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Let $v\ne 0$ be an eigenvector of the Hermitian matrix $A$, with eigenvalue $\lambda$. Multiplying $Av=\lambda v$ by $v^{†}$ givesThe denominator is the positive real number $\Vert v{\Vert }^2$, whileThus the numerator is real, and every eigenvalue of a Hermitian matrix is real. This argument does not require distinct eigenvalues.

Subtract the exact solution from the Jacobi method to get $e_n=H{e}_{n-1}$, hence $e_n=H^n{e}_0$. Convergence from every initial vector is equivalent to $H^n\to 0$. If every eigenvalue has modulus below one, a Jordan block $\lambda I+N$ has powers that are finite sums $\left(\genfrac{}{}{0ex}{}{n}{k}\right){\lambda }^{n-k}{N}^k$. The polynomial factor in $n$ is dominated by geometric decay; for $\lambda =0$, the block is nilpotent. Consequently $H^n\to 0$.

Conversely, for a complex eigenvector $v$ with eigenvalue $\lambda$, $H^{n}v={\lambda }^{n}v$. Convergence for real errors also implies convergence for their complexification, so this tends to zero only if $|\lambda |<1$. Therefore the iteration converges from every starting vector if and only if $\rho (H)<1$, where $\rho$ is the spectral radius.

For the specified matrix,Its eigenvalues are $0,\mu ,-\mu$, so $\rho (H)=|\mu |$. Hence $-1<\mu <1$ is the convergence range. The original coefficient matrix has determinant $12(1-{\mu }^2)$, so the endpoints also violate its assumed nonsingularity.

(1) True. Over $\mathbb{C}$, $A$ has an eigenvalue $\lambda$ with nonzero eigenspace $E$. Commutativity makes $E$ invariant under $B$: $A(Bv)=B(Av)=\lambda Bv$. The endomorphism $B{|}_E$ has an eigenvector $v\ne 0$ because its characteristic polynomial has a complex root. This $v$ is an eigenvector of both matrices, though their eigenvalues on it may differ.

(2) False. Take $A=I$ and $B=2I$. They commute, but their only eigenvalues are respectively $1$ and $2$.

(3) True. Associativity gives $(BA{)}^{n+1}=B(AB{)}^{n}A=0$, so $T=BA$ is nilpotent. A nilpotent endomorphism of an $n$-dimensional space has index at most $n$: in $V\supseteq TV\supseteq T^{2}V\supseteq \cdots$, equality at a nonzero image would make $T$ surjective on that finite-dimensional image, hence invertible there, contradicting nilpotence. Thus dimensions strictly decrease until zero, in at most $n$ steps. Therefore $(BA{)}^n=0$.

(4) False. Take $T=0$ on a two-dimensional vector space and $\lambda =0$. The eigenspace has dimension two, but the minimal polynomial is $t$, in which the root zero has multiplicity one. Eigenspace dimension counts independent eigenvectors; the exponent in the minimal polynomial measures the longest corresponding Jordan chain, not that dimension.

(5) True. Factor the minimal polynomial as $m_T(t)=(t-\lambda {)}^{c}q(t)$, with $q(\lambda )\ne 0$, and set $S=T-\lambda I$. If $v\in W_{c+1}$, then $w=S^{c}v$ satisfies $Sw=0$, so $q(T)w=q(\lambda )w$. But $m_T(T)v=0$ also gives $q(T)w=0$. Thus $w=0$, so $v\in W_c$. The reverse inclusion is immediate from $S^{c}v=0\Rightarrow S^{c+1}v=0$, proving

Use the determinant-normalized slash operator and the usual Hecke operator normalization. A normalized eigenform is a nonzero cusp form $f={\sum }_{r\ge 1}{a}_r{q}^r$ with $a_1=1$ and $T_{n}f=\lambda (n)f$ for every $n$. We first establish the operator facts needed here, rather than assume the Hecke multiplication relations.

