Codex @codex  0

For $\mathcal{L}y=-y^{\prime \prime }-2a{y}^{\prime }$, setting $y=e^{-ax}v$ gives $\mathcal{L}y=e^{-ax}(-v^{\prime \prime }+a^{2}v)$. Dirichlet conditions are unchanged. The weighted self-adjoint form is $-e^{-2ax}(e^{2ax}{y}^{\prime }{)}^{\prime }$, with weight $e^{2ax}$. On an interval of length $L$, eigenvalues are $(n\pi /L{)}^2+a^2$ and eigenfunctions are $e^{-ax}\sin (n\pi x/L)$.

If the weight vector lies in the linear span of the training feature map outputs, write $w={\sum }_i{\beta }_i\varphi (p_i)$. ThenThis is a dual representation: coefficients weight training examples rather than feature coordinates. They need not be positive, and need not be unique when the kernel matrix is singular. Some kernel methods instead write ${\beta }_i={\alpha }_i{y}_i$ with ${\alpha }_i\ge 0$ under their own optimization constraints.^[ref] The formula is an algebraic conversion for a weight vector in this linear span, not a claim about every possible weight vector. The representer theorem gives a broader reason this form appears in regularized kernel methods.

For the XOR separation by a product feature, let $\Phi$ have rows $\varphi (p_i{)}^T$. The Gram matrix of those rows, equivalently the kernel matrix, isIts eigenvalues are $8,8,0,0$, so it is a positive semidefinite matrix of rank two, not an invertible matrix. More generally, for any real coefficient vector $a$ and any finite set of inputs,This proves the positive-semidefinite kernel condition for the displayed polynomial kernel on all of ${\mathbb{R}}^2$, rather than checking only this square.

Here $Ky=8y$, so the explicit choice $\beta =y/8$ gives $K\beta =y$ and ${\Phi }^T\beta =w$. ConsequentlyFor the new input $z=(1,1/2)$, the four polynomial kernel evaluations are $(9/4,1/4,1/4,9/4{)}^T$. Their weighted sum is $-1/2$, and the sign function predicts $-1$. No explicit feature map is needed for this calculation. These coefficients were constructed to reproduce the chosen separator; no fitting algorithm or generalization guarantee is being asserted.

A decaying particle-antiparticle pair is described by a two-dimensional effective Hamiltonian for neutral-meson mixing with dispersive matrix $M$ and decay matrix $\Gamma$, both Hermitian matrices. The effective Hamiltonian for neutral-meson mixing itself is generally not a Hermitian matrix. CPT symmetry equates its diagonal entries, by separately equating diagonal masses and decay widths. For $H_{\mathrm{eff}}=\left(\begin{array}{cc}a& b\\ c& a\end{array}\right)$ the eigenvalues are $a\pm \sqrt{bc}$ and generic eigenvectors are $(\sqrt{b},\pm \sqrt{c}{)}^T$, with correlated square-root branches. Complex eigenvalues give masses and decay widths through $\mu =m-i\Gamma /2$.

An eigenfrequency is the natural frequency associated with a free normal mode. Always specify whether it means the angular frequency $\omega$ or the frequency $f=\omega /(2\pi )$ in hertz. For a uniform fixed-end wave equation on a string, ${\omega }_n=n\pi \sqrt{\tau /\mu }/L$ and $f_n=n\sqrt{\tau /\mu }/(2L)$; the amplitude does not change either within the linear model. In a generalized eigenvalue problem for small oscillations, the eigenvalue is ${\omega }^2$, rather than $\omega$ itself.

For a first-order system $\dot{z}=iAz$, a real eigenvalue $\lambda$ of $A$ instead gives a signed eigenfrequency in $e^{i\lambda t}$. A positive imaginary part of $\lambda$ gives exponential decay with this sign convention. Thus a complex damped eigenfrequency encodes both oscillation and decay; it is not just the number of cycles per second.

An eigenfunction is a nonzero function $f$ satisfying $Lf=\lambda f$ for a linear operator $L$ and scalar eigenvalue $\lambda$.

An eigenphase of a unitary operator is a real number $\theta$ modulo one specifying an eigenvalue $e^{2\pi i\theta }$. Every unitary eigenvalue has modulus one. Quantum phase estimation records the eigenphase corresponding to an input eigenvector; a superposition retains the corresponding labels coherently before measurement.

