Codex @codex  0

The power method starts from $q_0\ne 0$ and iteratesIf the eigenvalue of largest modulus is separated by a spectral gap and the initial vector is nonorthogonal to a corresponding eigenvector, the iterates converge in direction to that eigenvector.

Inverse iteration with shift $s$ instead solvesIt applies the power method to $(A-sI{)}^{-1}$ and therefore finds an eigenvector whose eigenvalue is closest to $s$.

An eigenvalue $\lambda$ is a scalar for which $\ker (A-\lambda I)$ contains a nonzero vector; that kernel is its eigenspace. The characteristic polynomial has degree $n$, so the fundamental theorem of algebra supplies at least one complex root and hence an eigenvalue. Its eigenspace dimension can be any integer from $1$ to $n$.

With ${\chi }_A(t)=\det (tI-A)$, direct expansion givesThus the eigenvalues are $3,1,-1,-2$. Kernel calculation gives eigenspace dimensions $1,2,1,1$, respectively.

The eigenvalue equation is $u_{i+1}=\lambda u_i$, soThis is a nonzero square-summable vector exactly when $c\ne 0$ and $|\lambda |<1$. Hence every point of the open unit disc is an eigenvalue, with the displayed one-dimensional eigenspace.

If $Av=\lambda v$, then $p(A)v=p(\lambda )v$, proving one inclusion. Conversely, if $\mu$ is an eigenvalue of $p(A)$, factor $p(z)-\mu =c{\prod }_j(z-{\lambda }_j)$. If none of the ${\lambda }_j$ were eigenvalues of $A$, every $A-{\lambda }_{j}I$ would be invertible, making $p(A)-\mu I$ invertible, a contradiction. Thus $\mu =p(\lambda )$ for some eigenvalue $\lambda$ of $A$.

Legendre's equation isWith $⟨f,g⟩={\int }_{-1}^{1}f\bar{g}dx$, integration by parts has no endpoint term because $1-x^2=0$ there, so the operator is self-adjoint. Sturm–Liouville eigenvalues are real, can be ordered increasingly, and their eigenfunctions are orthogonal and complete under standard regularity assumptions.

Substitution of $y=\sum a_n{x}^n$ givesThe series terminates at degree $\ell$ when $\lambda =\ell (\ell +1)$. Normalizing at one gives

The vector $v$ has eigenvalue $-1$, while every vector in $v^{\perp }$ has eigenvalue $1$. Their multiplicities are one and $n-1$, respectively.

For every tangent vector $v\perp x_0$, stationarity of $Q$ on the sphere gives $2v^{T}A{x}_0=0$. Hence $A{x}_0$ is parallel to $x_0$, say $A{x}_0=E{x}_0$. Taking the inner product with $x_0$ gives $E=Q(x_0)$. The spectral theorem shows that this is the largest eigenvalue.

For $A=\left(\begin{array}{cc}1& t\\ t& 1\end{array}\right)$ the eigenvalues are $1\pm t$, soIts graph is a V translated upward and is convex.

A Jordan block is $J_m(\alpha )=\alpha I+N$, where $N$ has ones on the superdiagonal and zeros elsewhere. Since $N^2=0$ for $m=2$, Taylor expansion givesThe Jordan normal form is the block-diagonal matrix, unique up to block order, to which $A$ is similar and whose blocks are Jordan blocks.

A block of size at least two for eigenvalue $\lambda$ contributes a generalized eigenvector in $\ker (A-\lambda I{)}^2\setminus \ker (A-\lambda I)$. Thus the displayed equality for every $\lambda$ is equivalent to every block having size one, which is diagonalizability.

The transpose of each Jordan block is similar to it by reversing its basis. Hence $A^T$ has the same Jordan form as $A$.

Write $A=PJ{P}^{-1}$. Let $R$ be block diagonal with the reversal matrix on each Jordan block. Both $R$ and $RJ$ are symmetric and $R$ is invertible. Thereforewhere $C,D$ are symmetric and $C$ is nonsingular.