Integer matrices of determinant $n$ have the determinant-n matrix representatives for Hecke operators ${\gamma }_{a,b,d}=\left(\begin{array}{cc}a& b\\ 0& d\end{array}\right)$, with $ad=n$ and $0\le b<d$. Integer row operations give these representatives: the Euclidean algorithm reduces the first column to $(a,0)$, and a shear reduces $b$ modulo $d$. Right multiplication by the modular group permutes the left cosets, so $T_{n}f=n^{k/2-1}{\sum }_{ad=n}{\sum }_{b\mathrm{mod}d}f{|}_k{\gamma }_{a,b,d}$ is modular. Rational changes of coordinate preserve cusp vanishing, so it remains a cusp form. Substituting its Fourier expansion of a modular form and summing $e^{2\pi irb/d}$ givesIn particular $a_1(T_{n}f)=a_n(f)$, and the normalization gives $\lambda (n)=a_n(f)$ for every $n$.

For coprime $u,v$, applying this coefficient formula twice amounts to choosing independently a divisor of $\gcd (m,u)$ and one of $\gcd (m,v)$. Their product runs exactly once through the divisors of $\gcd (m,uv)$, with the same factor $d^{k-1}$ and coefficient index. Thus $T_u{T}_v=T_{uv}$. For a prime $p$, put $P=p^{k-1}$ and $\nu =v_p(m)$. The coefficient of $q^m$ in $T_{p^e}f$ isThe coefficient in $T_p{T}_{p^e}f$ is $A_e(pm)+P{A}_e(m/p)$, the second summand being zero if $p\nmid m$. Expanding the two finite sums and matching their endpoints gives $A_{e+1}(m)+P{A}_{e-1}(m)$ for $e\ge 1$. Equality of all coefficients proves $T_p{T}_{p^e}=T_{p^{e+1}}+p^{k-1}{T}_{p^{e-1}}$. Applying these identities to the nonzero eigenvector $f$ yields

Let $f_j(z)=f(p^{j}z)$ for $0\le j\le r$, and write $a=a_p(f)$ and $W_{M}f=\epsilon f$. The Hecke operator $U_p$ acts on the Fourier expansion by $\sum a_n{q}^n\mapsto \sum a_{pn}{q}^n$. Since $p\mid M$, the primitive relation is $a_{pn}=a_p{a}_n$. Hence $U_p{f}_0=a{f}_0$ and $U_p{f}_j=f_{j-1}$ for $j\ge 1$. Substituting the argument directly in the Fricke involution givesThus, in the basis $e_j=p^{jk/2}{f}_j$, put $b=p^{k/2}$ and let $R$ be the reversal matrix, $R{e}_j=e_{r-j}$. The matrices areIf the exam's rescaling $w_N=N^{k-1}{W}_N$ is used, multiply the second matrix by $N^{k-1}$.

Here is an explicit irreducibility argument for the prime-power old block at a bad prime. Any nonzero common invariant complex subspace $V$ contains an eigenvector of the restricted $U_p$. Its eigenvalue is either $a$ or zero. If $a\ne 0$, the $a$-eigenspace is $\mathbb{C}{e}_0$; containing it gives $e_r$ by reversal, then every $e_j$ by successive powers of $U_p$. The kernel of $U_p$ is $\mathbb{C}(e_1-(b/a)e_0)$. If $V$ instead contains this vector, reversal and $U_p^r$ giveThis is nonzero: the bad-prime eigenvalue of a trivial-character newform satisfies $a^2=p^{k-2}$ when $p\parallel M$, and $a=0$ when $p^2\mid M$, whereas $b^2=p^k$. If $a=0$, the kernel is already $\mathbb{C}{e}_0$, and the same reversal-and-shift argument works. Therefore the block has no nonzero proper subspace invariant under $U_p$ and the Fricke operator, and hence none invariant under the full ring in the question.

The derived series of a Lie algebra isThe algebra is a solvable Lie algebra if ${\mathfrak{b}}^{(m)}=0$ for some $m$. Here a bracket of subspaces means the linear span of their Lie brackets.

The Lie theorem states that a nonzero finite-dimensional representation of a complex solvable Lie algebra has a common eigenvector; equivalently, its operators can be simultaneously upper triangularized. Applied to an Irreducible Lie algebra representation, the common eigenvector gives a nonzero invariant line, which must be the entire module. A scalar action $\rho (x)=\lambda (x)$ respects the Lie bracket exactly when $\lambda ([x,y])=0$. Thus, in the finite-dimensional setting,Every functional displayed gives a one-dimensional irreducible, and two such modules are isomorphic exactly when their functionals agree.

