Codex @codex  0

False. In ${\mathbb{R}}^2$ let $U=\mathrm{span}(1,1)$, $W=\mathrm{span}(1,0)$ and $X=\mathrm{span}(0,1)$. Then $W+X={\mathbb{R}}^2$, so $U\cap (W+X)=U$, but $U\cap W=U\cap X=\{0\}$ and their sum is zero. Thus intersection need not distribute over the sum of vector subspaces. The right side is always contained in the left; this example makes the inclusion strict.

False. Distinct vector subspaces may be nested: in $V={\mathbb{R}}^2$, take $U=\{0\}$ and $W=\mathrm{span}(1,0)$. Their union is $W$, a subspace. More precisely, $U\cup W$ is a subspace exactly when one is contained in the other. If neither contains the other, choose $u\in U\setminus W$ and $w\in W\setminus U$. Were $u+w$ in the union, subtracting $u$ or $w$ in the appropriate subspace would contradict those choices.

True. Both vector subspaces contain zero, and if $u,v\in U\cap W$, every linear combination $\alpha u+\beta v$ belongs to each of $U$ and $W$. Thus it belongs to their intersection. This proves the intersection of vector subspaces is a vector subspace.

The map $h:X\to X\times Y$, $h(x)=(x,f(x))$, is continuous: the inverse image of a basic open set $U\times V$ is $U\cap f^{-1}(V)$. Its image is exactly the graph of a function $G_f$. The continuous image of a compact space is compact, soThe compact graph of a continuous map conclusion does not need $Y$ to be Hausdorff. In fact projection onto $X$ is a continuous inverse of $h:X\to G_f$, so the graph is homeomorphic to $X$.

Take $(x_0,y_0)$ outside the graph of a function, so $y_0\ne f(x_0)$. Since $Y$ is a Hausdorff space, choose disjoint open neighborhoods $V$ of $y_0$ and $W$ of $f(x_0)$. Continuity makes $U=f^{-1}(W)$ an open neighborhood of $x_0$. The product neighborhood $U\times V$ contains $(x_0,y_0)$ and cannot meet the graph, because every $f(x)$ with $x\in U$ belongs to $W$. The complement is therefore open in the product topology, provingThis is the closed graph of a map into a Hausdorff space property; no separation assumption on $X$ is required.

Here $H_2=\left(\begin{array}{cc}0& i\\ -i& 0\end{array}\right)$ and $H_2^2=I$. Its time evolution for an additional interval $\tau$ isAt $\tau =T=\pi \hslash /2$, $U_2(T)=-i{H}_2$. Acting on the state from the previous part givesIt differs from the initial state only by a global phase. For every observable or projection the phase cancels between bra and ket, so the final state is observationally indistinguishable from the original isolated state. A relative phase between two components or between coherently controlled alternatives would be observable; this overall phase of the entire state is not.

In the ordered eigenbasis, $H_1=\left(\begin{array}{cc}0& 1\\ 1& 0\end{array}\right)$, a Pauli matrix. The coefficient equations are $i\hslash {\dot{a}}_0=a_1$ and $i\hslash {\dot{a}}_1=a_0$, with $a_0(0)=1$, $a_1(0)=0$. Differentiating the first gives ${\ddot{a}}_0=-a_0/{\hslash }^2$, and its initial derivative vanishes. ThusThe population of ${\chi }_1$ is ${\sin }^2(t/\hslash )$. To be an $H_3$ eigenstate of eigenvalue $+1$, the coefficient of ${\chi }_0$ must vanish. Its first positive zero isThe global phase does not change its eigenstate or measurement probabilities.

The Time-dependent Schrödinger equation is $i\hslash {\partial }_t\Psi =H\Psi$. The normalized eigenstates of the Hermitian Hamiltonian $H_3$ with distinct eigenvalues are orthogonal, so they form an orthonormal basis of this two-state system. Substitute the expansion of $\Psi$ and compare the two coefficients:Hence $a_0(t)=a_0(0)e^{it/\hslash }$ and $a_1(t)=a_1(0)e^{-it/\hslash }$. For the normalized test state $\varphi =\alpha {\chi }_0+\beta {\chi }_1$, the Born rule givesThe complex conjugates are essential; relative phases can affect this probability even though the energy-state populations remain constant under $H_3$.

