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We first prove the needed Fredholm alternative for a compact operator without assuming self-adjointness. If $C$ is compact, the finite-rank approximation theorem for compact operators on a Hilbert space follows directly by covering the compact closure of $C$'s unit-ball image by finitely many balls of radius $\epsilon$, spanning their centers by a finite-dimensional space $E$, and using the orthogonal projection $P_E$. Then $\Vert C-P_{E}C\Vert \le \epsilon$.

Choose a finite-rank $F$ with $\Vert C-F\Vert <1$, put $E_0=C-F$, and factorThe first factor is invertible by the Neumann series, and $G$ has finite rank. With $V=\mathrm{ran}G$ and $H=V\oplus V^{\perp }$, the second factor has the triangular formIt is injective precisely when the finite-dimensional block $I_V-G{|}_V$ is injective, and onto precisely when that block is onto. These properties are equivalent in finite dimension. This proves that $I-C$ is injective if and only if it is onto; in that case its inverse is bounded.

For $\lambda \ne 0$, apply this result to $C=K/\lambda$. If $\lambda$ lies in the spectrum of a bounded operator $K$, then $K-\lambda I$ is not invertible, hence is not injective. Therefore every nonzero spectral point of $K$ is an eigenvalue. On its eigenspace $E_{\lambda }$, the operator $K$ equals $\lambda I$. Compactness of $K$ would make the closed unit ball of $E_{\lambda }$ compact after dividing by $\lambda$; part (a) forces $\dim E_{\lambda }<\infty$.

It remains to rule out infinitely many distinct eigenvalues away from zero. Suppose $|{\lambda }_n|\ge \epsilon >0$ are distinct, with respective eigenvectors $v_n$. Distinct eigenvalues give linearly independent eigenvectors. Let $V_n=\mathrm{span}(v_1,\dots ,v_n)$ and choose a unit vector $h_n\in V_n\cap V_{n-1}^{\perp }$. Since $V_n$ is invariant and $(K-{\lambda }_{n}I)V_n\subset V_{n-1}$,For $m<n$, $K{h}_m\in V_{n-1}$, and orthogonality gives $\Vert K{h}_n-K{h}_m\Vert \ge |{\lambda }_n|\ge \epsilon$. This contradicts compactness. Thus for each $\epsilon >0$ there are only finitely many distinct spectral points with modulus at least $\epsilon$.

Taking $\epsilon =1/j$ shows that the nonzero spectrum is countable; if it is infinite, its distinct elements can be enumerated as a sequence tending to zero. Since the spectrum is closed, zero then also belongs to it. In an infinite-dimensional space zero belongs to the spectrum even if there are only finitely many nonzero spectral points: an invertible compact $K$ would make $I=K^{-1}K$ compact, contradicting part (a). The only possible accumulation point is zero, and every nonzero eigenspace is finite-dimensional. Zero need not be an eigenvalue. The argument uses invariant spans, not an orthogonal eigenbasis, which is unavailable for a general non-self-adjoint compact operator.

Use the stated Fredholm definition of the essential spectrum, which applies to general bounded operators:For each $z$, part (f) applied to $L-zI$ says that $L-zI$ is Fredholm exactly when $L+K-zI$ is Fredholm. Taking the complement of these equivalent Fredholm conditions givesThis is the general Weyl theorem for bounded compact perturbations. Individual isolated eigenvalues can change; the Fredholm obstruction, and hence this essential spectrum, cannot.

Adding the two homogeneous reaction rates gives $a-u$, so the unique positive Brusselator equilibrium isThe reaction Jacobian matrix at that state isA real two-dimensional linear system has strict asymptotic stability exactly when $\tau <0$ and $\Delta >0$: the eigenvalues solve ${\lambda }^2-\tau \lambda +\Delta =0$. Real eigenvalues then have positive product and negative sum; a complex conjugate pair has real part $\tau /2$. Conversely, stable eigenvalues require those signs.

