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If $CD=DC$ and $Dx=\lambda x$, then $D(Cx)=C(Dx)=\lambda Cx$. Therefore each eigenspace of $D$ is invariant under $C$. The vector $Cx$ may be zero; invariance does not require it to be a nonzero eigenvector. This observation is the starting point for simultaneous diagonalization.

If $W$ is an invariant subspace for a finite-dimensional diagonalizable linear map $T$, its restriction is diagonalizable over the same field. The minimal polynomial of $T{|}_W$ divides that of $T$: an annihilating polynomial for $T$ still annihilates the restriction, and polynomial division by its minimal polynomial proves divisibility. A divisor of a product of distinct linear polynomials remains a product of distinct linear polynomials, proving the assertion. This is the key step in simultaneous diagonalization of commuting linear maps.

On each nonzero eigenspace $V_{\lambda }$, the restricted linear map $S_{\lambda }=S{|}_{V_{\lambda }}$ is diagonalizable. Indeed $m_S(S_{\lambda })=0$, so the minimal polynomial of $S_{\lambda }$ divides $m_S$. A divisor of a product of distinct linear polynomials has the same property, proving the diagonalizable restriction lemma.

Choose a basis of $S_{\lambda }$-eigenvectors in each $V_{\lambda }$. Every chosen vector is simultaneously a $T$-eigenvector, with eigenvalue $\lambda$. The direct sum in part (ii) shows that joining these bases gives a basis of $V$. Therefore $S$ and $T$ have a common eigenbasis, which is exactly simultaneous diagonalization.

Suppose a single basis gives the diagonal matrices $D_S,D_T$. Entrywise multiplication gives $D_S{D}_T=D_T{D}_S$. If $P$ is the change-of-basis matrix, thenThus simultaneous diagonalization implies commutation.

Yes, provided “commute” means pairwise commute. Necessity follows because three diagonal matrices commute pairwise.

For sufficiency, first choose the common eigenbasis for the first two linear maps, say $S,T$, obtained in part (b). Their joint eigenspacesgive a direct sum decomposition of $V$. The third map $U$ preserves each $W_{\lambda ,\mu }$, since $SUv=USv=\lambda Uv$ and $TUv=UTv=\mu Uv$. By the diagonalizable restriction lemma, $U{|}_{W_{\lambda ,\mu }}$ has a basis of eigenvectors there. Every such vector already has fixed eigenvalues for $S$ and $T$.

Joining these bases therefore gives a common eigenbasis for all three maps. The same induction proves simultaneous diagonalization for any finite pairwise commuting family of diagonalizable linear endomorphisms.

We use the supplied weight decomposition $V={⨁}_{\lambda \in \Psi (\mathfrak{a})}{V}_{\lambda }$. There is a small distinction to justify: the definition of a toral subalgebra uses the Adjoint representation, while these weight spaces use the defining representation. For the symplectic Lie algebra, adjoint semisimplicity in the symplectic Lie algebra implies that each $a\in \mathfrak{a}$ is diagonalizable on $V$ as well. Here is a direct verification.

Define the symplectic adjoint $a^{\ast }$ by $\omega (av,w)=\omega (v,a^{\ast }w)$. The symplectic Lie algebra condition is $a^{\ast }=-a$. Write the Jordan–Chevalley decomposition in $\mathrm{End}(V)$ as $a=S+N$, with $S$ diagonalizable, $N$ a nilpotent linear map, and $SN=NS$. Uniqueness of the Jordan–Chevalley decomposition, applied to $a^{\ast }=-a$, gives $S^{\ast }=-S$ and $N^{\ast }=-N$. Thus $S,N\in \mathfrak{sp}(V)$. On $\mathrm{End}(V)$, $\mathrm{ad}S$ is diagonalizable by using a matrix unit basis in an eigenbasis of $S$, and $\mathrm{ad}N$ is a nilpotent linear map, sincevanishes for large $m$. These operators commute and preserve $\mathfrak{sp}(V)$, so their restrictions are the Jordan–Chevalley decomposition of $\mathrm{ad}a$. Because $\mathrm{ad}a$ is diagonalizable, its nilpotent part is zero: $\mathrm{ad}N=0$ on $\mathfrak{sp}(V)$.

