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A damped pendulum is a simple pendulum whose mechanical energy is removed by a resisting torque. Here the model is undriven, with a fixed pivot, a point mass $m>0$ at distance $\ell >0$, uniform gravitational acceleration $g>0$, and resisting torque $-bd\theta /dt$ with constant $b>0$. The angle $\theta$ is measured from the downward vertical in radians. An ideal viscous rotary dashpot realizes this torque law; tangential linear drag $-\alpha v$ on the bob gives the same equation with $b=\alpha {\ell }^2$. The moment of inertia is $m{\ell }^2$, soHere $t$ is physical time, ${\omega }_0$ is the undamped small-amplitude angular frequency, and $\gamma$ has units of inverse time. In the normalized equation $\ddot{\theta }+c\dot{\theta }+\sin \theta =0$, dots mean differentiation with respect to $\tau$, not $t$. The parameter $c$ is dimensionless. This separates the chosen damping law from the small-angle approximation: retaining $\sin \theta$ allows finite angles even when the resisting torque is linear in angular velocity.

To motivate the model, release a pendulum without a push and record how its swing shrinks. A published example by Gintautas and Hubler uses a rotary sensor built from a computer mouse to measure the frequency and damping coefficient of a physical pendulum. ^[ref] The following is a proposed experiment with a chosen viscous rotary dashpot; the plotted observations are predictions, not a reproduction of their measurements.

Mount a compact bob on a light rigid rod with a fixed pivot, couple the pivot to an adjustable viscous rotary dashpot, and measure $\ell$ from the pivot to the bob's centre. First record decay with the dashpot disconnected, then repeat with it connected at several settings; this establishes how much decay is already due to other losses. Keep the bob, rod, camera, and release angle unchanged between these trials. The simple model requires rod and damper-rotor moments of inertia to be negligible compared with $m{\ell }^2$. Otherwise measure the total moment of inertia $I$ and the distance $d$ to the assembly's centre of mass: the physical pendulum equation becomes $I{d}^2\theta /d{t}^2+bd\theta /dt+Mgd\sin \theta =0$, where $M$ is the total moving mass. A damper that also exerts an elastic restoring torque would require another term.

Place a camera perpendicular to the swing plane, include the pivot and bob centre in every frame, and determine their coordinates relative to the resting vertical. For horizontal and downward separations $x,y$, use $\theta =atan2(x,y)$; record the frame timestamps rather than assuming a frame rate. Release from rest at about $10^{\circ }$, record through several complete swings, and repeat the release. For every same-side turning point, tabulate its time $t_n$ and angular amplitude $A_n=|\theta (t_n)|$. Measure the oscillation period from differences $t_{n+1}-t_n$; comparing several periods exposes drift. Interpolating around each observed extremum reduces frame discretization error. Check that late peaks still exceed the angular resolution; taking the natural logarithm of tiny, uncertain peaks amplifies their relative errors.

In the small-angle approximation, the damped pendulum becomes a damped harmonic oscillator. For $0<\gamma <2{\omega }_0$ and release from rest at ${\theta }_0$, its exact linear-model solution isThe decaying sinusoidal envelope has coefficient $|{\theta }_0|{\omega }_0/{\omega }_d$; because release is from rest, its successive same-side extrema occur at $t_n=n{T}_d$ and have $A_n=|{\theta }_0|e^{-\gamma n{T}_d/2}$. Thus the envelope and the sequence of actual peak heights are different descriptions of the same trace. The linear damping regimes and this solution are derived in Richard Fitzpatrick, Damped Oscillatory Motion, equations 82–93.

The experiment tests two predictions independently: nearly equal same-side peak spacing and a straight line when plotting $\ln (A_n/A_0)$ against $t_n-t_0$. Its predicted slope is $-\gamma /2$, givingUse several peaks and repeated trials rather than a single ratio, and compare the frequency identity with independently measured $\ell$ and $g$. For the illustrative choices $\ell =1.00\mathrm{m}$, $g=9.81m/{\mathrm{s}}^2$, $c=0.20$, and ${\theta }_0=10^{\circ }$, the linear model gives $\gamma \approx 0.6264{\mathrm{s}}^{-1}$, $T_d\approx 2.0162\mathrm{s}$, and $A_{n+1}/A_n\approx 0.5318$. The first two positive peaks after release are therefore approximately $5.318^{\circ }$ and $2.828^{\circ }$. The undamped small-amplitude oscillation period is about $2.0061\mathrm{s}$: substantial loss per swing need not imply a large change in period.