Since $a>0$, the Brusselator is homogeneously stable precisely forAt $b_H=1+a^2$, the linear eigenvalues are $\pm ia$, so strict decay is lost through a homogeneous Hopf bifurcation threshold. For larger $b$ the equilibrium is unstable. The source's matrix label $A$ is here written $J$ to distinguish it from the filament modulus in the preceding question.

We prove independence of eigenvectors for distinct eigenvalues by induction on the number of vectors. A single eigenvector is nonzero and hence independent. Suppose the first $d-1$ are independent, and let ${\sum }_{j=1}^d{c}_j{v}_j=0$. Applying $A-{\lambda }_{d}I$ givesBy the induction hypothesis every coefficient in this shorter relation vanishes. The eigenvalues are distinct, so $c_j=0$ for $j<d$. The original relation becomes $c_d{v}_d=0$, and $v_d\ne 0$ gives $c_d=0$. Thus the entire family of eigenvectors is linearly independent. The argument works over either the real or complex field.

Represent the quadratic form by the symmetric matrixThe $xy$ block has characteristic polynomial $(2-\mu )(5-\mu )-4=(\mu -1)(\mu -6)$. Corresponding orthonormal eigenvectors, together with the $z$ direction, areThe eigenvalues are $1,6,-1$. To eliminate the linear term, solve $H\tilde{O}=-(0,3\sqrt{5},0{)}^T$, obtainingIndeed the first equation gives $x=y$, and the second then gives $3y=-3\sqrt{5}$. Define the new coordinates by $X=\tilde{O}+\tilde{x}{\tilde{e}}_1+\tilde{y}{\tilde{e}}_2+\tilde{z}{\tilde{e}}_3$. The original polynomial takes the value $-15$ at the center, so its translated and orthogonally diagonalized form isThusThe surface is an elliptic one-sheet hyperboloid, centered at $\tilde{O}$, with axis parallel to ${\tilde{e}}_3$. Its waist at $\tilde{z}=0$ is an ellipse with semiaxes $\sqrt{15}$ and $\sqrt{5/2}$. Every constant-$\tilde{z}$ section is a nonempty ellipse, so the two signs of $\tilde{z}$ belong to the same connected sheet. This is an application of principal-axis reduction of a quadric.

By the given matrix similarity, $A^{\prime }=P^{-1}AP$ for an invertible $P$. HenceThe characteristic polynomials are identical. Therefore the characteristic equations and eigenvalues, including algebraic multiplicities, agree. The conclusion is independent of the choice between $\det (tI-A)$ and $\det (A-tI)$ conventions.

Let the repeated eigenvalue be $a$, and choose an eigenvector $v_1\ne 0$. Complete it to a basis $(v_1,v_2)$ of ${\mathbb{C}}^2$. In this basis the matrix is an upper triangular matrix, with first column $(a,0{)}^T$, say $T=\left(\begin{array}{cc}a& c\\ 0& d\end{array}\right)$. Its characteristic polynomial is $(t-a)(t-d)$. By part (a)(ii), this must equal $(t-a{)}^2$, so $d=a$.

If $c=0$, the matrix is already $aI$. If $c\ne 0$, rescale the first basis vector by $k=c$ and apply part (a)(iii); the upper-right entry becomes one. ThusThe second possibility is a size-two Jordan block. It has only one independent eigenvector, whereas $aI$ has a two-dimensional eigenspace, so the two types cannot be related by matrix similarity. This repeated-eigenvalue classification in dimension two proves the required result without assuming the general Jordan normal form theorem.

Use the orthonormal right-handed basisDirect multiplication gives $B(r)u=u$, $B(r)v=rv+w/\sqrt{2}$ and $B(r)w=-v/\sqrt{2}+rw$. Thus the matrix in this basis isThe last two columns are orthogonal and both have squared norm $r^2+1/2$. Therefore $B(r)$ is an orthogonal matrix precisely when $r^2+1/2=1$, giving the unique positive valueIts determinant is $r_0^2+1/2=1$, so it is a proper rotation matrix, not a reflection. The transverse block has cosine and sine both $1/\sqrt{2}$. ConsequentlyWith the oriented axis $u$, the angle is positive by the right-hand rule, since $u\times v=w$. The axis also follows from the unit-eigenvalue eigenspace. As a check, $\mathrm{tr}B(r_0)=1+\sqrt{2}=1+2\cos (\pi /4)$.