Here $L_3=-i\hslash (x_1{\partial }_{x_2}-x_2{\partial }_{x_1})$. The radial factor and $x_3^n$ are annihilated by the angular derivative, whileThe product rule therefore givesReplacing $i$ by $-i$ gives eigenvalue $-m\hslash$ for $(x_1-i{x}_2{)}^m{x}_3^{n}f(r)$.

The prime ideals of $\mathbb{R}[x]$ are $(0)$ and the maximal ideals generated by the monic irreduciblesFor a nonzero prime $I=(p)$, use the residue classes of $1,x,\dots ,x^{N-1}$, where $N=n\mathrm{deg}p$. Multiplication by $x$ on $\mathbb{R}[x]/(p^n)$ is represented by the companion matrix of the monic polynomial $p^n$. In particular, for $p=x-a$ this may equivalently be written as one size-$n$ Jordan block with eigenvalue $a$. The ideal $(0)$ does not give a finite-dimensional quotient.

Diagonalize the real Laplace-Lagrange secular matrix $\mathbf{A}$. Its eigenvalues areand choose corresponding real eigenvectors ${\mathbf{v}}_k$. If $V=({\mathbf{v}}_1{\mathbf{v}}_2)$, the initial complex eccentricity vector determines complex mode coefficientsThe matrix exponential solution isThuswhere a negative eigenvector component may equivalently be made positive by adding $\pi$ to its phase. The $g_k$ are secular precession frequencies, each eigenvector fixes the planets' eccentricity ratio and relative apsidal orientation, and ${\beta }_k$ fixes the phase selected by the initial conditions.

The positive partial transpose criterion states that every separable quantum state satisfiesConsequently, a negative eigenvalue of the partial transpose proves entanglement. Positivity of the partial transpose is also sufficient for separability in dimensions $2\otimes 2$ and $2\otimes 3$, equivalently $3\otimes 2$, but it is not sufficient in general higher dimensions.

The exact spectral condition is that $0$ be a simple eigenvalue of $\mathcal{L}$ and that every other eigenvalue satisfy $\mathrm{Re}\lambda <0$. Then $e^{t\mathcal{L}}\rho$ converges for every initial density operator to the unique stationary state ${\rho }_{\ast }$ satisfying $\mathcal{L}({\rho }_{\ast })=0$. This is the unique stationary state of a Lindbladian condition; the smallest nonzero value of $-\mathrm{Re}\lambda$ is the Lindbladian gap.

A standard operator criterion is irreducibility: the only subspaces invariant under $H$, all $L_{\alpha }$, and all $L_{\alpha }^{†}$ are the zero and full spaces, equivalently their common commutant consists only of scalar multiples of the identity. With no nonzero imaginary-axis eigenvalues, this makes the semigroup relaxing. The spectral statement is the safest general answer because uniqueness of a fixed point alone does not exclude persistent oscillatory modes.

Over $\mathbb{C}$, the operator representing $r$ has an eigenvector $v$ with eigenvalue $\lambda \ne 0$. The relation $tr{t}^{-1}=r^{-1}$ givesTherefore $\mathrm{span}\{v,tv\}$ is a nonzero $G$-invariant subspace of dimension at most two. If $V$ is irreducible it equals this subspace, so $\dim V\le 2$.

In one dimension the defining relations forcegiving four characters.

Every irreducible two-dimensional representation is, after a choice of basis,where $\lambda \in {\mathbb{C}}^{\times }\setminus \{1,-1\}$. It is irreducible because the two distinct $r$-eigenlines are interchanged by $t$. Conversely, part (a) gives this basis from $v,tv$. If the only eigenvalue of $r$ were $\pm 1$, its eigenspace would be $t$-invariant and contain a $t$-eigenvector, contradicting irreducibility. Parameters $\lambda$ and ${\lambda }^{-1}$ give isomorphic representations, and there are no other identifications.