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A perturbation of a renormalization-group fixed point is relevant if it grows as the coarse-graining length increases. For a continuous flow $d\delta g/d\ell =M\delta g$, an eigenvector of the linearized matrix with eigenvalue $\sigma$ grows as $e^{\sigma \ell }$; a positive real part is relevant, a negative real part is irrelevant and a zero real part requires nonlinear analysis. For a discrete transformation with length factor $b>1$, its Jacobian matrix multiplier $\lambda$ instead gives $\delta g_n={\lambda }^n\delta g_0$: $|\lambda |>1$ is relevant, $|\lambda |<1$ is irrelevant and $|\lambda |=1$ requires nonlinear analysis. For a positive multiplier the scaling exponent is $y=\log \lambda /\log b$. Thus a positive discrete eigenvalue smaller than one is an irrelevant operator direction, while the sign criterion belongs to the continuous generator.

Representation theory describes abstract symmetries by linear maps on vector spaces. A group representation assigns an invertible linear map to each group element, respecting multiplication. The payoff is that a symmetry problem becomes a problem about matrices, invariant subspaces, and eigenvalues. Decomposing the action can reveal combinations of variables that a symmetric system cannot mix.

Take the cyclic group $C_3=\{e,g,g^2\}$, with $g^3=e$. On three real coordinates, let $g$ send $(a,b,c)$ to $(c,a,b)$:This permutation representation does more than relabel a list: it acts linearly on every possible assignment of values to the three positions. The vector $(1,1,1)$ stays fixed. The vector subspace $W=\{(a,b,c):a+b+c=0\}$ is also preserved, since a permutation does not change the sum.

Every vector separates uniquely into these two types:For example, $(4,1,1)=2(1,1,1)+(2,-1,-1)$. The first term belongs to a trivial representation; the second is a contrast with zero coordinate sum. This is a direct sum decomposition ${\mathbb{R}}^3=\mathbb{R}(1,1,1)\oplus W$. The symmetry preserves the mean and rotates the contrasts.

The rotation is literal after choosing the orthonormal basis $u=(2,-1,-1)/\sqrt{6}$, $v=(0,1,-1)/\sqrt{2}$ of $W$. In this basis,It is a rotation through $120$ degrees, and three applications return to the starting point. There is no nonzero invariant real line, because this rotation has no real eigenvector. Thus the two-dimensional action is an irreducible representation over the real numbers. Over the complex numbers it instead splits into two eigenlines with eigenvalues $e^{2\pi i/3}$ and $e^{-2\pi i/3}$. The coefficient field is part of an irreducibility claim.

Why would this decomposition matter for a physical calculation? Consider three equal masses with scalar displacements $x_i$ and the symmetric quadratic coupling energyThe equations of motion are $m\ddot{x}=-kLx$, whereis a Graph Laplacian and $J$ is the all-ones matrix. The constant vector has eigenvalue zero: translating all coordinates equally costs no coupling energy. On $W$, $J$ vanishes and $L=3I$, so both independent contrast directions have angular frequency $\sqrt{3k/m}$. A three-coordinate coupled problem reduces to a free mean and two identical harmonic oscillators.

For initial displacement $x(0)=(4,1,1)$ and initial velocity zero, the exact solution is $x(t)=2(1,1,1)+\cos (\sqrt{3k/m}t)(2,-1,-1)$. The fixed mean and oscillating contrast can be read directly from the decomposition. More generally, a linear operator commuting with the group action carries its fixed-vector subspace into itself; with an invariant inner product, it preserves the corresponding orthogonal complement too.

For a finite group, averaging over the action constructs an invariant inner product over the real numbers or complex numbers. The resulting invariant complements explain Maschke's theorem: a finite-dimensional group representation decomposes into irreducible representations. Characters compress the action into traces and help identify those pieces. The assumptions matter; in modular representation theory the coefficient field can have characteristic dividing the group order, and such decompositions can fail. Chapter 4 of Pavel Etingof and coauthors' notes develops these mechanisms. ^[ref].

For a real symmetric positive definite matrix $A$, the error in each eigenvector direction is multiplied by $1-\tau \lambda$. Convergence from every initial vector is therefore equivalent to $0<\tau <2/{\lambda }_{\max }$. At either endpoint there is a nondecaying error direction. The optimal constant parameter is $2/({\lambda }_{\min }+{\lambda }_{\max })$, giving contraction factor $({\lambda }_{\max }-{\lambda }_{\min })/({\lambda }_{\max }+{\lambda }_{\min })$.

The right Cauchy-Green tensor measures stretch on the reference configuration. Its eigenvalues are the same squared principal stretches as those of the left Cauchy-Green deformation tensor, but its eigenvectors are reference directions.

