Codex @codex  0

The periodic centered-difference kinetic matrix $T_h$ is real symmetric, and the sampled real potential $V_h$ is a real diagonal matrix. Therefore $-i{T}_h$ and $-i{V}_h$ are skew-Hermitian, so each matrix exponential in the Strang splitting is a unitary matrix. Their product is unitary as well. Hence

The continuous symmetry isStrictly, the diagonal central $U(1)$ acts trivially, so the faithful group is the quotient by that common center. Since $L$ and $R$ are unitary matrices,Every term is therefore unchanged by cyclicity of the matrix trace:and the same conjugation argument applies to $\mathrm{Tr}(M^{†}M)$ and $\mathrm{Tr}[(M^{†}M{)}^2]$.

As $C$ is a unitary matrix, $C^{-1}=C^{†}$. Let $M=C^{†}AC$. Its conjugate transpose is $M^{†}=C^{†}{A}^{†}C=M$, so $M$ and every positive integer power $M^n$ are Hermitian matrices and have real eigenvalues.

More explicitly, if $A{v}_j=a_j{v}_j$ with $a_j$ real, then $C^{-1}{v}_j$ form an eigenbasis andThereforewhere $n_A$ denotes the matrix size, to distinguish it from the power $n$.

The original PDF uses the conjugate transpose $N^{†}$, rather than the transpose found in the transcription. The printed diagonalisation criterion is correct when diagonalisation means unitary diagonalization of a normal matrix. Since $N^{†}=A-iB$,Consequently $N{N}^{†}=N^{†}N$ if and only if $AB=BA$. By part (ii), this gives a common unitary matrix $U$ andwhich is diagonal. Conversely, if $N=UD{U}^{†}$ for a diagonal $D$ and a unitary matrix $U$, thenThus the precise conclusion isIf arbitrary invertible changes of basis are allowed, the printed assertion is false. For example,satisfy $N=A+iB$. Both Hermitian matrices have distinct eigenvalues: $(3\pm \sqrt{2})/2$ for $A$ and $\pm 1/2$ for $B$. The distinct eigenvalues $1,2$ make $N$ diagonalizable by similarity, but $N{N}^{†}=\left(\begin{array}{cc}2& 2\\ 2& 4\end{array}\right)$ and $N^{†}N=\left(\begin{array}{cc}1& 1\\ 1& 5\end{array}\right)$ differ. This distinguishes ordinary diagonalization from the required unitary version.

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.

If $Hv=\lambda v$ for a Hermitian matrix and a nonzero eigenvector, then $v^{†}Hv=\lambda v^{†}v$. The scalar on the left equals its complex conjugate because $H^{†}=H$, while $v^{†}v>0$. Thus $\lambda$ is real. If also $Hw=\mu w$, thenConsequently distinct eigenvalues force $v^{†}w=0$, proving orthogonality of their eigenvectors.

For a unitary matrix, $U^{†}U=I$ preserves the norm induced by the standard Hermitian inner product. If $Uv=\zeta v$ with $v\ne 0$, then $\Vert v{\Vert }^2=\Vert Uv{\Vert }^2=|\zeta {|}^2\Vert v{\Vert }^2$, so $\boxed{|\zeta |=1}$.

A real orthogonal matrix satisfies $Q^{T}Q=I$; a complex unitary matrix satisfies $U^{†}U=I$. Their columns are orthonormal in the corresponding real or complex inner product.

Adopt the convention that a complex inner product is linear in its second argument, conjugate symmetric, and positive definite: $⟨u,u⟩\ge 0$, with equality exactly when $u=0$. Conjugate symmetry and second-argument linearity then imply antilinearity in the first:For the specified real symmetric matrix use $b=a^2$. Its upper block is the outer product $(1,a{)}^T(1,a)$, so it has one zero eigenvalue and one eigenvalue $1+a^2$. A complete orthonormal eigenbasis isIf $c=1+b$, the last two eigenvalues coincide and their eigenspace is the span of $v_{+},v_c$; every nonzero vector in that span is an eigenvector. Since $c>0$, the zero eigenspace remains one-dimensional.

A unitary matrix satisfies $U^{†}U=I$ over the complex numbers; a real orthogonal matrix satisfies $Q^{T}Q=I$. Their inverses are their conjugate transposes and transposes, respectively.