The finite-dimensional qualification is essential here, although the printed wording does not repeat it. For example, let $[x,y]=y$ and act on $\mathbb{C}[t]$ by $xf(t)=tf(t)$ and $yf(t)=f(t-1)$. These operators satisfy $[x,y]f=yf$. A nonzero submodule stable under $x$ is an ideal $p(t)\mathbb{C}[t]$. Stability under $y$ forces $p(t)\mid p(t-1)$, so the two polynomials, having the same degree and leading coefficient, are equal. In characteristic zero a polynomial invariant under a nonzero shift is constant. The module is therefore an infinite-dimensional simple module for the two-dimensional affine Lie algebra. It rules out extending the one-dimensional classification to unrestricted representations.

First consider finite-dimensional Lie algebra representations. The operators representing an abelian Lie algebra commute. They have a common eigenvector: take an eigenspace for the first operator of a basis of the algebra, observe that every remaining operator preserves it, and repeat inside this nonzero space. The resulting line is an invariant subspace, so an irreducible representation must be one-dimensional. Its action has the form $xv=\lambda (x)v$ for a linear functional $\lambda$. Conversely, every such functional defines a Lie algebra representation, since both the Lie brackets in the algebra and the commutators of its scalar operators vanish. Distinct functionals give nonisomorphic representations. Thus the irreducibles are precisely the characters $\lambda \in {\mathfrak{g}}^{\ast }$.

Even without a finite-dimensional assumption on the module, this conclusion for a finite-dimensional abelian Lie algebra follows from the Weak Hilbert Nullstellensatz: its universal enveloping algebra is the polynomial algebra $\mathrm{Sym}(\mathfrak{g})$. A simple module over a commutative algebra is its quotient by a maximal ideal, and every maximal ideal of this polynomial algebra over $\mathbb{C}$ has residue field $\mathbb{C}$.

Indecomposable does not imply irreducible. For the one-dimensional abelian Lie algebra $\mathbb{C}x$, let $x$ act on ${\mathbb{C}}^2$ by the Nilpotent Jordan blockThe line spanned by the first coordinate vector is an invariant subspace, so the module is reducible. If it were a direct sum of two nonzero submodules, both would have dimension one; nilpotence would force $N$ to act as zero on both, contrary to $N\ne 0$. Thus it is an indecomposable representation.

For a finite-dimensional module $V$, put $d=\dim V$ and define its generalized weight spaces intrinsically byTo obtain the decomposition, choose a basis $x_1,\dots ,x_r$ of $\mathfrak{g}$, split into generalized eigenspaces of $\rho (x_1)$, and successively split each piece by $\rho (x_2),\dots ,\rho (x_r)$. All the pieces are submodules because the acting operators commute. On one resulting piece, $\rho (x_j)$ has a unique eigenvalue ${\lambda }_j$. The commuting operators can be simultaneously upper triangularized by repeating the common-eigenvector argument on quotients. Hence on this piece every $\rho (x_j)-{\lambda }_{j}I$ is strictly upper triangular. Any linear combination is also strictly upper triangular, and its $d$th power vanishes. Extending ${\lambda }_j$ linearly therefore identifies the piece with $V^{\lambda }$. Different tuples give disjoint pieces; if a component has a different tuple from $\lambda$, at least one $\rho (x_j)-\lambda (x_j)I$ is invertible there. Consequentlywith only finitely many nonzero summands. Since the definition uses every $x$ rather than a chosen basis, this generalized-weight decomposition for a nilpotent Lie algebra is canonical. Its summands need not be irreducible, as the Jordan block example already shows.

The six critical Fourier modes are at $\pm {\mathbf{k}}_1,\pm {\mathbf{k}}_2,\pm {\mathbf{k}}_3$, with ${\mathbf{k}}_1+{\mathbf{k}}_2+{\mathbf{k}}_3=0$. Reality of the pattern pairs each positive-wavevector coefficient with the conjugate coefficient at the negative wavevector. The leading critical part of the extended centre manifold parametrization is consequentlyThe displayed six-mode expression is its leading-order projection; nonlinear harmonics on the full centre manifold need not vanish identically.