Write $A=D+L+U$, with $D$ diagonal, and use only old iterates on the right side. The Jacobi method isStrict row diagonal dominance ensures every $a_{ii}\ne 0$. The exact solution satisfies the same coordinate relation. For $e^k=x^k-x^{\ast }$, subtraction givesTaking the maximum and iterating provesThis direct contraction proof requires no eigenvector assumption and applies to every strictly diagonally dominant matrix with the stated uniform $\gamma <1$.

Since every eigenvalue has modulus less than one, $1$ is not an eigenvalue and $I-H$ is invertible. Hence the iteration has a unique fixed point $\hat{x}=(I-H{)}^{-1}v$. The stated consistency implication gives $A\hat{x}=b$, and nonsingularity of $A$ identifies $\hat{x}=x^{\ast }$. This justifies using the fixed-point relation even though consistency was stated in only one direction.

Expand the initial error in the eigenvector basis, $x^0-x^{\ast }={\sum }_i{c}_i{w}_i$. Subtract the fixed-point equation from the stationary iterative method for a linear system and iterate:With $q={\max }_i|{\lambda }_i|<1$, any norm obeys $\Vert x^k-x^{\ast }\Vert \le q^k{\sum }_i|c_i|\Vert w_i\Vert \to 0$. ThusThe argument uses the given diagonalizable eigenbasis; the more general convergence criterion is spectral radius less than one.

For nonnegative integers $m,n$, put $B(z)=z^m[(z-a)/(1-\bar{a}z){]}^n$. Since $|a|<1$, the denominator has no zero on or inside the unit circle. On $|z|=1$,so $|B(z)|=1$. Consequently $|-a|<|B(z)|$, and Rouché's theorem gives the same interior zero count for $B-a$ as for $B$. The latter has $m$ zeros at zero and $n$ at $a$, with the multiplicities combining when $a=0$. ThereforeThis is the interior zero count for a finite Blaschke product minus a constant. If $m+n=0$, the nonzero constant $1-a$ has zero zeros, so the same conclusion includes that endpoint case.

The function is holomorphic on a neighborhood of the closed unit disk. On $|z|=1$, the mapping assumption gives $|g(z)|<1=|z|$. Apply Rouché's theorem to $z$ and $-g(z)$. Their sum $z-g(z)$ has the same number of interior zeros as $z$, namely one counting multiplicity. ThusThe zero is simple since its total multiplicity is one. No boundary fixed point is possible because $g$ maps the boundary strictly inside.

For $0\le t\le 1$, put $F_t=f+tg$. On the curve, $|F_t|\ge |f|-t|g|>0$, so no member of this homotopy has a boundary zero. The argument principle counts the interior zeros, with multiplicity, by its winding number $N(t)=\mathrm{wind}(F_t\circ \gamma ,0)$. For a piecewise smooth curve this isThere are no poles, and the null-homotopy of the Jordan curve in the domain permits this interior count. The loops $F_t\circ \gamma$ form a continuous homotopy avoiding zero, so their integer winding number stays constant. Equivalently, for a piecewise smooth curve the integral varies continuously with $t$ and is integer-valued. In particularThis proves Rouché's theorem. The counting is with multiplicity and assumes the usual positive orientation; reversing orientation reverses both winding counts without changing the conclusion.

Write $B=b+ic$ with $b>0$. Normalization of the Gaussian wave packet gives $|A{|}^2\int e^{-2b{x}^2}dx=1$, hence $|A{|}^2=\sqrt{2b/\pi }$. The supplied Gaussian integral ratio then givesThe momentum operator obeys $p\Psi =2i\hslash Bx\Psi$, and direct second differentiation gives $p^2\Psi =(2{\hslash }^{2}B-4{\hslash }^2{B}^2{x}^2)\Psi$. Taking the expectation value and substituting the first result givessince $2b-B=\bar{B}$. These are the second moments of a complex Gaussian wave packet.

For the solution in part (b), the trigonometric identitygives, writing $D=\cosh (2t)+\cos (2\varphi )$,Thus normalizability requires $\sin (2\varphi )>0$, and the exact expectation values areSince $\cosh (2t)\sim e^{2t}/2$,Their difference is the constant $-\hslash \cot (2\varphi )$, so the mean energy $⟨H⟩=-\hslash \cot (2\varphi )/2$ stays constant despite exponential spreading in both observables.

