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For $\varphi (x)=x$ and centered data, the objective is the Rayleigh quotientThis is the first-direction optimization in principal component analysis. Therefore $\hat{u}$ is any unit eigenvector of the sample covariance matrix $S$ corresponding to its largest eigenvalue.

The Karhunen–Loève expansion has the following Hilbert-space form. Let $X$ be a square-integrable Random element of a Hilbert space $H$, with mean $\mu$ and covariance operator $C$. There are nonnegative eigenvalues ${\lambda }_1\ge {\lambda }_2\ge \cdots$ and orthonormal eigenvectors ${\varphi }_k$ spanning the closure of the range of $C$ such thatin $L^2(\Omega ;H)$. The functional principal component scoressatisfyIf $X$ is a Gaussian random element, the scores are independent normal random variables.

For the proof, $C$ is positive, self-adjoint, and trace-class, withIt is therefore compact, so the spectral theorem for compact Hermitian operators supplies the eigenpairs. Their score covariance isFor the residual $R_m=X-\mu -{\sum }_{k=1}^m{\xi }_k{\varphi }_k$, Parseval identity and the trace formula giveThe centered variable's projection onto $\ker C$ has zero second moment and is therefore zero almost surely, which completes the mean-square expansion. When $C$ has a continuous covariance kernel, Mercer's theorem additionally expands that kernel as $c(s,t)={\sum }_k{\lambda }_k{\varphi }_k(s){\varphi }_k(t)$.

Because $A$ is a real symmetric positive-definite matrix, the finite-dimensional spectral theorem supplies an orthonormal eigenbasis $u_1,\dots ,u_n$ withIts eigendecomposition is also its singular value decomposition. If $e=y^{(\delta )}-y$, thensoThe operator norms satisfy $\Vert A{\Vert }_2={\lambda }_n$ and $\Vert A^{-1}{\Vert }_2=1/{\lambda }_1$. Therefore the worst-case relative perturbation bound isThe ratio ${\kappa }_2(A)$ is the spectral condition number of a positive-definite matrix. A large ratio means that data noise aligned with an eigenvector for the smallest eigenvalue is strongly amplified, so the inverse problem is ill conditioned.

Extend a grid vector $v$ by zero to the Dirichlet boundary. Pairing contributions along undirected nearest-neighbour edges gives the discrete energy identitywhere $E$ includes edges from an interior node to a boundary node. This is the three-dimensional version of summation by parts for the seven-point Dirichlet Laplacian.

The right side is nonpositive. If it vanishes, every pair of neighbouring values agrees; connectivity of the grid and the zero boundary values then imply $v=0$. Thus $v^{T}Av<0$ for every nonzero $v$, so $A$ is negative definite. In particular, zero is not an eigenvalue, and therefore

A complex representation of a finite group $G$ is a homomorphismfor a finite-dimensional complex vector space $V$. It is an irreducible representation when its only invariant subspaces are $0$ and $V$, and its degree is $\dim V$. Two representations are isomorphic when there is an invertible linear map intertwining their $G$-actions.

Write the dihedral group asIn an irreducible representation choose an eigenvector $0\ne v\in V$ of $\rho (r)$, which is possible because $r^n=1$. If its eigenvalue is $\lambda$, thenThus $W=\mathrm{span}\{v,\rho (s)v\}$ is invariant under both $r$ and $s$. Irreducibility forces $W=V$, proving $\dim V\le 2$ as in irreducible complex representations of a finite dihedral group.

Let $\omega =e^{2\pi i/n}$. Besides the one-dimensional representations described below, define for each indicated $j$These matrices satisfy the defining relations. Since ${\omega }^j\ne {\omega }^{-j}$, the only one-dimensional $r$-invariant subspaces are the two coordinate axes, and $s$ interchanges them. Hence ${\rho }_j$ is irreducible. Their characters satisfywhose values are distinct in the ranges below, so the ${\rho }_j$ are pairwise nonisomorphic. This is the family of two-dimensional representations of a finite dihedral group.

If $n$ is odd, a one-dimensional character must send $r$ to a scalar $\epsilon$ satisfying ${\epsilon }^n=1$ and $\epsilon ={\epsilon }^{-1}$. Thus $\epsilon =1$, while $s$ may independently map to $1$ or $-1$. These give two one-dimensional irreducibles. Takinggives $(n-1)/2$ two-dimensional irreducibles, for a total of

If $n$ is even, $r$ may map to either $1$ or $-1$, and again $s$ may map independently to either sign. This gives four pairwise nonisomorphic one-dimensional irreducibles. Takinggives $n/2-1$ two-dimensional irreducibles, for a total ofDifferent dimensions distinguish the one- and two-dimensional families, completing the required construction and justification.

