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For a diagonalizable matrix, the active spectrum consists of eigenvalues whose eigenspaces have nonzero components of the starting vector in its spectral decomposition. These components are orthogonal projections for a self-adjoint operator, but need not be orthogonal projections for a general diagonalizable matrix. With a unique largest modulus in this active spectrum, the power method selects its eigendirection; an exactly absent dominant component is never created in exact arithmetic.

If a normal matrix satisfies $Ax=\lambda x$, then $A^{†}x=\overline{\lambda }x$. Indeed, $M=A-\lambda I$ is normal and $\Vert Mx{\Vert }^2=x^{†}{M}^{†}Mx=x^{†}M{M}^{†}x=\Vert M^{†}x{\Vert }^2$. Since $Mx=0$, also $M^{†}x=0$. This identity proves orthogonality of distinct eigenspaces and gives spectral restrictions for Hermitian matrices, skew-Hermitian matrices and unitary matrices.

The Born rule connects a quantum state to measurement probabilities. Let $\{|j⟩\}$ be a complete orthonormal basis, finite or countably infinite, for a Hilbert space, and letAn ideal measurement distinguishing these basis vectors returns label $j$ with probabilityThis assignment is a physical measurement rule. Its compatibility with total probability one follows from orthonormality and normalization:The probabilities are the squared moduli, not the amplitudes themselves or the squares $a_j^2$. ^[ref].

If the basis is an eigenbasis of an observable, the reported numerical value is its eigenvalue. Several vectors can share one eigenvalue; the probability of that value is then the sum over the corresponding basis labels. More generally, a projective measurement with orthogonal projections $P_r$ assignsfor a pure state and a general density matrix $\rho$, respectively. This also covers mixed states and degenerate eigenspaces. The Born rule determines outcome statistics; the ideal conditional state change is specified separately by the Lüders rule. See projective measurement for that formulation and its assumptions. ^[ref].

For a worked example, take the normalized qubit quantum stateThe squared moduli add to $3/4+1/4=1$. A computational-basis measurement, equivalently a measurement of the $Z$ Pauli matrix, hasThese probabilities do not depend on the relative quantum phase $\varphi$. Now choose the Hadamard basiswhose vectors are eigenstates of the $X$ Pauli matrix, also used by the Pauli X gate. The new probability amplitudes and probabilities areBoth probabilities are nonnegative and their sum is one. The relative quantum phase becomes visible because the two probability amplitudes add before their squared modulus is taken:

A Hadamard gate followed by a computational-basis measurement realizes these $X$ statistics, since $H|\pm ⟩=|0⟩,|1⟩$, respectively. Compare the two bases using fresh, identically prepared copies: a preceding measurement may change the state. The quantum superposition section shows why replacing the preparation by the mixed state $\mathrm{diag}(3/4,1/4)$ keeps the $Z$ statistics but gives $p_{+}=p_{-}=1/2$ for every $\varphi$.

The usefulness of the Born rule is this conversion from linear algebra to experimental predictions: calculate overlaps in the basis actually measured, then take squared moduli. Its appealing geometry is that changing basis redistributes probability while preserving total squared norm. The same rule relates an energy eigenstate expansion to an energy measurement and a position wavefunction to a probability density.

Born's 1926 paper on quantum scattering interpreted the outgoing-wave coefficients in terms of the probabilities of collision outcomes; a corrective footnote specified proportionality to the square of the amplitude. This is a historical scattering formulation, rather than the general density matrix formula above. ^[ref]; ^[ref].

In a finite-dimensional short exact sequence $0\to N\to E\to \mathbb{C}\to 0$ of representations of a complex semisimple Lie algebra, the generalized zero eigenspace of the Casimir operator maps onto the trivial quotient. All its irreducible composition factors have zero Casimir eigenvalue, so they are trivial. The action is therefore strictly upper triangular and has solvable image; because a semisimple Lie algebra is a perfect Lie algebra, that image is zero. Every lift in this generalized zero eigenspace is invariant, giving a split short exact sequence.

For a diagonal matrix $D=\mathrm{diag}(d_1,\dots ,d_n)$ over a field, matrix multiplication gives $(BD-DB{)}_{ij}=(d_j-d_i)b_{ij}$. Hence a matrix $B$ commutes with $D$ exactly when $b_{ij}=0$ wherever $d_i\ne d_j$. If all diagonal entries are distinct, the centralizer of a diagonal matrix consists exactly of diagonal matrices. When entries repeat, the allowed entries form unrestricted blocks on the corresponding eigenspaces.

