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Iterate $q^{(k+1)}=A{q}^{(k)}/\Vert A{q}^{(k)}\Vert$. For a self-adjoint operator with a unique dominant eigenvalue separated in modulus from the others, a starting vector that is nonorthogonal to its dominant eigenvector gives convergence of the eigendirection at the spectral-ratio rate. For a general diagonalizable matrix, the condition is a nonzero coefficient in the dominant component of its eigenbasis, detected by the corresponding left eigenvector; an ordinary dot product with the right eigenvector does not generally detect that coefficient.

The power method is power iteration; its convergence depends on the dominant spectral component of the starting vector.

For example, $B=\left(\begin{array}{cc}2& 1\\ 0& 1\end{array}\right)$ has dominant right eigenvector $(1,0{)}^T$ and a dominant left eigenvector $(1,1{)}^T$. The starting vector $q^{(0)}=(-1,1{)}^T/\sqrt{2}$ has nonzero dot product with the right eigenvector, but $B{q}^{(0)}=q^{(0)}$: it stays forever in the eigenspace for the smaller eigenvalue $1$. Its coefficient along the dominant direction is zero, as the left eigenvector detects. An orthonormal eigenbasis makes the two tests agree in the self-adjoint case.

The principal stresses are the eigenvalues of the symmetric Cauchy stress tensor. In their orthonormal eigenbasis, each traction is normal to its corresponding principal plane.

A quantum Gibbs state is the density operator of a canonical ensemble with inverse temperature $\beta =1/(k_{B}T)$. It exists when the canonical partition function $Z=\mathrm{Tr}(e^{-\beta H})$ is finite. The spectral theorem assigns probabilities $e^{-\beta E_j}/Z$ to an orthonormal eigenbasis of the Hamiltonian. An additive constant in the Hamiltonian cancels between numerator and denominator. For $H=-\mu B{\sigma }_z$, the exponential of a Pauli vector, continued to an imaginary argument, gives ${\rho }_{\beta }=[I+\tanh (\beta \mu B){\sigma }_z]/2$.

A quantum quench changes a quantum system's control parameters rapidly and studies the subsequent unitary time evolution. In the ideal sudden limit, write an initial energy eigenstate $|m_i⟩$ using a complete final orthonormal eigenbasis:The Born rule probabilities for a nondegenerate final energy measurement are $|c_n{|}^2$, with ${\sum }_n|c_n{|}^2=1$. For a degenerate energy eigenvalue, sum the probabilities over its orthonormal energy eigenstates. Subsequent relative quantum phases can alter other measurement probabilities even though these final energy populations stay fixed. A quench need not begin in an energy eigenstate; the same expansion applies to any initial pure state.

The Rayleigh quotient measures the quadratic form of a self-adjoint operator per unit squared norm. For a nonzero vector $x$ and a finite-dimensional Hermitian matrix $A$,For a real symmetric matrix, replace the conjugate transpose $x^{\ast }$ by $x^T$. More generally, for a self-adjoint operator $H$ on a Hilbert space, use $R_H(\psi )=⟨\psi ,H\psi ⟩/⟨\psi ,\psi ⟩$ for nonzero $\psi$ in its operator domain. On an eigenvector it returns the exact eigenvalue. Multiplying the vector by any nonzero scalar leaves the quotient unchanged, so it measures a direction rather than a chosen length.

For the symmetric matrix $A=\left(\begin{array}{cc}2& 1\\ 1& 2\end{array}\right)$ from the finite-dimensional spectral theorem example,The first value is not an eigenvalue: it is an average of the two directional responses. To see every direction, write the unit vector $q(\theta )=\cos \theta e_{+}+\sin \theta e_{-}$ in the orthonormal eigenbasis. ThenThe largest value is $3$ along $e_{+}$; the smallest is $1$ along $e_{-}$. A direction halfway between them has quotient $2$. In the left panel, the color on the unit circle represents the quotient; in the right panel, the same values are plotted against the angle measured from $e_{+}$.

The Hermitian Rayleigh quotient range explains why this is useful: evaluating one trial vector gives an upper bound on the smallest eigenvalue and a lower bound on the largest one, without solving a characteristic polynomial. At an extreme eigenvector, a small angular error changes the quotient only quadratically. Near $e_{+}$ in this example, $3-R_A(q(\theta ))=2{\sin }^2\theta =2{\theta }^2+O({\theta }^4)$. This is the geometric reason for quadratic Rayleigh-quotient improvement for the power method and motivates the Rayleigh-Ritz method.