A right singular vector $v_j$ is a unit vector in the input space satisfying $A^{\ast }A{v}_j={\sigma }_j^2{v}_j$. For positive singular values, $A{v}_j={\sigma }_j{u}_j$ determines the matching left singular vector. In the ellipse example, $A{v}_1=(5,0{)}^T$ is not parallel to $v_1$: a right singular vector need not be an eigenvector of $A$. Indeed, the characteristic polynomial of $A$ is ${\lambda }^2-(18/5)\lambda +5$, with eigenvalues $(9\pm i\sqrt{44})/5$ and no real eigenvectors. Nevertheless, its real right singular vectors give a complete orthonormal basis. Eigenvectors describe preserved directions; right singular vectors describe orthogonal input channels of stretching.^[ref]

An arbitrary input mixes these channels. For $x=c_1{v}_1+c_2{v}_2$, orthogonality gives $\Vert Ax{\Vert }_2^2=25|c_1{|}^2+|c_2{|}^2$, while $\Vert x{\Vert }_2^2=|c_1{|}^2+|c_2{|}^2$. Equal contributions, $c_1=c_2=1/\sqrt{2}$, produce output Euclidean norm $\sqrt{13}$ from a unit vector. Concentrating the entire input along $v_1$ produces the maximum $5$. This is why the right singular vectors give a direct reading of how any input will behave. For a compact operator, the same definition describes the positive terms of its singular value system.

Consider the diagonal matrix $D=2I_2$ and a Jordan blockBoth have characteristic polynomial $(t-2{)}^2$, trace $4$, determinant $4$, and eigenvalue $2$ with algebraic multiplicity two. But $\ker (D-2I)$ is the whole plane, whereas $(J-2I)(a,b{)}^T=(b,0{)}^T$ vanishes only when $b=0$. Their eigenspaces have different dimensions, so eigenvectors under similarity rules out matrix similarity. There is an even shorter proof: $S^{-1}(2I)S=2I$ for every invertible matrix $S$.

The missing information is how the linear operator acts beyond its eigenvectors. Here $D-2I=0$, while $J-2I\ne 0$ but $(J-2I{)}^2=0$. Their minimal polynomials are respectively $t-2$ and $(t-2{)}^2$. Over a field where the characteristic polynomial splits, the full sizes of the Jordan blocks, encoded by the Jordan normal form, determine the matrix similarity class.

Let $f$ have continuous second partial derivatives near an interior critical point $x_{\ast }$ of an open subset of ${\mathbb{R}}^n$, with $\nabla f(x_{\ast })=0$. The Hessian matrix gives the following sufficient conditions:

A positive semidefinite or negative semidefinite Hessian matrix with a zero eigenvalue does not settle the classification. The higher-order test for separated extrema gives two quartic examples with exactly the same zero Hessian. These conditions apply to an interior critical point; a boundary optimum or a constrained optimization problem requires attention to the allowed directions. ^[ref]

The proof outline is the Taylor expansionFor a positive-definite matrix, the smallest eigenvalue supplies a positive bound proportional to $\Vert h{\Vert }^2$, which dominates the remainder for small enough $h$. Negate the function for the local maximum case. For opposite-sign eigenvalues, choose the two corresponding eigenvectors: the leading terms have opposite signs along their lines.

There are also necessary conditions. At an interior local minimum, the Hessian matrix is positive semidefinite; at an interior local maximum, it is negative semidefinite. Restriction to each straight line reduces this to the one-variable second derivative condition. Since the Laplacian is the trace of the Hessian, it is nonnegative at such a local minimum and nonpositive at such a local maximum. The sign of the Laplacian alone is insufficient: $x^2-2y^2$ has negative Laplacian but a saddle point of a scalar function at the origin.

Under power iteration convergence for a diagonalizable matrix, take $v_1=u_1$ to be a unit vector. The displayed formula describes both the dominant eigenvector direction and its rotating complex unit factor. With real data and a positive dominant eigenvalue, $x_k$ converges to $u_1$ or $-u_1$ according to the sign of $c_1$. With a negative dominant eigenvalue, consecutive vectors approach opposite signs even though their power iteration direction error tends to zero.

For complex data write ${\lambda }_1=|{\lambda }_1|e^{i\varphi }$, where $\varphi$ is its complex argument. The leading factor turns by $e^{i\varphi }$ per step, so the vectors generally do not converge. For example, $A=\mathrm{diag}(2i,1)$ and $x_0=(1,1{)}^T/\sqrt{2}$ give $A^k{x}_0=((2i{)}^k,1{)}^T/\sqrt{2}$: the first component cycles through four phases while the ratio of component moduli tends to zero. Its power iteration direction error is $2^{-k}/\sqrt{1+2^{-2k}}$. Multiplying $x_k$ by $e^{-i(\arg c_1+k\varphi )}$ gives a sequence converging to $u_1$.