To conclude that $N=0$ without an assumption on the center of a Lie algebra, observe that each map $T_v(w)=\omega (v,w)v$ lies in $\mathfrak{sp}(V)$. If $v\ne 0$, its image is $kv$, because $\omega$ is a nondegenerate bilinear form. Since $N$ commutes with $T_v$, it preserves $kv$. Hence $Nv\in kv$ for every $v$, forcing $N$ to be a scalar linear map: compare its action on two linearly independent vectors and their sum. A scalar nilpotent linear map is zero. For $V=0$ the assertion is immediate. This proves the needed diagonalizability. Commuting diagonalizable linear maps admit simultaneous diagonalization, giving the displayed weight decomposition.

Now let $v\in V_{\lambda }$ and $w\in V_{\mu }$. For every $a\in \mathfrak{a}$, preservation of the alternating bilinear form givesIf $\lambda +\mu \ne 0$ as a linear functional, choose $a$ on which it is nonzero; then $\omega (v,w)=0$. ThereforeFor $v\ne 0$ in $V_{\lambda }$, the fact that $\omega$ is a nondegenerate bilinear form supplies a $w\in V$ with $\omega (v,w)\ne 0$. Decomposing $w$ into weight spaces, only its $V_{-\lambda }$ component can contribute. Thus $V_{-\lambda }\ne 0$, and $\Psi (\mathfrak{a})=-\Psi (\mathfrak{a})$. The same argument on both sides shows that $\omega :V_{\lambda }\times V_{-\lambda }\to k$ is a perfect pairing, so $\dim V_{\lambda }=\dim V_{-\lambda }$. On $V_0$, the restriction of $\omega$ is a nondegenerate bilinear form that is an alternating bilinear form, so $\dim V_0$ is even.

Choose one representative ${\lambda }_i$ from each pair of nonzero weights of a representation, for $1\le i\le r$. HereThe linear map $\mathfrak{a}\to k^r$ sending $a$ to $({\lambda }_1(a),\dots ,{\lambda }_r(a))$ is injective: an element in its kernel acts by zero on every nonzero weight space, and also on $V_0$, so it is the zero element of $\mathfrak{a}\subseteq \mathrm{End}(V)$. Taking dimensions yieldsThe final inequality follows because each pair contributes at least two to $\dim V=2n$. Equivalently, the nonzero weights of a representation span the dual space ${\mathfrak{a}}^{\ast }$; they need not be linearly independent at this stage.

The second gradient is ${\nabla }_{v}b=[A+B-(\alpha +\beta )I]v$. Combining its vanishing with $(A-\alpha I)v=0$ givesThe zero vector always works. A nonzero common stationary vector exists exactly when $A,B$ have a common eigenvector with respective eigenvalues $\alpha ,\beta$; both stationary values are then zero. Commuting symmetric matrices have a common orthonormal eigenbasis, so commutation is sufficient to find such vectors for the paired eigenvalues in that basis. It does not guarantee an arbitrary prescribed pair.

Commutation is also not necessary for one common eigenvector. For example,are real symmetric matrices sharing $e_1$ with eigenvalues $1,\beta$, but $AB\ne BA$. This separates the question's common-vector condition from the stronger simultaneous diagonalization of the entire space.

For positive frequency take $p^0=E=\sqrt{|\mathbf{p}{|}^2+m^2}>0$, with $m\ge 0$. Lowering the index in $p_{\mu }{\sigma }^{\mu }$ is important: putThese are commuting Hermitian matrices with $AB=(E^2-|\mathbf{p}{|}^2)I_2=m^2{I}_2$. Their eigenvalues are $E\pm |\mathbf{p}|$, so they are positive definite for $m>0$ and positive semidefinite matrices for $m=0$. Simultaneous diagonalization defines their positive square roots of a matrix and gives $\sqrt{A}\sqrt{B}=m{I}_2$.