There is a concrete historical link with normal modes. In The Theory of Sound (1877), Volume I, Sections 88–89, Lord Rayleigh studies the period of a constrained trial vibration. He shows that the period has a stationary point at a normal mode and that a trial type bounds the longest natural period. These energy ratios are a mechanical setting for the modern Rayleigh quotient and generalized Rayleigh quotient; the book describes the result through vibration periods, rather than today's matrix notation.^[ref]

The real case of the finite-dimensional spectral theorem says that a self-adjoint operator on a finite-dimensional real inner product space has a real orthonormal eigenbasis. In orthonormal coordinates this is the spectral theorem for real symmetric matrices, $A=Q\Lambda Q^T$. The worked ellipse under finite-dimensional spectral theorem shows why both orthogonal diagonalization and orthogonality matter: the new coordinates separate the quadratic form into squares without distorting lengths or angles.

A real normal matrix need not have real eigenvalues. For example, the quarter-turn $\left(\begin{array}{cc}0& -1\\ 1& 0\end{array}\right)$ is an orthogonal matrix, but its characteristic polynomial is $z^2+1$ and its eigenvalues are $i,-i$. Over the complex numbers it has an orthonormal eigenbasis; over the real numbers it has no eigenvector. Self-adjointness supplies the reality conclusion in the real spectral theorem.

The displayed parts of a unitary matrix are skew-Hermitian matrices. Expanding $U{U}^{†}=U^{†}U=I$ gives $AB=BA$ and $A^2+B^2=-I$. They admit a common orthonormal eigenbasis. For a unitary eigenvalue $u$, the corresponding part eigenvalues are $i\mathrm{Im}u$ and $-i\mathrm{Re}u$. If $U^T=-U$, then $U^{†}=-\overline{U}$, so these are exactly its real and imaginary parts and both are real skew-symmetric matrices. Such an invertible antisymmetric matrix has even dimension, by its determinant. If one part has distinct eigenvalues, commutation alone shows its one-dimensional eigenspaces are preserved by the other part.

On a closed manifold with a Riemannian metric, an orthonormal eigenbasis for the positive Laplace-Beltrami operator gives this formula, with eigenvalues counted with multiplicity. Polynomial elliptic bounds and exponential time decay give convergence of every derivative away from time zero. The conjugate is necessary for a complex basis.

For a finite irreducible Markov chain satisfying detailed balance with $\pi$, its transition matrix is a self-adjoint operator for $⟨f,g{⟩}_{\pi }={\sum }_x\pi (x)f(x)g(x)$. For a real orthonormal eigenbasis $(f_j)$ with eigenvalues $({\lambda }_j)$,Indeed, apply the spectral decomposition of $P^t$ to the function $1_{\{y\}}/\pi (y)$, whose coefficient against $f_j$ is $f_j(y)$. For $t=0$, every eigenvalue multiplier is interpreted as one.

For a finite-dimensional Hermitian matrix $A$, its spectral projection $P_{\lambda }$ is the orthogonal projection onto the eigenspace for $\lambda$. In an orthonormal eigenbasis, the displayed sum is over every basis vector with that eigenvalue. The projection is independent of the choice of basis within a repeated eigenspace. For distinct eigenvalues $\lambda ,\mu$,This is the spectral decomposition grouped by distinct eigenvalues: decompose a vector into orthogonal pieces, then scale each piece. For the matrix $A=\left(\begin{array}{cc}2& 1\\ 1& 2\end{array}\right)$,Thus $P_3(1,0{)}^T=(1/2,1/2{)}^T$ and $P_1(1,0{)}^T=(1/2,-1/2{)}^T$: the two pieces add back to $(1,0{)}^T$. Squaring the matrix simply squares each scalar, $A^2=9P_3+P_1$, and similarly the matrix exponential is $e^{tA}=e^{3t}{P}_3+e^t{P}_1$. Spectral projections of a self-adjoint matrix turn repeated matrix calculations into scalar calculations on fixed vector subspaces.