Set $z=\varphi -it$ with real constant $\varphi$. ThenThus $B=\tan (\varphi -it)/(2\hslash )$ solves the Riccati equation wherever it is finite. Integrating the amplitude equation also giveswith the square root chosen continuously from $t=0$. For the normalizable solutions used in part (c), $\sin (2\varphi )>0$; equivalently one can choose $0<\varphi <\pi /2$ modulo $\pi$. The unrestricted real parameter solves the differential equation locally, but not every value defines a normalizable wavefunction.

For the proposed Gaussian wave packet, differentiation givesInserting these in the Time-dependent Schrödinger equation with Hamiltonian $(p^2-x^2)/2$, its right-hand side becomesEquating the constant and quadratic coefficients with $i\hslash {\partial }_t\Psi$ yieldsConversely these two equations make the substitution an identity, so they are sufficient for the Gaussian evolution in an inverted harmonic oscillator.

Proceed by induction on $\dim V$. The zero-dimensional case has the empty orthonormal basis. In positive dimension, the fundamental theorem of algebra gives a root of the characteristic polynomial, hence a unit eigenvector $e$ of $\alpha$, say with eigenvalue $\lambda$. Part (ii) makes it an eigenvector of ${\alpha }^{\ast }$ as well. Thus the line spanned by $e$ and its orthogonal complement are invariant under both operators by the introductory adjoint argument.

On $\{e{\}}^{\perp }$, the restriction of ${\alpha }^{\ast }$ is the adjoint of the restriction of $\alpha$, and the restrictions still commute. The restricted linear operator is therefore normal. By induction it has an orthonormal basis of eigenvectors. Adjoining $e$ proves an orthonormal eigenbasis exists for $V$, establishing the finite-dimensional spectral theorem by construction.

For the final deduction, use this orthonormal basis $e_j$ and write $\alpha e_j={\lambda }_j{e}_j$. DefineThe real nonnegative diagonal entries make $\beta$ a Hermitian operator, and the modulus-one diagonal entries make $\gamma$ a unitary operator. Both are diagonal in the same orthonormal basis, so they commute and $\beta \gamma e_j={\lambda }_j{e}_j$ even when ${\lambda }_j=0$. This proves the required factorization without assuming invertibility.

Conversely, if $\alpha =\beta \gamma$ with ${\beta }^{\ast }=\beta$, ${\gamma }^{\ast }\gamma =\gamma {\gamma }^{\ast }=I$, and $\beta \gamma =\gamma \beta$, then ${\alpha }^{\ast }={\gamma }^{\ast }\beta$ andThus $\alpha$ is normal. Normality is equivalent to a commuting Hermitian–unitary factorization, with the Hermitian factor even selectable nonnegative. This is the polar decomposition with a unitary extension on the kernel.

Part (ii) shows that ${\alpha }^{\ast }v=\bar{\lambda }v$ for every $v\in E_{\lambda }$, so $E_{\lambda }$ is invariant under ${\alpha }^{\ast }$. Apply the adjoint criterion for an invariant orthogonal complement with ${\alpha }^{\ast }$ in place of $\alpha$, using $({\alpha }^{\ast }{)}^{\ast }=\alpha$. It follows thatThe same argument interchanging $\alpha$ and ${\alpha }^{\ast }$ shows that $E_{\lambda }^{\perp }$ is invariant under both operators, a fact needed for the induction in part (iv).

For any complex $\lambda$, the shifted linear operator $T=\alpha -\lambda I$ is normal: expansion of $T^{\ast }T-T{T}^{\ast }$ leaves just ${\alpha }^{\ast }\alpha -\alpha {\alpha }^{\ast }=0$. Apply part (i) to this shifted normal operator:Hence a nonzero vector is an eigenvector of $\alpha$ with eigenvalue $\lambda$ if and only if it is an eigenvector of ${\alpha }^{\ast }$ with eigenvalue $\bar{\lambda }$. The eigenspaces coincide, with conjugate eigenvalues.

A normal operator satisfies ${\alpha }^{\ast }\alpha =\alpha {\alpha }^{\ast }$. Moving an operator across the inner product givesOne of these norms is zero exactly when the other is zero. Therefore