Repeat at a larger release angle to distinguish two limitations. The relative error in replacing $\sin \theta$ by $\theta$ is about ${\theta }^2/6$ for small angles: about $0.51\%$ at $10^{\circ }$, but about $17.3\%$ at $60^{\circ }$ when compared with $\theta$. Finite-amplitude swing spacing can therefore change as the swing shrinks even if $b$ is constant. Separately, friction at the pivot or quadratic drag can invalidate the assumed linear resisting torque; exponential peak decay should be checked, not imposed on all recorded motion. Peters discusses why real pendulum damping can require nonlinear terms. ^[ref] Repeatability, residual pivot motion, changes in damper temperature, and finite camera resolution are useful checks on this proposed measurement.

The normalized equilibrium points are $(\theta ,\dot{\theta })=(k\pi ,0)$. Even multiples of $\pi$ are attracting: their linearization has eigenvalues $(-c\pm \sqrt{c^2-4})/2$, giving a stable focus for $0<c<2$, repeated negative eigenvalues at $c=2$, and a stable node for $c>2$. These are the underdamped, critically damped, and overdamped small-displacement regimes. Odd multiples of $\pi$ have eigenvalues $(-c\pm \sqrt{c^2+4})/2$ of opposite signs and are saddle equilibria. The energy dissipation of a damped pendulum explains why trajectories in the phase portrait spiral inward near a stable focus instead of closing into the recurring orbit of an undamped simple pendulum.

If a symmetric matrix $H$ has largest eigenvalue ${\lambda }_1$ separated from ${\lambda }_2$ by $\gamma >0$, and $P=v{v}^{\top }$ is its leading rank-one orthogonal projection matrix, then every unit vector $u$, with $Q=u{u}^{\top }$, satisfies the displayed inequality. Expanding $u$ in an orthonormal eigenbasis gives ${\lambda }_1-u^{\top }Hu\ge \gamma (1-|u^{\top }v{|}^2)$, while $\Vert P-Q{\Vert }_F^2=2(1-|u^{\top }v{|}^2)$. The analogous statement for a smallest eigenvalue follows by replacing $H$ by $-H$.

A matrix is defective when an eigenvalue has geometric multiplicity smaller than its algebraic multiplicity. Over a field in which the characteristic polynomial splits, this is precisely the obstruction to an eigenbasis: at least one Jordan block has size greater than one, and generalized eigenvectors are needed to complete a basis. The shear mapping in diagonalizable matrix is the simplest example.

If two principal moments of inertia coincide, for example $I_1=I_2\ne I_3$, every direction in their two-dimensional eigenspace is a principal axis. The perpendicular axis is determined, but there is no privileged orthogonal pair within that plane. Rotating the pair in the plane leaves the diagonal inertia tensor unchanged. If all three principal moments of inertia coincide, ${\mathbf{I}}_O=I_{0}1$ and every axis through $O$ is principal. This follows from the eigenvalue multiplicity and eigenspace, rather than from a failure of diagonalization. ^[ref]

Equal principal moments of inertia do not prove that a body's outline has continuous rotational symmetry. For example, six equal point masses at $(\pm a,0,0)$, $(0,\pm a,0)$ and $(0,0,\pm a)$ have ${\mathbf{I}}_O=4m{a}^{2}1$. This discrete layout has the same axial moment of inertia in every direction, although its actual mass distribution changes under most rotations. Conversely, even when all three principal moments of inertia are distinct, principal axes are unoriented lines: reversing an eigenvector's sign gives the same axis. Near repeated eigenvalues, small measurement errors can rotate fitted principal axes substantially within the nearly degenerate eigenspace, so quoted directions need an uncertainty assessment as well as quoted principal moments of inertia.

A finite-dimensional density operator $\rho$ represents a quantum state on a Hilbert space. It is Hermitian, positive semidefinite, and has trace one:The density matrix is its matrix in a chosen orthonormal basis. In dimension $N$ it has $N^2-1$ real parameters: a Hermitian matrix has $N^2$, and fixing its trace removes one; positivity restricts the allowed values.

A quantum state ensemble of normalized pure states $|{\psi }_i⟩$ with positive probabilities $p_i$ givesZero-weight terms can be omitted. Conversely, the spectral theorem gives such an ensemble for every density operator, using its orthonormal eigenvectors and nonnegative eigenvalues, with zero eigenvalues omitted. A pure state has $\rho =|\psi ⟩⟨\psi |$; a mixed state needs at least two nonzero eigenvalues. ^[ref]

For a projective measurement with orthogonal projections ${\Pi }_a$, the Born rule isThis is the average of the component pure state probabilities. Positivity makes each outcome probability nonnegative, and ${\sum }_a{\Pi }_a=I$ makes them sum to one. The same density operator therefore predicts every measurement on the system without requiring a unique preparation history. ^[ref]