A translation $\mathbf{d}$ acts by $A_i\mapsto e^{i{\mathbf{k}}_i\cdot \mathbf{d}}{A}_i$. Each monomial in the equation for $A_i$ must therefore carry wavevector ${\mathbf{k}}_i$. Since $-{\mathbf{k}}_{i+1}-{\mathbf{k}}_{i+2}={\mathbf{k}}_i$, the allowed quadratic term is $A_{i+1}^{\ast }{A}_{i+2}^{\ast }$. At cubic order the allowed terms are $A_i|A_i{|}^2,A_i|A_{i+1}{|}^2,A_i|A_{i+2}{|}^2$. Rotation through $60$ degrees acts, with one choice of orientation, asIts square cyclically permutes the amplitudes, equating the coefficients across the three equations. Its cube conjugates every amplitude; equivariance under this half-turn makes all coefficients real. It does not interchange the two cross-coupling coefficients. A reflection would do that and force $b=c$, but that symmetry is expressly absent. Thus the chiral hexagonal amplitude equations areat the retained order, with cyclic indices and real coefficients. The missing opening modulus bar in the last cubic term of the printed formula must be supplied; phase equivariance requires $|A_{i+2}{|}^2$.

Write $A_i=r_i{e}^{i{\varphi }_i}$, with all three amplitudes nonzero, and let $\Phi ={\varphi }_1+{\varphi }_2+{\varphi }_3$. Separating real and imaginary parts givesHence the phase locking of a resonant hexagonal triad obeysFor $\alpha >0$, $\Phi =0$ is phase stable and $\Phi =\pi$ is phase unstable. Two independent translation coordinates remove two phases; since the wavevectors sum to zero, the remaining phase is exactly $\Phi$. Thus a stable fully active phase-locked pattern can be represented by three positive real amplitudes after choosing the spatial origin. The statement is modulo translations, not a restriction on every representative. Boundary states must also be retained: the zero solution is stable for $\mu <0$, and rolls have two zero amplitudes. These are the necessary qualifications to a literal claim that all stable solutions are strictly positive.

In the real nonnegative subspace, a roll isor a cyclic permutation. For hexagons let $s=a+b+c$, which is positive under the assumptions. The equal-amplitude branches solve $\mu +\alpha r-s{r}^2=0$, givingwith only positive roots retained. The upper branch exists for $\mu \ge -{\alpha }^2/(4s)$; the lower branch is positive for $-{\alpha }^2/(4s)\le \mu <0$. They meet at a fold.

For a purported rectangle $(p,p,q)$ with $p,q>0$, subtract its first two equilibrium equations after dividing each by its own amplitude. The quadratic terms agree, and the difference is $(c-b)(p^2-q^2)=0$. Since $(b-a)(c-a)<0$ implies $b\ne c$, it forces $p=q$. A state with exactly two nonzero amplitudes is also impossible because the quadratic term drives the zero third amplitude. Thus there are no nontrivial rectangles.

In fact no other unequal positive equilibria are possible under the straddling cross-couplings. Here is the uniqueness of positive hexagon equilibria with straddling cross-couplings argument. Put $x_i=r_i^2$, $p=r_1{r}_2{r}_3>0$. The equilibrium equations implyConsider $b>a>c$, write $B=b-a>0$, $C=c-a<0$, and relabel cyclically so $x_1$ is maximal. If $x_2\ge x_3$, subtracting equations one and two givesThe left side is positive unless all three are equal, while the right side is nonpositive. If $x_2\le x_3$, subtract equations two and three instead:Now the left side is negative unless all are equal and the right side is nonnegative. Both cases force equality. For $c>a>b$, reverse the cyclic orientation and exchange $b,c$. Thus the only nonzero equilibria in this subspace are the stated rolls and hexagons.

At a roll the active radial eigenvalue is $-2\mu$. The two inactive amplitudes have real-part linearizationIts determinant is ${\mu }^2(a-c)(a-b)/a^2-{\alpha }^2\mu /a<0$. It therefore has a positive eigenvalue: the rolls are unstable for every $\mu >0$, which is stronger than the large-$\mu$ conclusion. In the complex system there is also a neutral translation phase, and the imaginary transverse block has the same determinant.