The plane-wave ansatz reduces the Dirac equation to $(/p-m)u=0$, whereFor $u_L=\sqrt{A}\xi$ and $u_R=\sqrt{B}\xi$, commutativity gives $A{u}_R=m{u}_L$ and $B{u}_L=m{u}_R$. ThereforeThis is the Hermitian square-root construction of Dirac plane waves.

For negative frequency retain the future-directed momentum label $p$ and use ${\psi }_{-}(x)=v(\mathbf{p})e^{+ip\cdot x}$. The equation is now $(/p+m)v=0$. Its general solution iswith arbitrary two-spinor $\zeta$. The minus between its two blocks gives $A{v}_R=-m{v}_L$ and $B{v}_L=-m{v}_R$. For $m>0$ these parametrizations span the two-dimensional positive- and negative-frequency solution spaces. At nonzero null momentum the square roots have complementary rank-one supports and still give the full two-dimensional kernels. If ${\xi }^{†}\xi ={\zeta }^{†}\zeta =1$, both spinors have norm squared $2E$; the positive one also has $\bar{u}u=2m$.

Suppose $C=S{\Lambda }_C{S}^{-1}$ and $D=S{\Lambda }_D{S}^{-1}$ for an invertible $S$ and diagonal matrices ${\Lambda }_C,{\Lambda }_D$. Since diagonal matrices commute,Therefore simultaneous diagonalization implies commutation. This argument works over either $\mathbb{R}$ or $\mathbb{C}$ and explains why simultaneous diagonalization produces commuting operators.

Direct multiplication givesTake $D=B$ and $C=A$ in the preceding result: $B$ has the three distinct eigenvalues $2,4,6$ and the matrices commute. The common eigenbasis from part (i) givesIndeed $AS=S\mathrm{diag}(1,2,2)$ and $BS=S\mathrm{diag}(2,4,6)$, and $\det S=2\ne 0$. These equalities directly verify the simultaneous diagonalization.

The simultaneous diagonalization from part (a) applies to both $B$ and $C$. With ${\theta }_k=k\pi h$,andMultiplying the $j$th block equation by $Q^{-1}$ and writing $v_j=Q^{-1}{u}_j$ and $c_j=Q^{-1}{b}_j$ giveswith the absent boundary terms omitted. This is exactly system (3).

Bloch theorem states that the energy eigenstates of a Hamiltonian invariant under translations by a Bravais lattice can be chosen as Bloch statesIndeed, the unitary operators $T_r$ commute with one another because $T_r{T}_s=T_{r+s}=T_s{T}_r$, and they commute with $H$ by hypothesis. The simultaneous diagonalization theorem therefore lets us diagonalize all translations within each energy eigenspace. Their eigenvalues form a unitary character $\chi$ of the additive lattice:Writing $\chi (a_j)=e^{-ik\cdot a_j}$ on a primitive basis gives $\chi (r)=e^{-ik\cdot r}$. In position space, with $(T_r\psi )(x)=\psi (x-r)$, this implies ${\psi }_k(x+r)=e^{ik\cdot r}{\psi }_k(x)$. Hence $u_k(x)=e^{-ik\cdot x}{\psi }_k(x)$ is lattice-periodic. Adding a reciprocal lattice vector to $k$ leaves the character unchanged, so the crystal momentum lies in a Brillouin zone.

Forthe equations $a_i\cdot b_j=2\pi {\delta }_{ij}$ give the reciprocal basisThe reciprocal lattice is triangular, so the first Brillouin zone is its Wigner-Seitz cell, a regular hexagon. Its six corners areReciprocal-lattice translations identify these corners in two classes of three. Representatives areFor example, $(b_1-b_2)/3=K^{\prime }-b_2$, while the other equivalences follow by symmetry and reciprocal translations. This gives the requested sketch: a regular hexagon with the vertical edge from $K^{\prime }$ to $K$ at $k_x=2\pi /(\sqrt{3}a)$ and alternating $K,K^{\prime }$ corner classes.