For a closed system, Schrödinger picture evolution under a unitary operator iswhen ${\rho }_H={\rho }_S(0)$ denotes the time-independent Heisenberg picture state and $U(0)=I$. In the Schrödinger picture the density operator evolves; in the Heisenberg picture the observables evolve instead, with identical measurement predictions. This unitary operator preserves the eigenvalues and hence the purity of a density operator. ^[ref]

A historical source for the probabilistic formulation of quantum mechanics is von Neumann's 1927 paper Wahrscheinlichkeitstheoretischer Aufbau der Quantenmechanik. ^[ref]

Perturb each diagonal position of a Jordan normal form matrix by a different arbitrarily small scalar, avoiding the finitely many collisions between diagonal entries. The resulting upper triangular matrix has distinct eigenvalues, hence is diagonalizable. Conjugate back to approximate any complex matrix. Continuous identities proved for diagonalizable matrices can therefore extend to all matrices.

For real $-1\le \alpha \le 1$, this two-qubit density operator has equal $00$ and $11$ populations and off-diagonal coherences $\alpha /2$. It is a pure state exactly when $\alpha =\pm 1$, and a separable quantum state exactly when $\alpha =0$. Its partial transpose has eigenvalues $1/2,1/2,\alpha /2,-\alpha /2$, directly witnessing entanglement for every nonzero admissible coherence.

A finite-dimensional linear operator is diagonalizable over a specified field when it has an eigenbasis. Equivalently, its matrix $A$ admitswhere $P$ is an invertible matrix whose columns are eigenvectors, and $D$ is a diagonal matrix of the corresponding eigenvalues. This change of basis reveals a linear map as independent one-dimensional scalings. The matrix may look coupled in the original coordinates, but each coordinate in the eigenbasis evolves on its own.

For the symmetric matrix $A=\left(\begin{array}{cc}2& 1\\ 1& 2\end{array}\right)$ computed in eigenvector, takeThen $AP=PD$, hence $P^{-1}AP=D$. The orthonormal basis version is $Q=P/\sqrt{2}$, an orthogonal matrix satisfying $Q^{T}AQ=D$. Every real symmetric matrix has an orthonormal eigenbasis by the spectral theorem; a general diagonalizable matrix need not have orthogonal eigenvectors.

The immediate payoff is repeated application: adjacent factors $P^{-1}P$ cancel, giving $A^n=P{D}^n{P}^{-1}$. In this example,The component in $E_3$ grows while the component in $E_1$ stays fixed. After division by its Euclidean norm, this particular sequence approaches $(1,1{)}^T/\sqrt{2}$. This is the mechanism behind the power method; an initial vector entirely in $E_1$ would stay there and never find the dominant eigenvalue $3$.

Two different obstructions deserve comparison. The planar rotation $R$ in eigenvector is diagonalizable over $\mathbb{C}$ but not over $\mathbb{R}$, because its eigenvalues are $i$ and $-i$. By contrast, the shear mappinghas only $\lambda =1$, of algebraic multiplicity two, and $E_1=\mathrm{span}\{(1,0{)}^T\}$, of geometric multiplicity one. There is no second independent eigenvector, even over $\mathbb{C}$. This is a defective matrix, already a two-dimensional Jordan block. Its powers areThus even though both eigenvalues are one, repeated application can produce growth. Eigenvalues alone miss the coupling recorded by a Jordan block. A repeated eigenvalue is harmless when it has enough independent eigenvectors, as $I$ demonstrates. More generally, a matrix is diagonalizable exactly when its characteristic polynomial splits over the chosen field and every eigenvalue has equal geometric multiplicity and algebraic multiplicity. The proof outline is to join bases of the distinct eigenspaces; their direct sum fills the space exactly under these conditions.

A physical motivation is the separation of normal modes. Consider two equal masses $m$ between three identical springs of stiffness $k$, with the outer ends fixed. By Hooke's law and Newton's second law, the displacement vector $q=(q_1,q_2{)}^T$ obeysThe eigenvectors $(1,1{)}^T$ and $(1,-1{)}^T$ have eigenvalues $1$ and $3$: the masses move together in the first normal mode and oppositely in the second. Writing $q=a(t)(1,1{)}^T+b(t)(1,-1{)}^T$ separates the coupled equations into two harmonic oscillators,Their angular frequencies are $\sqrt{k/m}$ and $\sqrt{3k/m}$; the eigenvalues give squared angular frequencies after multiplication by $k/m$. This is the beauty of diagonalization: finding the right directions turns interacting motions into independent patterns. For a physical treatment, see Yen-Jie Lee's "Coupled Oscillators, Normal Modes" (MIT 8.03SC, 2016), and the local article normal modes of two equal masses between three springs.^[ref]