A stopping rule based only on $\Vert x_{k+1}-x_k{\Vert }_2$ therefore rejects successful sign-alternating iterations. A comparison invariant under complex unit factors isEven this comparison can be small for a slowly changing wrong mixture; the eigenpair residual supplies a separate diagnostic.

Suppose $v_i=\pm 1/\sqrt{n}$ encodes the true partition and $\hat{v}$ is a unit estimated eigenvector. Every coordinate of the wrong sign contributes at least $1/n$ to ${\min }_{\sigma =\pm 1}\Vert \hat{v}-\sigma v{\Vert }_2^2=2(1-|{\hat{v}}^{\top }v|)$. Since this is at most $2(1-|{\hat{v}}^{\top }v{|}^2)$, which is the squared Frobenius distance between the rank-one orthogonal projection matrices, the bound follows. Coordinates estimated as zero may be assigned consistently to either group.

A similarity transformation replaces a square matrix by a similar matrix. With the change-of-basis matrix convention $x=Sy$, it sends $A$ to $B=S^{-1}AS$: reconstruct the input in old coordinates, apply $A$, then extract the output's new coordinates. These three operations explain the order of the factors.

The equivalent formula $A\mapsto SA{S}^{-1}$ uses $S$ as the conversion from old to new coordinates, reversing the role assigned to $S$ above. For the opening example's numerical $S$, this opposite conversion gives $SA{S}^{-1}=\left(\begin{array}{cc}2& 3\\ 0& 5\end{array}\right)$. If the old basis is $\mathcal{E}$, the new basis vectors in that convention are the columns of $S^{-1}$, namely $e_1$ and $-e_1+e_2$. Both formulas are similarity transformations; the direction of the coordinate conversion decides which one describes a specified basis.

An especially useful form is the intertwining operator equationIt says that applying $B$ before reconstruction by $S$ gives the same physical output as reconstructing first and applying $A$. Given an invertible matrix $S$, multiplying this equation by $S^{-1}$ proves matrix similarity. Without invertibility, the equation alone does not prove matrix similarity.

Suppose $D$ has $n$ distinct eigenvalues over a field and $C$ commutes with $D$. Each eigenspace of $D$ is one-dimensional. Since commuting maps preserve eigenspaces, $C$ acts as a scalar on each of them. The eigenvectors of $D$ give a common eigenbasis and hence simultaneous diagonalization. For real matrices, the result gives a real eigenbasis only if the eigenvalues of $D$ are real; otherwise it applies over $\mathbb{C}$.

An ideal-MHD simple wave traces a one-parameter state-space curve whose tangent is a right eigenvector of the characteristic matrix. Its parameter obeys $q_t+v(q)q_x=0$ and its characteristic speed obeys the Inviscid Burgers equation. Nonlinear Alfvén waves are a linearly degenerate example.

A Lie algebra representation is simultaneously upper triangular precisely when it preserves a complete flag, with $\dim V_j=j$. A basis adapted to the invariant flag makes every representing matrix upper triangular. For a complex finite-dimensional representation of a solvable Lie algebra, the Lie theorem provides a common eigenvector; iteration on the quotient representations constructs the flag.

For a nonzero real three-dimensional skew-symmetric matrix, choose real orthogonal vectors $u,v$ of equal length with $Au=\theta v$, $Av=-\theta u$ and $\theta >0$, and a real unit vector $e_3$ in its kernel. The matrix exponential acts by $e^{A}u=u\cos \theta +v\sin \theta$, $e^{A}v=v\cos \theta -u\sin \theta$ and $e^A{e}_3=e_3$. Its matrix in this orthonormal basis is a planar rotation matrix with an unchanged axis. A purely imaginary nonzero eigenvalue and its complex eigenvector give $u,v$ by taking real and imaginary parts; antisymmetry proves their equal norms and orthogonality.

Spectral graph theory studies graphs through eigenvalues and eigenvectors of associated matrices, including the adjacency matrix and Laplacian matrix. A graph eigenvalue usually refers to the adjacency-matrix spectrum unless another matrix is specified.

Let $X\ne \{0\}$ be a Banach space over the complex numbers, and let $T:X\to X$ be a bounded linear operator defined on all of $X$. Its spectrum consists of those $\lambda \in \mathbb{C}$ for which $T-\lambda I$ has no everywhere-defined bounded inverse. Its complement is the resolvent set of an operator $\rho (T)$. By the bounded inverse theorem, a bijective $T-\lambda I$ automatically has a bounded inverse. The spectrum is a nonempty compact subset of the complex plane, withThe bound uses the operator norm, and nonemptiness uses the complex field.^[ref] For a real linear operator, complex spectral values are understood by extending it to a complex vector space; restricting $\lambda$ to the real field can give an empty set.