Let a Hermitian matrix have distinct real eigenvalues with multiplicities $n_1,\dots ,n_k$. A unitary matrix commuting with it must preserve each eigenspace: if $Av=\lambda v$, then $A(Uv)=UAv=\lambda Uv$. Conversely, any unitary transformation acting independently on these mutually orthogonal eigenspaces commutes with $A$. The centralizer in $U(n)$ is therefore ${\prod }_{j}U(n_j)$. Imposing total determinant one gives the displayed subgroup inside the special unitary group $SU(n)$.

For multiplicities $(2,1)$ in SU(3), it is $S(U(2)\times U(1))\cong U(2)$, embedded by $U\mapsto \mathrm{diag}(U,(\det U{)}^{-1})$. Three distinct eigenvalues leave the diagonal maximal torus $U(1{)}^2$. This determines the unbroken group for a diagonal adjoint scalar field vacuum, including its global determinant constraint.

Use the standard polytabloids for tableaux with rows $12\mid 34$ and $13\mid 24$, as an ordered basis $e_1,e_2$. Adjacent transpositions act byIn characteristic three their common fixed line is $F(e_1+e_2)$. On its quotient every transposition acts by minus one, giving the sign representation. Thus the displayed sequence has the trivial representation and sign as its composition factors, each once. It is nonsplit: the minus-one eigenspace of $\rho (12)$ is $F{e}_2$, which is not preserved by $\rho (23)$.

The natural tabloid form restricts to Gram matrix $\left(\begin{array}{cc}4& 2\\ 2& 4\end{array}\right)$. Its radical in characteristic three is exactly that fixed line. Therefore $D^{(2,2)}$ is the sign quotient, while the trivial factor is $D^{(4)}$. This gives an explicit small failure of semisimplicity in modular symmetric-group representation theory.

Over an algebraically closed field of characteristic zero, every three-dimensional perfect Lie algebra is isomorphic to the sl2 Lie algebra. Indeed the trace of a commutator gives $\mathrm{tr}(\mathrm{ad}x)=0$ for all $x$, because each element is a sum of Lie brackets. If every $\mathrm{ad}x$ were a nilpotent linear map, the Engel theorem would make the Lie algebra nilpotent, contradicting its constant nonzero lower central series of a Lie algebra. Some $\mathrm{ad}x$ therefore has a nonzero eigenvalue $\lambda$. Its other eigenvalues are $0,-\lambda$, since it kills $x$ and has zero trace. They are distinct, so its eigenvectors give a basis $h,e,f$ after scaling $x$ to $h=2x/\lambda$, with $[h,e]=2e$ and $[h,f]=-2f$. The Jacobi identity puts $[e,f]$ in the zero eigenspace $kh$. Its coefficient cannot vanish, since otherwise the derived algebra would lie in $ke\oplus kf$. Rescaling $f$ gives $[e,f]=h$. Thus this basis is an sl2 triple, proving the claimed Lie algebra isomorphism.

Two real symmetric matrices are coaxial when they share an orthonormal eigenbasis. Equivalently they commute. Degenerate eigenspaces can have several choices of shared axes.

If $CD=DC$ and $Dx=\lambda x$, then $D(Cx)=C(Dx)=\lambda Cx$. Therefore each eigenspace of $D$ is invariant under $C$. The vector $Cx$ may be zero; invariance does not require it to be a nonzero eigenvector. This observation is the starting point for simultaneous diagonalization.

For commuting real symmetric matrices $A,B$, each eigenspace of $A$ is invariant under $B$: $A(Bv)=B(Av)=\lambda Bv$. The restriction of $B$ to that eigenspace is again symmetric. The spectral theorem therefore supplies an orthonormal eigenbasis for $B$ inside each eigenspace of $A$. Combining these bases gives a common orthonormal eigenbasis of the whole space. Conversely, two symmetric matrices diagonal in the same orthonormal basis commute.

For a strongly continuous unitary representation of a compact group, this Bochner integral exists in the Hilbert-Schmidt norm and is a positive compact operator commuting with the representation. It is nonzero if $\xi \ne 0$, because $⟨K_{\xi }\xi ,\xi ⟩={\int }_G|⟨U_g\xi ,\xi ⟩{|}^{2}dg>0$. A nonzero eigenspace is finite-dimensional and invariant. Applying this on orthogonal complements yields an orthogonal direct sum of finite-dimensional irreducible representations, with no separability assumption on the represented Hilbert space.