Let $v$ be a real unit vector in $d$-dimensional Euclidean space. Then $P=v{v}^{\mathsf{T}}$ is a rank-one orthogonal projection: $Pv=v$, while $Pw=0$ for $w$ orthogonal to $v$. Consequently the symmetric matrix $A=\alpha I+\beta P$ has eigenvalue $\alpha +\beta$ on the line spanned by $v$ and eigenvalue $\alpha$ on its orthogonal complement. Any orthonormal basis of that complement completes $v$ to an orthonormal eigenbasis. ThusThe trace and determinant formulas remain valid when one of these eigenvalues is zero. When $\beta =0$, all nonzero vectors have the same eigenvalue, and the chosen splitting is optional.

For random walk on $\{-1,1{\}}^n$ that chooses a coordinate uniformly and flips it with probability $1/2$, the Walsh characters $f_J(x)={\prod }_{j\in J}{x}_j$ form an orthonormal eigenbasis under the uniform distribution on a finite set. Flipping a coordinate in $J$ negates $f_J$, so $P{f}_J=(1-|J|/n)f_J$. Hence $1-k/n$ has multiplicity $\left(\genfrac{}{}{0ex}{}{n}{k}\right)$, and the ordinary and absolute spectral gaps both equal $1/n$.

For a real symmetric matrix $X$ with eigenvalues ${\lambda }_1\ge \cdots \ge {\lambda }_n$, the sum of its $k$ largest eigenvalues is the support function of the fantope:Here $⪯$ is the Loewner order. In an orthonormal eigenbasis for $X$, the objective depends only on the diagonal entries of $Y$, which lie in the capped simplex. Thus the identity reduces to the sum of the largest components of the eigenvalues.

For a real symmetric matrix $M$, real $\mu$ and nonzero vector $u$, some eigenvalue of $M$ lies within the stated ratio of $\mu$. In an orthonormal eigenbasis, write $u={\sum }_j{c}_j{q}_j$ and $M{q}_j={\theta }_j{q}_j$. Then $\Vert Mu-\mu u{\Vert }_2^2={\sum }_j({\theta }_j-\mu {)}^2{c}_j^2\ge {\min }_j|{\theta }_j-\mu {|}^2\Vert u{\Vert }_2^2$, proving the bound.

For a real orthogonal matrix $A$, $A^{\mathsf{T}}=A^{-1}$. If $Av=\lambda v$ for a nonzero complex eigenvector, then $A^{\mathsf{T}}v={\lambda }^{-1}v$. Therefore the same eigenvector has eigenvalue $(\lambda +{\lambda }^{-1})/2$ for the symmetric part of a matrix $(A+A^{\mathsf{T}})/2$. Orthogonality implies $|\lambda |=1$, so writing $\lambda =e^{i\theta }$ gives $\cos \theta$. The complete complex orthonormal eigenbasis of an orthogonal matrix shows that all the eigenvalues of its symmetric part lie in $[-1,1]$.

For a symmetric positive-definite matrix $A$, solving $A{x}_{\ast }=b$ is equivalent to minimizing the quadratic form with linear termWriting the error vector $e=x-x_{\ast }$ and expanding gives the exact energy norm identityThe cross term vanishes because $A{x}_{\ast }=b$. With $\eta =Se$ and $B=S^{-1}A{S}^{-1}$, the same quantity is ${\eta }^{T}B\eta /2$. Symmetric positive-definite preconditioning changes coordinates without changing the energy assigned to a corresponding error vector.

In an orthonormal eigenbasis of $A$, the level set $e^{T}Ae=t>0$ has semiaxis lengths $\sqrt{t/{\lambda }_i}$. Its longest-to-shortest axis ratio is $\sqrt{{\kappa }_2(A)}$. A large spectral condition number of a positive-definite matrix therefore appears geometrically as a thin ellipsoid. Good scaling makes the transformed ellipsoid rounder, allowing a simple search direction to improve all coordinates together.