For the tight-binding model, introduce the normalized Bloch sumEach hop by $\pm a_j$ multiplies this state by $e^{\mp ik\cdot a_j}$. Thus the two-direction nearest-neighbour tight-binding dispersion isAlong the boundary edge from $K^{\prime }$ to $K$, $k_x=2\pi /(\sqrt{3}a)$ and $|k_y|\le 2\pi /(3a)$, soFor $\lambda >0$, it rises from $E_0+2\lambda$ at either corner to $E_0+4\lambda$ at the midpoint; for $\lambda <0$ the ordering reverses. Globally the two cosines in the first expression can simultaneously equal $1$ or $-1$, and therefore

Each orbital band contains two one-electron states per lattice site because an electron has two spin states. By band filling, a valency of one leaves this band half-filled and the material conducts, whereas a valency of two fills it. A filled band can be insulating only if it is separated from every empty band by a positive band gap.

The first band's maximum is $E_0+4|\lambda |$. If the second band's minimum is $E_0+\Delta$, a gap exists precisely when $\Delta >4|\lambda |$. Consequently, among the nontrivial fillings described here,For valency one the first band is partially filled. If $\Delta \le 4|\lambda |$, the bands overlap or touch, so even at valency two the overlapping energy bands prevent a band insulator.

Use $[AB,C]=A[B,C]+[A,C]B$ and the canonical commutation relation. With $L_i={\epsilon }_{iab}{x}_a{p}_b$,Using these two transformation laws on $L_j={\epsilon }_{jab}{x}_a{p}_b$ givesThese are the orbital angular momentum commutation relations.

NowThe factor $[L_j,L_i]$ is antisymmetric in $j$ and its remaining product is symmetric after the two terms are combined, so the contraction vanishes:

The same commutators show that $p^2=p_j{p}_j$ and $r^2=x_j{x}_j$ are rotational scalars: their commutators with every $L_i$ vanish by contraction of the antisymmetric ${\epsilon }_{ijk}$ with a symmetric product. Therefore $U(r)$ also commutes with every $L_i$, and forwe have $[H,L_i]=0$, henceIn particular $H,L^2,L_3$ are pairwise commuting Hermitian operators. By simultaneous diagonalization, they admit a common eigenbasis, subject to the usual spectral-domain qualifications for unbounded operators. This is the rotational invariance of a central-potential Hamiltonian.

Eliminating the intermediate vector gives the amplification matrixThe commuting real symmetric matrices admit simultaneous diagonalization in the orthonormal basis $v^{(p,q)}$ from the question. On this vector,Since the basis is orthonormal, stability in the discrete Euclidean norm is equivalent to $|c_{pq}|\le 1$ for every $p,q$.

PutThen $-\beta \le {\lambda }_j\le -\alpha$. The smallest denominator is $1+\mu \alpha$, while the largest numerator modulus occurs at an endpoint of this interval. The condition involving $1-\mu \alpha$ is automatic, and the other isFor $m>1$, this is equivalent toSince $\beta -\alpha =4\cos (\pi h)$, the exact finite-grid stability condition isFor $m=1$ there is only one mode and the method is stable for every $\mu >0$. A simple grid-independent condition, and the limiting condition as $h\to 0$, isThis is the stability limit of one-implicit-direction diffusion splitting.

Schur lemma says that an intertwining linear map between irreducible finite-dimensional complex representations is either zero or an isomorphism. In particular, every endomorphism of an irreducible complex representation is a scalar multiple of the identity.

A continuous representation of a topological group is a continuous homomorphismfor a finite-dimensional complex vector space $V$. It is a unitary representation if $V$ has a positive-definite Hermitian inner product for whichfor every $g\in G$ and $v,w\in V$.

For $G=S^1$, start with any positive-definite Hermitian form and average it using normalized Haar measure:Translation invariance makes this form $S^1$-invariant, and positivity is preserved, proving unitarity. Since $S^1$ is abelian, the operators $\rho (z)$ commute; since they are unitary, they are normal. Simultaneous diagonalization therefore decomposes $V$ into common one-dimensional eigenspaces. Thus every representation of the circle group is a direct sum of one-dimensional representations.