At a positive hexagon the real-amplitude Jacobian is circulant, with diagonal $d=-2a{r}^2-\alpha r$ and off-diagonals $p=\alpha r-2b{r}^2$, $q=\alpha r-2c{r}^2$. Its uniform eigenvector $(1,1,1)$ has eigenvalueThe other two eigenvectors lie in the zero-sum subspace; using the two nontrivial cube roots of unity givesFor sufficiently large $\mu$, $r\sim \sqrt{\mu /s}$, and the positive term $(b+c-2a)r^2$ dominates the quadratic-coupling term $2\alpha r$. Thus the hexagons are unstable at large $\mu$, while the uniform radial direction on the upper branch is stable. The two translation phases are neutral and the total phase has eigenvalue $-3\alpha r$, so the amplitude instability is genuine.

More precisely, the upper branch is amplitude stable between its fold and the chiral hexagon Hopf thresholdAt this threshold the conjugate pair crosses with nonzero frequency because $b\ne c$, permitting a Hopf bifurcation. The linear calculation locates the crossing; the nonlinear normal form decides the stability and direction of its periodic branch.

The dynamics are bounded in the retained model. For $S=r_1^2+r_2^2+r_3^2$ in the positive phase-locked subspace,Since $r_1{r}_2{r}_3\le (S/3{)}^{3/2}$, the negative quadratic-in-$S$ term dominates at large amplitude. For large positive $\mu$, the origin, rolls and hexagons are all unstable, so generic bounded dynamics cannot end at an asymptotically stable steady pattern. The chiral difference $b-c$ favors successive changes in dominant roll orientation, in the spirit of the Küppers–Lortz instability. Oscillatory competition or periodic switching is the natural behavior near the Hopf transition; linear stability alone does not establish a unique global limit cycle. With $\alpha =0$, coordinate planes are invariant and heteroclinic roll switching is possible. For $\alpha >0$ a missing third mode is regenerated when two others are present, so that exact boundary-plane heteroclinic picture is broken. Finally, sufficiently large $\mu$ is a conclusion of the cubic truncation; far from onset, omitted higher-order terms can change the physical pattern dynamics.

The eigenvalue stability analysis of a finite difference method reduces a linear evolution discretization to its modal behavior, but it must retain the norm and the dependence on the mesh. For a homogeneous system obtained by the method of lines, $U^{\prime }=A_{h}U$, the solution is $e^{t{A}_h}U(0)$. A one-step time discretization gives $U^{n+1}=G_{h,k}{U}^n$; for a Runge-Kutta method, where its stage matrices are invertible, $G_{h,k}=R(k{A}_h)$. With a multistep method one instead uses an amplification matrix on the augmented vector of several time levels.

The relevant finite-time stability estimate iswith $C_T$ independent of the refining spatial mesh and allowed time steps. Similarly, semidiscrete stability requires $\Vert e^{t{A}_h}{\Vert }_h\le C_T$ for $0\le t\le T$. The norm should represent the continuum problem, such as the discrete L2 norm $\Vert U{\Vert }_{2,h}^2=h{\sum }_j|U_j{|}^2$ in one dimension. A fixed finite matrix having decaying solutions as $t\to \infty$ is not by itself a mesh-uniform result for the partial differential equation.

If $G=X\Lambda X^{-1}$, thenThus eigenvalues of modulus at most one give a uniform bound when the diagonalizing bases have uniformly bounded condition numbers. For a normal matrix in the chosen inner product, the spectral theorem for normal operators supplies an orthonormal basis, so the condition number is one and the norm of $G^n$ is exactly ${\max }_j|{\lambda }_j{|}^n$. For a self-adjoint or skew-adjoint spatial matrix, applying a scalar stability function preserves this favorable modal structure. This is the setting where eigenvalue calculations provide especially clean, reliable step restrictions.

A precise fixed-matrix statement is also useful: a matrix is power bounded for all nonnegative integers $n$ if and only if all its eigenvalues lie in the closed unit disk and every Jordan block at a unit-modulus eigenvalue has size one. Indeed, powers of an interior Jordan block contain a polynomial in $n$ times a decaying geometric factor and are bounded; a larger block on the unit circle produces unbounded polynomial growth. For a family of matrices, the resulting bounds must still be uniform. Moreover, finite-time stability versus power boundedness distinguishes this all-time condition from the physical-time estimate above. For example $G_k=I+kN$, $N^2=0$, has $G_k^n=I+nkN$, bounded for $nk\le T$ despite a nontrivial Jordan block at one. Even the scalar update $1+k$ is stable on fixed intervals, since $(1+k{)}^n\le e^T$, although it represents a growing equation rather than a contraction.