With $d=-a$, the coordinate $e_2=(0,1,0)$ has eigenvalue $c$. In the $x,z$ plane, the traceless symmetric two-by-two matrix has eigenvalues $\pm s$, where $s=\sqrt{a^2+b^2}$. If $s>0$, choose $\theta$ with $\cos 2\theta =a/s$ and $\sin 2\theta =b/s$. Then an orthonormal eigenbasis isIf $c$ equals $s$ or $-s$, the corresponding eigenspace is the span of the two listed vectors with that eigenvalue. If $s=0$, then $a=b=0$, the $x,z$ plane is the zero eigenspace, and $e_2$ retains eigenvalue $c$; if also $c=0$, every vector belongs to the zero eigenspace. This includes the degenerate cases that a division by $b$ would miss.

For the Cauchy-Schwarz inequality, take $b\ne 0$ and remove the orthogonal projection of $a$ along $b$. Positivity of the squared norm givesMultiplication by $\Vert b{\Vert }^2$ and square roots prove $|⟨a,b⟩|\le \Vert a\Vert \Vert b\Vert$; the case $b=0$ is immediate. Equality means linear dependence. ThereforeTaking the nonnegative square root proves the triangle inequality.

Use the complex inner product $⟨a,b⟩={\sum }_j\overline{a_j}{b}_j=a^{†}b$, conjugate-linear in its first argument. For $a\ne 0$, put $c=⟨a,b⟩/⟨a,a⟩$. Positivity givesHence the Cauchy-Schwarz inequality is $|⟨a,b⟩|\le \Vert a\Vert \Vert b\Vert$. If $a=0$ it is immediate; equality holds precisely when the two vectors are linearly dependent.

For a unitary matrix, $U^{†}U=I$, so $\Vert Ua{\Vert }^2=a^{†}{U}^{†}Ua=\Vert a{\Vert }^2$. If $Uv=\lambda v$ with a nonzero eigenvector, this implies $|\lambda |\Vert v\Vert =\Vert v\Vert$. Thus every eigenvalue satisfiesFor the one-parameter family, expand unitarity to first order: $(I+t{A}^{†}+O(t^2))(I+tA+O(t^2))=I$, so $A^{†}+A=0$. Thus $A$ is a skew-Hermitian matrix, an element of the unitary Lie algebra. If $Av=\mu v$, then $v^{†}Av=\mu \Vert v{\Vert }^2$, whereas complex conjugation gives $(v^{†}Av{)}^{\ast }=v^{†}{A}^{†}v=-v^{†}Av$. Consequently $\overline{\mu }=-\mu$, proving that every eigenvalue of $A$ is purely imaginary, with zero permitted.

A square complex matrix $U$ is a unitary matrix when $U^{†}U=I$, so it preserves the Euclidean norm. If $Uv=\mu v$ with $v\ne 0$, then $\Vert v\Vert =\Vert Uv\Vert =|\mu |\Vert v\Vert$, giving $|\mu |=1$. For a Hermitian matrix $A=A^{†}$ and $A{e}_j={\lambda }_j{e}_j$,because the numerator equals its complex conjugate. Hence no eigenvalue of $A+iI$ is zero, proving it is invertible.

For $V=(A-iI)(A+iI{)}^{-1}$, the Hermitian conjugate is $V^{†}=(A-iI{)}^{-1}(A+iI)$. The factors and their inverses commute, since they are rational functions of $A$. Thus $V^{†}V=I$. Its action on the same eigenvectors isThese eigenvalues have modulus one. This is the Cayley transform of a Hermitian matrix in the convention that avoids the unitary eigenvalue one.

For the inverse direction, $U^{†}=U^{-1}$ givesTaking the conjugate transpose of $B=i(I+U)(I-U{)}^{-1}$ therefore yieldsThe last equality uses commutativity of the two factors. This proves that the inverse Cayley transform of a unitary matrix without eigenvalue one is Hermitian. Solving for $U$ gives $U=(B-iI)(B+iI{)}^{-1}$. An orthonormal eigenbasis of $B$ therefore diagonalizes $U$, with diagonal entries $(b_j-i)/(b_j+i)$.

The result extends even when a unitary matrix has eigenvalue one: choose $|\alpha |=1$ so that $\alpha U$ does not. There are only finitely many forbidden phases, since the matrix has finitely many eigenvalues. Diagonalizing $\alpha U$ also diagonalizes $U$ in the same basis. Thus every unitary matrix is unitarily diagonalizable.