For a prime $p>2$, a diagonal matrix with two distinct nonzero eigenvalues exists. The same common-eigenspace argument makes the normalizer of a diagonal subgroup of GL2 precisely the monomial matrices. There are $(p-1{)}^2$ choices of nonzero entries for each of two permutation patterns, givingAt $p=2$, however, the invertible diagonal subgroup is trivial. Every invertible matrix normalizes it, soThe first factor chooses a nonzero first column and the second chooses a column outside its one-dimensional span. The exceptional answer is necessary because the printed question permits every prime.

Use the drift-removing substitution for a Dirichlet Sturm-Liouville problem, $y=e^{-x}v$. Direct differentiation gives $\mathcal{L}y=e^{-x}(-v^{\prime \prime }+v)$. The homogeneous boundary conditions become $v(0)=v(\pi )=0$, so the eigenvalues and eigenfunctions areThe self-adjoint weighted form isThus the weight and leading coefficient are both $e^{2x}$. The orthogonality of Sturm-Liouville eigenfunctions readsAfter multiplying the forcing equation by $e^x$, it becomes $-v^{\prime \prime }+v=x$. Expanding $x$ in the sine series of part (i) and dividing by $n^2+1$ givesEvery term satisfies the endpoint conditions. As a check, solving the transformed ODE directly gives $v=x-\pi \sinh x/\sinh \pi$, which has the same coefficients and forcing.

A matrix is in Jordan normal form if it is block diagonal with Jordan blocks $J_r(\lambda )=\lambda I+N$, where $N$ has ones immediately above the diagonal and zeros elsewhere. For a Jordan normal form matrix $J$, number its diagonal positions $1,\dots ,n$ and putChoose positive ${\epsilon }_m\to 0$ avoiding the finitely many values solving ${\lambda }_i+i\epsilon ={\lambda }_j+j\epsilon$ for $i\ne j$. The matrix remains upper triangular and now has distinct eigenvalues, so it is diagonalizable. Its entries converge to those of $J$. For arbitrary $A=SJ{S}^{-1}$, use $A_m=S{J}_m{S}^{-1}$. This proves density of diagonalizable complex matrices.

If $A$ is diagonalizable with eigenvalues ${\lambda }_1,{\lambda }_2$, conjugating $B$ by the same basis change conjugates $T_A$ to $T_{\mathrm{diag}({\lambda }_1,{\lambda }_2)}$. On the four matrix units $E_{ij}$, this acts by $({\lambda }_i+{\lambda }_j)E_{ij}$. Thus, writing $s=\mathrm{tr}A$ and $d=\det A$, its characteristic polynomial isThe entries of $T_A$ depend linearly on those of $A$, and the coefficients of its characteristic polynomial depend polynomially, hence continuously, on those entries. Apply the formula to the approximating $A_m$ and pass to the limit. The boxed formula therefore holds for every $A$, including non-diagonalizable matrices. This is the two-by-two anticommutator characteristic polynomial.

Applying the angular momentum operator directly givesFor an eigenvalue $\lambda$, comparison with $\lambda (f(x)+iy)z$ gives $\hslash x=\lambda f(x)$ and $\hslash f^{\prime }(x)=\lambda$. The first identity excludes $\lambda =0$; differentiating it and using the second gives ${\lambda }^2={\hslash }^2$. No additive constant in $f$ is permitted. The possibilities are $f(x)=x$, $\lambda =\hslash$, and $f(x)=-x$, $\lambda =-\hslash$.

For $f=x$, the polynomial $\psi =(x+iy)z$ is homogeneous of degree two and satisfies $\Delta \psi =0$. Expanding the three squared orbital angular momentum operators gives, with $D=x{\partial }_x+y{\partial }_y+z{\partial }_z$,Here $D\psi =2\psi$ and $D^2\psi =4\psi$. Therefore ${\hat{L}}^2\psi =6{\hslash }^2\psi$. This is the degree-two case of angular momentum of a homogeneous harmonic polynomial; its angular labels are $l=2$, $m=1$.