Applying the transform twice givesThe root-of-unity filter makes the inner sum equal to $N$ exactly when $z\equiv -x(\mathrm{mod}N)$ and zero otherwise. HenceApplying modular negation twice givesTherefore ${\mathrm{QFT}}_N^2$ is an involution. Every one of its eigenvalues $\lambda$ satisfies ${\lambda }^2=1$, so its spectrum is contained in $\{1,-1\}$, as summarized by the square of the quantum Fourier transform.

The matrix $C$ from part (a) is Hermitian positive definite, while the real diagonal matrix $D$ is Hermitian. The stated product theorem, or similarityshows that every eigenvalue of $CD$ is real. Hence every eigenvalue of $B=-CD$ is real; this is the real spectrum of a positive-Hermitian times Hermitian product.

Moreover, $C$ is invertible and $D$ has rank $2d$, so for $d\ge 1$ the matrix $B$ has nonzero real eigenvalues. If $\lambda \ne 0$ is one of them, the corresponding mode of the semidiscrete equation has eigenvalue $i\pi \lambda$. The explicit Euler method has amplification factorfor every $\Delta t>0$. Therefore the explicit Euler discretization is unstable, as in explicit Euler instability on a nonzero imaginary eigenvalue.

At an equilibrium, the second equation gives $x=y=:s$. The first equation then givesHence the fixed-point branches in the $(a,x)$ plane areandThe three branches meet at $a=0$. The branch $x=a^2$ meets the positive square-root branch again when $a^2=\sqrt{a}$, so the second bifurcation value is

The Jacobian matrix at a general point isOn the branch $P_0=(a^2,a^2)$ it is triangular, with eigenvaluesThus $P_0$ is unstable for $a<0$, asymptotically stable for $0<a<1$, and unstable for $a>1$.

On a square-root branch write $s=\pm \sqrt{a}$. Thenwhose trace is $-1$ and determinant is $2s^2(s^3-1)$. The negative branch $s=-\sqrt{a}$ is therefore a saddle equilibrium for every $a>0$. The positive branch is a saddle for $0<a<1$ and asymptotically stable for $a>1$; close to $a=1$ it is a stable node.

To resolve the nonhyperbolic point at $a=0$, make the prescribed substitutionand append $\dot{a}=0$. The equations becomeThe centre subspace is $Y=X$. Seek the extended centre manifold for a parameter asThe centre-manifold invariance equation isTo second order, $\dot{X}=-aX+O(3)$, so comparison of coefficients givesand hence $A=C=0$, $B=1$. ThereforeSubstitution into the $X$ equation gives the reduced flowIts central branch $X=0$ is unstable for $a<0$ and stable for $a>0$, while the two nonzero branches $X\sim \pm \sqrt{a}$ for $a>0$ are unstable. Thus the bifurcation at zero is a subcritical pitchfork bifurcation with reversed normal-form parameter $\mu =-a$, exactly as recorded by the extended centre manifold of the 2023 Cambridge quadratic-product system.

The complete bifurcation diagram is therefore as follows. For $a<0$, only $P_0$ exists and is unstable. For $0<a<1$, $P_0$ is stable while both $P_{+}$ and $P_{-}$ are saddles. For $a>1$, $P_{+}$ is stable while $P_0$ and $P_{-}$ are saddles. At $a=1$, $P_0$ and $P_{+}$ cross and exchange stability, so the bifurcation is transcritical. This gives the bifurcation diagram of the 2023 Cambridge quadratic-product system.

Finally consider the phase plane near $(x,y)=(1,1)$ with $|a-1|\ll 1$. If $a<1$, the lower equilibrium $(a^2,a^2)$ is a stable node and the upper equilibrium $(\sqrt{a},\sqrt{a})$ is a saddle. If $a>1$, the lower equilibrium $(\sqrt{a},\sqrt{a})$ is the stable node and the upper equilibrium $(a^2,a^2)$ is the saddle. In each case the saddle has one-dimensional stable and unstable manifolds; one unstable separatrix runs toward the nearby stable node, while the other runs out of the local neighbourhood. The two equilibrium branches and their local invariant manifolds exchange roles as $a$ passes through one, which is the standard local phase portrait of a transcritical bifurcation.

For $|J,J⟩$, tracing out subsystem 2 gives the reduced density matrixwhich is pure, so its entanglement entropy is zero.

For $|J,J-1⟩$, the two subsystem-2 factors in part (iii) are orthogonal. The partial trace therefore removes the cross terms and givesIts nonzero eigenvalues are $j_1/J$ and $j_2/J$, so the Von Neumann entropy isWhen $j_1=j_2$, both eigenvalues are $1/2$ and $S=\log 2$, the maximal entropy for a rank-two reduced state.