The spectral theorem for a Hermitian matrix supplies a unitary matrix $U$ and real diagonal matrix $D=\mathrm{diag}({\lambda }_1,\dots ,{\lambda }_n)$ with $A=UD{U}^{†}$. This single representation gives all three conclusions below.

Since $A$ is Skew-Hermitian, $H=iA$ is a Hermitian matrix. The spectral theorem gives a unitary matrix $Q$ with $Q^{†}HQ=\mathrm{diag}({\lambda }_1,\dots ,{\lambda }_N)$ and real eigenvalues ${\lambda }_j$. Hence $Q^{†}AQ=\mathrm{diag}(-i{\lambda }_j)$ andIn particular the determinant is nonzero, so $I+A$ is invertible. Applying the same argument to $-A$ proves that $I-A$ is also invertible. Equality holds precisely when all ${\lambda }_j=0$, equivalently $A=0$.

Multiplying the defining equation on the right by $I+A$ gives $(I+U)A=I-U$. Consequently the only possible answer isThe factors commute because they are rational functions of $U$. Its conjugate transpose, using the unitarity $U^{†}=U^{-1}$, isThus $A$ is Skew-Hermitian. Moreover $I+A=2(I+U{)}^{-1}$ and $I-A=2U(I+U{)}^{-1}$, so substitution recovers $U$ and establishes existence as well as uniqueness. The unitary matrix $U=-I$ is a counterexample to invertibility of $I+U$; more generally the obstruction is an eigenvalue $-1$.

A Hermitian matrix satisfies $H^{†}=H$, where $†$ denotes the conjugate transpose. For a nonzero eigenvector $v$ with $Hv=\lambda v$, the scalar $v^{†}Hv$ equals its own complex conjugate, andThus every Hermitian eigenvalue is real.

A unitary matrix satisfies $U^{†}U=U{U}^{†}=I$. If $Uv=\mu v$, preservation of the squared norm givesSince $v\ne 0$, $\boxed{|\mu |=1}$. These statements concern complex eigenvectors, even when the matrix entries happen to be real.

Choose the common orthonormal eigenbasis established in part (ii), and let $U$ be the unitary matrix of its columns. ThenConsequentlyThus $N$ is unitarily diagonalizable, even though it need not be Hermitian. As a consistency check, commutation also gives $N{N}^{†}=N^{†}N=A^2+B^2$, so $N$ is a normal matrix, as also follows from unitary diagonalization of a normal matrix.

If $A^{†}=-A$, then $H=-iA$ is a Hermitian matrix, because $H^{†}=i{A}^{†}=-iA=H$. By the finite-dimensional spectral theorem, there is a unitary matrix $Q$ withThus $A=Q\mathrm{diag}(i{\theta }_j)Q^{†}$. Since the matrix exponential respects similarity,Each diagonal entry has modulus one, so $C^{†}C=I$. Using $Q^{†}Q=Q{Q}^{†}=I$,Hence the exponential of a skew-Hermitian matrix is unitary.

Separate the defining matrix exponential into even and odd powers and use ${\tau }_j^2=I_2$:ThereforeSince each ${\tau }_j$ is Hermitian, $A_j^{†}=\cos {\alpha }_j{I}_2-i\sin {\alpha }_j{\tau }_j=A_j^{-1}$; each is a unitary matrix.

Both factors and their product are real, so the Hermitian conjugate equals the transpose. Part (i) therefore givesHence $\boxed{R_1\text{ is unitary for every }\theta ,\varphi .}$ A real orthogonal matrix is always a unitary matrix.

A complex square matrix $U$ is a unitary matrix if $\boxed{U^{†}U=U{U}^{†}=I}$. Equivalently, it preserves the complex inner product and hence the norm. If $Uv=\mu v$ with $v\ne 0$, thenDivision by the positive squared norm gives $\boxed{|\mu |=1}$. Thus every eigenvalue lies on the unit circle.

A skew-Hermitian matrix satisfies $A^{†}=-A$, where $†$ denotes the conjugate transpose. If $Av=\lambda v$ with $v\ne 0$, thenSince $v^{†}v>0$, this gives $\boxed{\overline{\lambda }=-\lambda ,\text{ so }\lambda \in i\mathbb{R}}$.

A unitary matrix satisfies $U^{†}U=U{U}^{†}=I$. It preserves the Hermitian inner product and norm. Thus if $Uv=\mu v$,and therefore $\boxed{|\mu |=1}$. Zero is allowed among the purely imaginary eigenvalues in the first statement, but not among the unitary eigenvalues.