A linear map $\alpha$ is a nilpotent endomorphism if ${\alpha }^q=0$ for some positive integer $q$, and a semisimple endomorphism if it is diagonalizable over $\mathbb{C}$. Decompose $V$ into its generalized eigenspaces $V_{\lambda }=\ker (\alpha -\lambda I{)}^d$, where $d=\dim V$. Define ${\alpha }_s$ on $V_{\lambda }$ to be $\lambda I$, and set ${\alpha }_n=\alpha -{\alpha }_s$. Then ${\alpha }_s$ is diagonalizable, ${\alpha }_n$ is nilpotent, and both preserve these vector subspaces and commute. ThusFor uniqueness, suppose $\alpha =S+N$ with $S$ semisimple, $N$ nilpotent and $SN=NS$. Both commute with $\alpha$, so preserve each $V_{\lambda }$. On an eigenspace of $S$ of eigenvalue $\mu$ inside $V_{\lambda }$, the map $\alpha$ is $\mu I+N$ and has only the eigenvalue $\mu$. Since the same subspace lies in $V_{\lambda }$, $\mu =\lambda$. Hence $S=\lambda I$ on $V_{\lambda }$, proving $S={\alpha }_s$ and $N={\alpha }_n$. This is the additive Jordan–Chevalley decomposition.

We need a polynomial consequence of this decomposition. Hermite interpolation supplies a polynomial $p$ with $p(\alpha )={\alpha }_s$, by prescribing $p(t)\equiv \lambda (\mathrm{mod}(t-\lambda {)}^d)$ for each eigenvalue. For any endomorphism $T$, its semisimple part can likewise be expressed as a polynomial in $T$ with zero constant term: if zero is an eigenvalue, its interpolation condition already forces this; if not, add the independent condition $p(0)=0$.

On $\mathrm{End}(V)$, the maps $\mathrm{ad}{\alpha }_s$ and $\mathrm{ad}{\alpha }_n$ commute. The first is diagonalizable, with eigenvalues $\lambda -\mu$ on $\mathrm{Hom}(V_{\mu },V_{\lambda })$. The second is nilpotent, sincewhich vanishes for $k\ge 2q-1$ when $N^q=0$. Uniqueness therefore proves adjoint compatibility of additive Jordan decomposition: $\mathrm{ad}{\alpha }_s=(\mathrm{ad}\alpha {)}_s$.

Now assume $\alpha \in M$. The condition $[\alpha ,W]\subseteq U\subseteq W$ implies that both $W$ and $U$ are invariant under $\mathrm{ad}\alpha$, and every polynomial in $\mathrm{ad}\alpha$ with zero constant term maps $W$ into $U$. Define $\beta$ to be multiplication by $\overline{\lambda }$ on $V_{\lambda }$. It commutes with $\alpha$. On $\mathrm{Hom}(V_{\mu },V_{\lambda })$, $\mathrm{ad}\beta$ acts by $\overline{\lambda -\mu }$. Polynomial interpolation on the finite set of differences gives $\mathrm{ad}\beta =q(\mathrm{ad}{\alpha }_s)$ with $q(0)=0$. The preceding paragraph then expresses $\mathrm{ad}\beta$ as a polynomial in $\mathrm{ad}\alpha$ with zero constant term. Consequently $[\beta ,W]\subseteq U$, so $\beta \in M$.

The assumed trace orthogonality nilpotence lemma now follows directly. On $V_{\lambda }$, $\alpha =\lambda I+{\alpha }_n$ and $\beta =\overline{\lambda }I$, while the matrix trace of the nilpotent restriction of ${\alpha }_n$ is zero. HenceEvery summand is nonnegative, so every eigenvalue of $\alpha$ is zero. Its Jordan–Chevalley decomposition therefore has ${\alpha }_s=0$, and $\alpha$ is nilpotent. Notice that neither $U$ nor $W$ was required to be a Lie subalgebra.