The negative gradient of $F$ is the linear-system residual $r$. Relative to the norm $\Vert v{\Vert }_M=\sqrt{v^{T}Mv}$, a steepest direction is $z=M^{-1}r$: the Cauchy-Schwarz inequality applied to $r^{T}v=(S^{-1}r{)}^T(Sv)$ shows that the greatest decrease per unit $M$-length occurs along this direction. Equivalently, ordinary gradient descent in $y=Sx$ becomes preconditioned steepest descent in $x$.^[ref]

The sum of the largest eigenvalues has the semidefinite program representationFor every feasible $Y$ in the fantope, positive semidefinite trace nonnegativity gives $\mathrm{tr}(XY)\le kt+\mathrm{tr}S$. To attain equality, use an orthonormal eigenbasis of $X$ and choose $S=\mathrm{diag}(({\lambda }_i-t{)}_{+})$ in that basis, with the same threshold choice as in the threshold formula for the sum of the largest components. This argument also covers $k=n$, without requiring strict feasibility of the maximization program.

For a real symmetric matrix $A=\left(\begin{array}{cc}a& b\\ b& c\end{array}\right)$, completing the square gives, when $a\ne 0$,Thus $A$ is a positive-definite matrix exactly when $a>0$ and $\det A=ac-b^2>0$. For necessity, test $(1,0)$ first, then $(-b/a,1)$; for sufficiency, the displayed two positive squares give the result. These are the first two leading principal minors in Sylvester's criterion. A positive determinant alone is insufficient: $-I_2$ has determinant $1$ but is a negative-definite matrix.

The eigenvalues provide an equivalent test under the same real-symmetric hypothesis:The spectral theorem for real symmetric matrices supplies coordinates in an orthonormal basis in which $q={\lambda }_{+}{u}^2+{\lambda }_{-}{v}^2$. More generally, a complex Hermitian matrix has an orthonormal eigenbasis and $q(z)={\sum }_j{\lambda }_j|w_j{|}^2$, so it is a Hermitian positive-definite matrix exactly when every eigenvalue is positive. This is a short proof outline: test the eigenvectors for necessity, and sum the positive squared absolute values for sufficiency.^[ref]

For a real positive semidefinite matrix, the condition is instead $q(x,y)\ge 0$ for all vectors, equivalent here to $a\ge 0$, $c\ge 0$, and $ac-b^2\ge 0$. If $a>0$, the same completing the square proves this; if $a=0$, nonnegativity forces $b=0$ and leaves $c{y}^2$ with $c\ge 0$. Merely replacing the strict inequalities on the leading principal minors by nonstrict ones fails: $\mathrm{diag}(0,-1)$ has both leading principal minors zero yet gives $q(0,1)=-1$. An indefinite quadratic form takes both positive and negative values; its real symmetric matrix has eigenvalues of both signs.

Let a real symmetric matrix have an orthonormal eigenbasis $w_1,\dots ,w_n$, withWrite two starting vectors asFor two-column subspace iteration to recover
$\mathrm{span}(w_{n-1},w_n)$, the two projections onto that dominant subspace must be independent. The exact condition isRequiring each of the four coefficients to be nonzero does not imply this determinant condition.

Take an isolated two-level system, or a spin one-half qubit, with Pauli matrices ${\sigma }_x,{\sigma }_z$. Let ${\sigma }_z|0⟩=|0⟩$ and ${\sigma }_z|1⟩=-|1⟩$. The initial and final Hamiltonian operators areThe energy gaps are ${\Delta }_i,{\Delta }_f$. Prepare the initial ground state $|0⟩$, with energy eigenvalue $-{\Delta }_i/2$. A final orthonormal eigenbasis isDirect matrix multiplication gives $H_f|g_f⟩=-{\Delta }_f|g_f⟩/2$ and $H_f|e_f⟩=+{\Delta }_f|e_f⟩/2$. The unchanged initial quantum state is $|0⟩=c|g_f⟩-s|e_f⟩$. Its new probability amplitudes are the inner products $c,-s$; the Born rule gives $p_g=c^2,p_e=s^2$. The angle between the real amplitude vectors is $\theta /2$, whereas the Hamiltonian axis on the Bloch sphere rotates through $\theta$.

At $\theta =\pi /3$, $c=\sqrt{3}/2$ and $s=1/2$, so the final energy measurement gives $-{\Delta }_f/2$ with probability $3/4$ and $+{\Delta }_f/2$ with probability $1/4$. At $\theta =0$ the initial ground state remains the final ground state; at $\theta =\pi$ it is the final excited energy eigenstate. These statements are about the new energy labels, not an instantaneous physical rotation of the quantum state.