Differentiate the eigenvalue equationat $t=0$, where $A^{\prime }(0)=E$. Suppressing the argument $0$ givesLeft-multiplication by the transpose of the unit left eigenvector $u$ and use of $u^{T}A=\lambda u^T$ cancel the terms involving $v^{\prime }$:Because the eigenvalue is simple, $u^{T}v\ne 0$, and hence the first-order perturbation of a simple eigenvalue isThe definition of the operator norm, followed by the Cauchy-Schwarz inequality, givesTherefore

The characteristic polynomial of $A$ isA scalar $\lambda \in \mathbb{C}$ is an eigenvalue when there is a nonzero vector $v$ such thatSuch a $v$ is an eigenvector belonging to $\lambda$. Equivalently, $\lambda$ is a root of ${\chi }_A$ and the corresponding eigenvectors are the nonzero elements of $\ker (A-\lambda I)$.

Eigenvectors belonging to distinct eigenvalues are linearly independent. Indeed, supposeis a shortest nontrivial relation, where $A{v}_j={\lambda }_j{v}_j$ and the ${\lambda }_j$ are distinct. Applying $A-{\lambda }_{m}I$ givesa shorter nontrivial relation, a contradiction.

Thus $n$ distinct eigenvalues provide a basis of $n$ eigenvectors, proving that $A$ is diagonalizable over $\mathbb{C}$. If $A$ is real and all its eigenvalues are real, each real singular matrix $A-\lambda I$ has a nonzero real null vector. Choosing one for each eigenvalue gives a real eigenbasis. Hence the real matrix is diagonalizable over $\mathbb{R}$. This is the distinct eigenvalues imply diagonalizability criterion.

The characteristic polynomial ${\chi }_A(t)=\det (tI-A)$ has degree $n$. By the fundamental theorem of algebra, it has a complex root $\lambda$. Henceso $\lambda I-A$ has a nonzero kernel. Any nonzero $v$ in that kernel satisfiesThus $\lambda$ is an eigenvalue and $v$ is a corresponding eigenvector.

A complex matrix $U$ is unitary matrix whenIf $Uv=\lambda v$ with $v\ne 0$, unitarity preserves norms, soTherefore every eigenvalue satisfies

The special orthogonal group isIfits first column is a unit vector, say $(\cos \theta ,\sin \theta {)}^T$. The second column is the unique unit vector perpendicular to it that gives positive determinant, namely $(-\sin \theta ,\cos \theta {)}^T$. Thusso $A$ is rotation through $\theta$ about the origin.

For $A\in SO(3)$, its real characteristic polynomial of odd degree has a real eigenvalue. Every eigenvalue of an orthogonal matrix has modulus one, so every real eigenvalue is $\pm 1$. Nonreal eigenvalues occur in conjugate pairs whose product is one, while $\det A=1$; if all eigenvalues are real, their product likewise forces at least one to be $+1$. Hence $A$ fixes a nonzero vector $u$. The plane $u^{\perp }$ is $A$-invariant, and the restriction to it is an orientation-preserving orthogonal map. By the $SO(2)$ result it is a planar rotation. Therefore $A$ is a rotation about the axis $\mathbb{R}u$.

A differential equation is stiff when it contains rapidly decaying modes on time scales much shorter than those of interest, forcing an explicit method to take very small steps for stability rather than accuracy.

Hereso the decay rates are $1$ and $100$. For a negative real eigenvalue, forward Euler requirestherefore the fast mode imposesFor backward Euler the amplification factors arewhose moduli are at most one for every $h\ge 0$. ThusThis is the stiff two-mode linear system.

The eigenvalue equation $D_{r}f=\lambda f$ isIf $\lambda =-1$, this forces $f(x+r)=0$ for every $x$, hence $f=0$, so $-1$ is not an eigenvalue. If $c=1+\lambda \ne 0$, choose arbitrary values on representatives of the cosets of $r\mathbb{Z}$ and extend byThis produces nonzero eigenfunctions. Choosing functions supported on distinct cosets gives infinitely many linearly independent eigenfunctions. Thereforeand every eigenspace is infinite-dimensional.

Direct expansion giveswhich is symmetric in $r,s$. Hence

Suppose a degree-$n$ polynomial $p$ were a sum of $n$ periodic functions,Apply the commuting product $D_{r_1}\cdots D_{r_n}$. Every term on the right is killed by its corresponding factor, whereas for leading coefficient $a_n$ the mixed finite difference of a polynomial givesThis contradiction proves