrepresentation is an invariant subspace of a group representation or Lie algebra representation, with the restricted action.

A $(j,k)$ tensor has one fundamental representation factor for every upper index and a dual factor for every lower index. Under $A\in SU(n)$, upper factors transform by $A$ and lower factors by $A^{-1}$; unitarity identifies the latter with conjugate fundamental matrices. Thus complex conjugation exchanges $(j,k)$ and $(k,j)$. Index permutations and contraction with an invariant tensor commute with the action, producing invariant subspaces and decomposition maps.

The symmetric square of a group representation on $V$ is the quotient vector spacewith the induced tensor product representation. If the characteristic of a field is different from two, $vw\mapsto (v\otimes w+w\otimes v)/2$ identifies it with the invariant subspace fixed by the factor-exchange map $\tau$. Over the complex numbers, the projection operator $(I+\tau )/2$ givesIn characteristic two, the quotient definition still works, but this averaging projection operator does not exist. Real coordinate-quadratic perturbations of a nonmagnetic electronic multiplet transform in this symmetric square.

A linear map on a finite-dimensional vector space over an algebraically closed field has both an upper triangular matrix and a lower triangular matrix in suitable bases. One proof uses invariant subspaces and annihilators to make the induction step entirely in the dual space.

Choose an eigenvector $v$ and let $U=\mathrm{span}\{v\}$. The dual map preserves the annihilator of a vector subspace $U^0$, of dimension one less. By induction, choose a basis of $U^0$ making the restriction upper triangular. Extend this to a basis of the whole dual space; its matrix $C$ remains upper triangular because $U^0$ is an invariant subspace. In the corresponding primal basis, the matrix is $C^T$, which is lower triangular. Reversing the primal basis makes it upper triangular. Dimension zero and one start the induction. This also allows the original eigenvector to be the first vector of the resulting upper triangular basis.

The trivial representation sends every group element to the identity operator. A one-dimensional constant-coordinate line in a permutation representation is an example: relabelling positions leaves a uniform assignment unchanged. A trivial action on a higher-dimensional vector space has many invariant subspaces and is therefore not an irreducible representation.