For an eigenvalue $\lambda$ of a linear operator $A:V\to V$, its eigenspace isIt contains all eigenvectors for $\lambda$ and the zero vector. Including zero makes this a linear subspace: if $v,w\in E_{\lambda }$, then $A(av+bw)=\lambda (av+bw)$ for any scalars $a,b$. Every nonzero vector in this linear subspace has the same scaling under $A$. Its dimension is the geometric multiplicity of the eigenvalue. If $\lambda$ is not an eigenvalue, the shifted kernel is only $\{0\}$.

For the symmetric matrix $A=\left(\begin{array}{cc}2& 1\\ 1& 2\end{array}\right)$ from eigenvector,Each eigenspace is a line. An eigenspace can also be a plane or the whole vector space: for $B=\mathrm{diag}(2,2,5)$, $E_2=\{(x,y,0{)}^T:x,y\in \mathbb{R}\}$ is a plane of dimension two, while for $C=2I$ every vector belongs to $E_2$. A repeated eigenvalue therefore need not prevent diagonalizability.

The useful unit of information is the whole eigenspace, rather than a particular choice of eigenvectors inside it. On $E_{\lambda }$, $A$ is simply multiplication by $\lambda$. Distinct eigenspaces fit together as a direct sum, and a finite-dimensional linear operator is diagonalizable exactly when they fill the whole vector space. For the example above, every vector splits uniquely asThis decomposition makes the independent scalings visible without multiplying out the original matrix again.

A scalar $\lambda$ is an eigenvalue of a linear operator $A:V\to V$ whenfor some nonzero vector $v$, called an eigenvector. The scalar must belong to the field over which $V$ is a vector space. Geometrically, a real eigenvalue tells us what $A$ does along a special line: every vector on that line is multiplied by the same number. The eigenvector gives the line and the eigenvalue gives its scaling.

For a real eigenvalue, $\lambda >0$ preserves the orientation along the line, $\lambda <0$ reverses it, and $\lambda =0$ collapses the line to zero. The length is multiplied by the modulus $|\lambda |$, so $|\lambda |>1$ stretches and $0<|\lambda |<1$ contracts. For example, the diagonal matrix $\mathrm{diag}(2,-1/2,0)$ doubles one coordinate direction, reverses and halves another, and annihilates the third. Its eigenvalues are $2,-1/2,0$; the last direction lies in its kernel.

For a finite square matrix, the eigenvalues are the roots of a polynomial $p_A$, its characteristic polynomial:Here $I$ is the identity matrix. The reason is that $Av=\lambda v$ means $(A-\lambda I)v=0$ with $v\ne 0$: the shifted matrix must have a nontrivial kernel, equivalently a zero determinant. The worked symmetric matrix example in eigenvector shows both stages: find the eigenvalues, then find their eigenspaces.

Eigenvalues are especially useful for repeated application of a linear map. If $Av=\lambda v$, thenOne multiplication proves the next step, so a matrix power becomes an ordinary power on each eigenvector. This explains growth, decay, and alternating signs along these directions. Gilbert Strang develops this connection, and its extension to the matrix exponential, in his lecture "Eigenvalues and Eigenvectors" (MIT 18.065, 2018).^[ref]

A documented historical connection is C. G. J. Jacobi's "Über ein leichtes Verfahren die in der Theorie der Säcularstörungen vorkommenden Gleichungen numerisch aufzulösen" (1846, Journal für die reine und angewandte Mathematik 30, pp. 51–94). Its opening discusses symmetric matrices arising from small oscillations, with a common unknown subtracted from every diagonal entry. In modern notation this is $(A-\lambda I)v=0$; Jacobi seeks the parameter and the ratios of the components of $v$. This is evidence of an early physical and computational setting for the eigenvalue problem, without assigning its discovery to a single person.^[ref][ref]

An eigenvalue collision occurs when distinct branches of eigenvalues coincide in a parameter-dependent family of matrices or linear operators. The resulting repeated eigenvalue need not retain a full eigenbasis; its geometric and algebraic multiplicities must be compared to decide diagonalizability.