WriteThe group law and inverse areA calculation givesEvery element of the centre $Z$ occurs by taking, for example, $x^{\prime }=y=0$ and $y^{\prime }=1$. Hence the commutator subgroup is exactly $Z$. The image of a one-dimensional representation is abelian, so the one-dimensional representation kills the commutator subgroup and its kernel contains $Z$.

Now let $(\rho ,V)$ be a complex representation of $G/Z_0$. Its restriction to the central subgroupis a representation of the circle group. Decompose it into its distinct weight spaces:Because $Z/Z_0$ is central, every $\rho (g)$ commutes with its action and preserves every common eigenspace. Thus the $V_i$ are $G/Z_0$-subrepresentations, as asserted by the central circle weight-space decomposition.

Let $n_i=\dim V_i$. The mapis a one-dimensional representation of $G/Z_0$; after composition with $G\to G/Z_0$, its kernel contains $Z$. On $z\in Z/Z_0$ its value isso ${\theta }_i^{n_i}=1$. A continuous character of $S^1$ has the form $z\mapsto z^m$; the displayed identity forces $m=0$. Therefore every ${\theta }_i$ is trivial. Since the ${\theta }_i$ were distinct, $d=1$ and ${\theta }_1=1$.

It follows that every finite-dimensional complex representation of $G/Z_0$ kills the entire nontrivial central circle $Z/Z_0$. Its kernel is therefore nontrivial, so no such representation is faithful. This is the real Heisenberg quotient has no faithful finite-dimensional representation.

Suppose $AB=BA$ and $A{v}_j=a_j{v}_j$. ThenBecause the eigenvalues of $A$ are distinct, the eigenspace for $a_j$ is one-dimensional. It follows that $B{v}_j=b_j{v}_j$, including the possibility $b_j=0$. Thus an orthonormal eigenbasis of $A$ is also an eigenbasis of $B$. The simplicity of both spectra means their eigenvectors coincide as directions, up to ordering, phases and nonzero scalar multiples.

Conversely, if they share a complete eigenbasis, choose the unitary matrix $U$ whose columns are these common normalized eigenvectors. Then $U^{†}AU=D_A$ and $U^{†}BU=D_B$ are diagonal, so $D_A{D}_B=D_B{D}_A$ and $AB=BA$. HenceThis is simultaneous diagonalization of the two Hermitian matrices.

Let $Av=\alpha v$. The commutation relation gives $A(Bv)=B(Av)=\alpha Bv$, so $B$ preserves the eigenspace of $A$ with eigenvalue $\alpha$. Because it is a nondegenerate eigenvalue, the eigenspace is one-dimensional, and $Bv=\beta v$ for some real $\beta$. This includes $Bv=0$. Thus the eigenvectors of $A$ form a common eigenbasis, andWriting the corresponding unit-modulus eigenvalue of $U$ as $\zeta =e^{i\theta }$ givesThe pairing refers to the same common eigenvector; independent reorderings of the two real spectra would lose this information. The simultaneous diagonalization can also be constructed without requiring simple eigenvalues by diagonalizing $B$ within each eigenspace of $A$.

With the two diagonal matrices from the preceding part, write $D_A=PA{P}^{-1}$ and $D_B=PB{P}^{-1}$. Their diagonal products agree entry by entry, so $D_A{D}_B=D_B{D}_A$. ThereforeHence simultaneous diagonalization implies that $A,B$ are commuting operators.

The $n$ independent eigenvectors $v_1,\dots ,v_n$ of $B$ from part (e) are also eigenvectors of $A$. Put $S=(v_1\cdots v_n)$ and $P=S^{-1}$. ThenThis supplies the required simultaneous diagonalization over the same field as in part (e).

For $n=2$, takeThey are commuting operators because every matrix commutes with the identity matrix. However $A$ has only the eigenvalue zero, and $\ker A=\mathrm{span}\{(1,0{)}^T\}$ has dimension one. It has no eigenbasis, so it cannot appear in any simultaneous diagonalization. Equivalently, a diagonal matrix similar to this nonzero nilpotent matrix would have to be zero, an impossibility. For any prescribed $n\ge 2$, use $A=E_{12}$ and $B=I_n$; its zero eigenspace has dimension $n-1$. For $n=1$ no such example exists.