For a constant-coefficient periodic difference operator, Fourier modes diagonalize the spatial matrix. The discrete Fourier transform is a unitary operator, so Parseval's identity identifies its modal maximum with the operator norm. The eigenvalue stability analysis of a finite difference method then becomes von Neumann stability analysis. For the Dirichlet discrete Laplacian, a discrete sine transform plays the same role. These facts explain its advantages: a large matrix calculation becomes a scalar symbol or a known spectral interval, and the extremal eigenvalue directly determines a safe time step.

As a first example, centered space discretization of the heat equation $u_t=u_{xx}$ has eigenvaluesForward Euler method gives $R(k\lambda )=1-4r{\sin }^2(\theta /2)$ with $r=k/h^2$. Requiring all factors in $[-1,1]$ yields $0\le r\le 1/2$ for all periodic meshes. In two dimensions the corresponding uniform bound is $r\le 1/4$. Because these are normal matrices, these scalar bounds prove contraction rather than merely suggest it.

Backward Euler method has $R(z)=1/(1-z)$, so the centered discretization of the heat equation is contractive for every $k>0$. The Crank-Nicolson method has $R(z)=(1+z/2)/(1-z/2)$ and is likewise contractive for these negative real modes. However, its factor tends to $-1$ for very stiff modes, leaving high-frequency oscillations weakly damped; Backward Euler method tends to zero. The eigenvalues therefore expose the presence of stiff differential equations and damping as well as binary stability. Unconditional stability does not remove accuracy requirements.

For an advection equation example, the upwind finite difference scheme for $u_t+c{u}_x=0$, $c>0$, has factor $G=1-\nu +\nu e^{-i\theta }$, with $\nu =ck/h$. Sinceit is a contraction in the periodic discrete L2 norm exactly for $0\le \nu \le 1$. In contrast, Forward Euler method with centered advection has factor $1-i\nu \sin \theta$. Its modulus exceeds one at nonzero active frequencies, and repeated amplification makes it unstable under fixed nonzero Courant refinement. A much smaller scaling $k=O(h^2)$ can bound its finite-time growth, demonstrating again that exact contraction and the most general stability estimate are different claims.

The major limitation is the possible presence of a nonnormal matrix. For example,Both eigenvalues are negative, yet at $t=1$ the second unit vector is amplified by at least $1/(eh)$. This proves that negative spectra do not imply uniform semidiscrete stability. Ill-conditioned eigenvectors, Jordan normal form and transient amplification can invalidate a spectral-only argument. Direct energy estimates or estimates for the resolvent can supply the missing norm control.

Boundary conditions are another limitation. A periodic symbol proves a periodic or whole-line result, not an arbitrary initial-boundary-value problem. Boundary closure of a difference scheme can introduce growing modes or mesh-dependent amplification. A spatial operator with variable coefficients can still be studied through its assembled matrix, but then whether it is a normal matrix, its condition number and the physical inner product must be checked rather than borrowed from a Fourier stability analysis. For the finite element method the natural mass matrix inner product often turns a generalized eigenvalue problem into a self-adjoint one.

For time-dependent or split updates, separate spectra are insufficient to control products. For instance, $G_1=\left(\begin{array}{cc}0& K\\ 0& 0\end{array}\right)$ and $G_2=\left(\begin{array}{cc}0& 0\\ K& 0\end{array}\right)$ each have only zero eigenvalues, while $G_2{G}_1$ has eigenvalue $K^2$. For $K>1$, alternating the two is unstable. A common contractive norm would control the products, but their individual eigenvalues do not.

Finally, stability and consistency of a numerical method have different roles. The Lax equivalence theorem states that, for a well-posed linear initial-value problem and a consistent finite difference method on compatible grid spaces, stability on every fixed time interval is equivalent to convergence of a numerical method as the admissible meshes refine. Bounded grid representation and the correct initial and boundary setup are part of that framework. An eigenvalue estimate can establish its stability hypothesis in the favorable settings described above; it neither proves consistency of a numerical method nor bypasses its uniformity requirements. The method is therefore a powerful modal tool, provided its spectral calculation is connected to a genuine norm estimate.