On a closed manifold with a Riemannian metric the positive Laplace-Beltrami operator is self-adjoint with compact resolvent. It has a complete smooth orthonormal eigenbasis, finite-dimensional eigenspaces, nonnegative eigenvalues tending to infinity, and polynomial eigenvalue-counting bounds. Elliptic regularity bounds derivatives of its eigenfunctions polynomially in their eigenvalues.

A complete set of commuting observables consists of mutually commuting quantum observables whose joint eigenvalues distinguish all vectors of a simultaneous eigenbasis, up to phase. For a finite-dimensional Hilbert space, commuting self-adjoint operators admit a simultaneous orthonormal basis of eigenvectors; completeness means that every joint eigenspace is one-dimensional. An additional quantum observable that is already a function of the others supplies no independent quantum number.

For the addition of angular momentum, the uncoupled labels $(j_1,j_2,m_1,m_2)$ correspond to $\{J_1^2,J_2^2,J_{1z},J_{2z}\}$, whereas the coupled labels $(j_1,j_2,j,m)$ correspond to $\{J_1^2,J_2^2,J^2,J_z\}$. Within a fixed pair of irreducible angular momentum spaces each choice gives a natural eigenbasis. The Clebsch-Gordan coefficients implement the change between them.

In a unit square with prescribed conductive boundary temperatures and impermeable boundaries, the stream function convention $\mathbf{u}=(-{\psi }_z,0,{\psi }_x)$ gives linear equations $\Delta \psi =R{\theta }_x$, ${\theta }_t={\psi }_x+\Delta \theta$, with $\psi =\theta =0$ on all sides. The velocity equation follows by taking the vertical-plane curl of Darcy law, and the temperature equation by linearizing the heat equation about $1-z$. Unlike a laterally unbounded Darcy layer, the conducting side boundaries produce a two-dimensional real marginal eigenspace at $R_c=8{\pi }^2$.

A continuous symmetric approximate identity on a compact group gives a compact self-adjoint convolution operator $T$ on $L^2(G)$ which maps into $C(G)$ and commutes with left translations. Its nonzero eigenspaces are finite-dimensional invariant spaces of continuous functions, hence consist of matrix coefficients of finite-dimensional unitary representations. If $S_N$ projects onto increasing finite sums of these eigenspaces, $S_{N}Tf\to Tf$ in $L^2$, and the bound $\Vert Th{\Vert }_{\infty }\le \Vert k{\Vert }_2\Vert h{\Vert }_2$ gives $T{S}_{N}Tf\to T^{2}f$ uniformly. Choosing the approximate identity narrow makes $T^{2}f$ uniformly close to $f$. This supplies uniform approximation without confusing an $L^2$ limit with a uniform limit.

Distinct squared traction-free eigenfrequencies in a bounded lossless elastic body have eigenfunctions orthogonal in $L^2(\mathcal{D},\rho dV{)}^3$, by the Betti identity for elastodynamic fields. Normalize one field from each eigenspace. This is an orthonormal set in a separable Hilbert space, so it is countable: disjoint balls of radius less than $1/\sqrt{2}$ around its unit vectors must contain distinct points of a countable dense set. Positive and negative frequency partners add at most a factor of two. This argument proves countability without requiring a separate compact-resolvent proof.

The Davis-Kahan theorem bounds the change of an invariant eigenspace of a symmetric matrix in terms of a matrix perturbation and an appropriate spectral gap. The rank-one eigenprojector perturbation bound below depends only on the unperturbed gap. Yu, Wang and Samworth's formulation gives population-gap bounds in greater generality.

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 $\lambda =-3$.

Distinct linearly independent normal modes are degenerate if they have the same eigenfrequency. For $K\mathbf{u}={\omega }^{2}M\mathbf{u}$, their eigenvectors span a common eigenspace; linear combinations in that space are also modes with the same frequency. A symmetry preserving both quadratic energies maps this space into itself. Higher-dimensional irreducible symmetry representations can enforce degeneracy, while perturbations breaking that symmetry can split it. Degeneracies can also occur accidentally without such symmetry protection.