In a finite-dimensional vector space, failure of inversion forces a nonzero kernel, hence an eigenvector. An infinite-dimensional vector space allows an injective shifted linear operator that cannot solve every equation $(T-\lambda I)u=g$. The spectrum therefore records both exact eigenvectors and failures of solvability or stable inversion. Searching only for eigenvalues can miss every spectral value of a perfectly simple bounded linear operator.

Consider the multiplication operator $M$ on the complex L2 space $L^2([0,1],dx)$, with Lebesgue measure:Its operator norm is one. The inequality $\Vert Mf{\Vert }_2\le \Vert f{\Vert }_2$ gives the upper bound; unit vectors concentrated near $x=1$ approach it. The result isThis is the interval example of the general connection between multiplication operators and the essential range of their multiplier.^[ref] Indeed, an eigenfunction would satisfy $(x-\lambda )f(x)=0$ almost everywhere. It vanishes except possibly on the singleton $\{\lambda \}$, a null set; such a function is the zero vector in L2 space. Concentrating at one point cannot produce a nonzero L2 function.

For $\lambda \notin [0,1]$, the inverse is completely explicit:The distance to a set is positive, so the inverse multiplier is bounded. Its essential supremum gives the displayed operator norm; arbitrarily small intervals near a closest point approach the bound. Our resolvent of an operator uses $(\lambda I-M{)}^{-1}$, the negative of this inverse, with the same operator norm. For $\lambda =-1/4$, $5/4$, and $1/2+i/10$, the inverse norms are respectively $4$, $4$, and $10$. The last parameter lies directly above the middle of the spectrum in the complex plane.

This calculation gives the spectrum a practical meaning: it marks where solving a shifted equation loses uniform control. At $\lambda =1/2+i\eta$, the largest amplification of an input error is $1/|\eta |$. Changing $\eta$ from $0.1$ to $0.01$ increases that factor from ten to one hundred. The equation remains uniquely solvable for each nonzero $\eta$, but its sensitivity grows as the parameter approaches the spectrum.

To see directly why every $\lambda \in [0,1]$ belongs to the spectrum, choose $0<\epsilon <1$ and define a clipped interval and a normalized indicator function:Its L2 norm is exactly one, including at both endpoints. On its support, $|x-\lambda |\le \epsilon$, givingThus any choice ${\epsilon }_n\to 0$ produces approximate eigenvectors. This specializes the localized-indicator construction in Nonnenmacher's spectral notes.^[ref] A hypothetical bounded inverse $B$ would imply $1=\Vert u_{\lambda ,\epsilon }{\Vert }_2\le \Vert B\Vert \epsilon$, impossible for arbitrarily small $\epsilon$.

For an interior point with $\epsilon \le \min (\lambda ,1-\lambda )$, $\ell =2\epsilon$ and the residual is exactly $\epsilon /\sqrt{3}$. At $\lambda =0$ or $1$, $\ell =\epsilon$ and the same residual formula holds, but the amplitude is ${\epsilon }^{-1/2}$ rather than $(2\epsilon {)}^{-1/2}$. At $\lambda =1/2$, taking $\epsilon =0.2,0.1,0.05$ gives amplitudes approximately $1.5811,2.2361,3.1623$ and residuals $0.11547,0.057735,0.028868$. The narrowing support and increasing amplitude keep total squared mass equal to one.

For every $\lambda \in [0,1]$, the continuous spectrum description is also concrete. Any L2 function $g$ can be approximated in L2 norm by its truncations to $|x-\lambda |\ge \delta$. Each truncation lies in the range of $M-\lambda I$, since division there is bounded. The range is therefore dense. However, the constant function $g=1$ has no L2 function preimage: $1/(x-\lambda )$ is not square-integrable, including at the endpoints.

These approximate eigenvectors do not approach an eigenvector in L2 norm. Any possible limit would have L2 norm one and satisfy $(M-\lambda I)u=0$ by continuity, contradicting the empty point spectrum. This is the striking feature: the linear operator acts increasingly like scalar multiplication on suitable unit vectors, although no direction realizes that scaling exactly.

A complementary example is the diagonal operator on sequence space $(Dx{)}_n=x_n/n$ on complex l2 sequence space, with $n=1,2,\dots$. The coordinate unit vectors satisfy $D{e}_n=(1/n)e_n$, soZero is not an eigenvalue, since $Dx=0$ forces every coordinate to vanish. Nevertheless $\Vert D{e}_n\Vert =1/n\to 0$: these are approximate eigenvectors at zero. More concretely, $g=(1/n{)}_{n\ge 1}$ belongs to l2 sequence space, but $Dx=g$ would require $x_n=1$ for all $n$, which does not. Outside the displayed closed set, division by $1/n-\lambda$ is uniformly bounded, proving that no further spectral values occur. Thus the spectrum contains the missing limit of genuine eigenvalues, even when that limit has no eigenvector.