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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.

If a real linear operator $A$ on an $n$-dimensional vector space has $n$ distinct real eigenvalues, any real square root $B$ commutes with $A$ and preserves each one-dimensional eigenspace. There $B$ is multiplication by a real $b_i$, hence ${\lambda }_i=b_i^2\ge 0$. Conversely, if all the eigenvalues are nonnegative, choosing $b_i=\sqrt{{\lambda }_i}$ on an eigenbasis supplies a real square root. Without simple spectrum the necessity fails: a real quarter-turn squares to $-I$.

A real symmetric matrix is a Hermitian matrix, so $u^{\ast }Au$ is real for every complex vector $u$. If $u$ is an eigenvector, its eigenvalue is $(u^{\ast }Au)/(u^{\ast }u)$ and is real. Then $u^{\ast }Av=(Au{)}^{\ast }v$ gives $(\mu -\lambda )u^{\ast }v=0$. Reality is used in changing conjugate transpose to transpose; complex symmetry alone does not imply real eigenvalues.

For normalized linearly independent pure states, their Gram matrix $G_{ij}=⟨{\psi }_i|{\psi }_j⟩$ is positive definite. The displayed reciprocal vectors satisfy $⟨{\chi }_i|{\psi }_j⟩={\delta }_{ij}$. The operator $R={\sum }_i|{\chi }_i⟩⟨{\chi }_i|$ has nonzero eigenvalues equal to those of $G^{-1}$: use the polar factorization of the matrix whose columns are ${\psi }_i$. Choosing $0<ϵ\le {\lambda }_{\min }(G)$ gives the POVM effects $E_i$ and $E_{?}=I-ϵR$. Each conclusive outcome has probability $ϵ$ on its own state and zero on every other input. The inconclusive effect preserves positivity, so the construction is a physical measurement in quantum mechanics.

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.

A renormalization-group transformation integrates out selected short-distance degrees of freedom and restores chosen cutoff and field-normalization conventions. It defines a map $R_b$ on couplings at a length factor $b$. Its discrete Jacobian matrix multipliers differ from eigenvalues of the generator of a continuous renormalization-group flow; the relation for a scaling direction is $\lambda =b^y$.

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 $L>0$, constant $\beta >0$ and $\mathrm{Re}p>0$, on $(0,L)$ put $b=\beta /2$, $\kappa =\sqrt{p+b^2}$ with the principal square root, andThese satisfy $(p-{\partial }_x^2-\beta {\partial }_x)u_p=(p-{\partial }_x^2-\beta {\partial }_x)v_p=0$ and the left and right homogeneous Neumann boundary conditions, respectively. The weighted Wronskian givesThe first derivative jumps by $-e^{-\beta y}$, so the resolvent integrates against $e^{\beta y}dy$. Its time Laplace transform interpretation is ${\mathcal{R}}_p={\int }_0^{\infty }{e}^{-pt}{K}_{\beta }(t)dt$. The factor $p$ records the zero eigenvalue and constant stationary eigenfunction.

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.

If two nearby inner planets couple much more strongly to each other than to a distant third planet, their slow secular eigenmode has a common complex inclination. With $L_{12}=L_1+L_2$, the effective inner-to-outer coupling isAngular momentum balance makes the reverse coupling $c{L}_{12}/L_3$. The reduced matrix has eigenvalues $0$ and $-c(1+L_{12}/L_3)$. If $L_3\gg L_{12}$, the nonzero frequency is approximately $-c$, while the fast inner differential mode has frequency $-(B_{12}+B_{21})$.

For the rotational hexagon amplitude equations, put $d={\nu }_2-{\nu }_1$ and $N={\nu }_1+2{\nu }_2$. A roll with $r^2=\mu /{\nu }_1$ has transverse eigenvalues $-d{r}^2\pm \sqrt{{\delta }^2{r}^4+r^2}$, each twice, radial eigenvalue $-2\mu$, and one neutral translation. A hexagon with $\mu =N{h}^2-h$ has real-amplitude eigenvalues $h-2N{h}^2$ and $-2h+2d{h}^2\pm 2i\sqrt{3}\delta h^2$; its imaginary-amplitude eigenvalues are $-3h,0,0$. These follow by diagonalizing the cyclic Jacobian matrix. Stability is understood modulo translations. If ${\nu }_2>{\nu }_1>0$ and $\delta \ne 0$, the upper positive hexagon branch reaches a Hopf bifurcation at $h=1/d$, $\mu =(2{\nu }_1+{\nu }_2)/d^2$.

Under rotation by a rotation matrix, a symmetric second-rank tensor transforms by conjugation, so its traces of powers are invariant. For a three-dimensional traceless second-rank tensor, the Cayley-Hamilton theorem implies $\mathrm{Tr}{Q}^4=(\mathrm{Tr}{Q}^2{)}^2/2$. In two dimensions its eigenvalues are $s,-s$, so $\mathrm{Tr}{Q}^3=0$. These identities organize the Landau-de Gennes free energy.

A conservative equilibrium with one real pair of linear eigenvalues $\pm \lambda$ and one imaginary pair $\pm i\omega$ has saddle and centre directions. Generic perturbations have an exponentially growing component, even though finely chosen initial conditions can remain in the centre-stable subspace. The local L2 Lagrange point dynamics is an example.