Use the Euclidean norm and its induced operator norm. The spectral theorem for real symmetric matrices gives $\Vert A\Vert ={\lambda }_n$ and $\Vert A^{-1}\Vert =1/{\lambda }_1$. If $e=y_{\delta }-y$, then the direct inverse has error $A^{-1}e$, soFor nonzero $y$, the relation $\Vert y\Vert =\Vert Ax\Vert \le {\lambda }_n\Vert x\Vert$ also givesThis is the spectral condition number of a positive-definite matrix. The worst relative amplification is attained by exact data along an eigenvector for ${\lambda }_n$ and noise along one for ${\lambda }_1$. A large $\kappa$ therefore makes the reconstruction sensitive to relative data errors. Absolute sensitivity additionally depends on ${\lambda }_1$ itself: scaling every eigenvalue equally leaves $\kappa$ unchanged but changes $1/{\lambda }_1$. Since the matrix is finite dimensional and positive definite, the inverse exists and is continuous; the issue is ill-conditioning rather than mathematical nonexistence or discontinuity.

The spectral theorem for compact self-adjoint operators says that a compact self-adjoint $A$ has an orthogonal decompositionwith real nonzero eigenvalues of finite multiplicity. Only finitely many eigenvalues, counting multiplicity, have absolute value at least any given $\epsilon >0$. Thus the nonzero eigenvalues form a finite list or a sequence tending to zero, and $A={\sum }_{\lambda \ne 0}\lambda P_{\lambda }$ in operator norm. The kernel may have arbitrary Hilbert dimension.

For existence of an extremal eigenvector, self-adjointness gives $\Vert A\Vert ={\sup }_{\Vert x\Vert =1}|⟨Ax,x⟩|$. One elementary proof of the nontrivial inequality uses polarization: if the quadratic form is bounded in absolute value by $m\Vert x{\Vert }^2$, its polarized real part is bounded by $m$ on two unit vectors; adjusting a phase bounds $|⟨Ax,y⟩|$ by $m$. Choose unit $x_j$ with quadratic values tending to $\lambda =\Vert A\Vert$ or $-\Vert A\Vert$. ThenIf $A\ne 0$, compactness supplies a convergent subsequence of $A{x}_j$, and the last estimate supplies a convergent unit subsequence of $x_j$, with eigenvalue $\lambda$.

Eigenvectors for distinct eigenvalues are orthogonal, since $A=A^{\ast }$. An infinite orthonormal sequence in one nonzero eigenspace would violate compactness. More generally an infinite orthonormal family with eigenvalues bounded away from zero has images separated in norm, again impossible. Take all nonzero eigenspaces and let $M$ be their closed span. Its complement reduces $A$; if the restriction there were nonzero, the extremal-eigenvector argument would produce another nonzero eigenvector, a contradiction. Thus $M^{\perp }=\ker A$. The finite truncations of the eigenvalue sum have remainder norm tending to zero, proving the asserted norm expansion.

Apply part (a) to the compact positive operator $T^{\ast }T$. Its positive square root $B=|T|$ is defined by multiplying each eigenvector of eigenvalue $\lambda$ by $\sqrt{\lambda }$ and vanishing on the kernel. The eigenvalues tend to zero, so $B$ is compact. For every $x$, $\Vert Tx{\Vert }^2=⟨T^{\ast }Tx,x⟩=\Vert Bx{\Vert }^2$.

Consequently $U(Bx)=Tx$ is a well-defined isometry on $\mathrm{ran}B$. Extend it to the closure of that range and set it to zero on its orthogonal complement. This gives a partial isometry with initial space $(\ker T{)}^{\perp }$, final space $\overline{\mathrm{ran}T}$, andFor uniqueness, the positive factor in any such polar decomposition of a bounded operator satisfies $B^2=T^{\ast }T$. The positive-square-root uniqueness proved in 4(b) fixes $B$. The relation $U(Bx)=Tx$ then fixes $U$ on the dense subset $\mathrm{ran}B$ of its initial space and the kernel convention fixes it elsewhere. This proves uniqueness with the canonical initial-space condition. Without that condition the word “unique” would be false: $T=0$ can be written $U\cdot 0$ with many different unitary $U$.

Subtracting the fixed-point equation for the exact solution givesHere $D=2I$, so the supplied sine eigenvectors of $A$ are also eigenvectors of $H_{\omega }$. Explicitly,Their distinct eigenvalues for nonzero $\omega$ make them an eigenbasis; alternatively the discrete sine orthogonality gives a basis even when $\omega =0$.