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The closest point theorem in a Hilbert space states that every nonempty closed convex subset $C$ of a Hilbert space has a unique point nearest to any given $x$.

Let $d={\inf }_{y\in C}\Vert x-y\Vert$ and choose $y_n\in C$ with $\Vert x-y_n\Vert \to d$. The parallelogram identity and convexity giveso $(y_n)$ is Cauchy. Completeness and closedness give a limit $y\in C$ with $\Vert x-y\Vert =d$. If $y,z$ were both minimizers, the same identity with their midpoint would force $\Vert y-z\Vert =0$.

Apply the theorem to a closed subspace $F$. For $x\in H$, let $y\in F$ be closest and put $z=x-y$. For every $u\in F$, minimality of $\Vert z-tu{\Vert }^2$ at $t=0$ for real and imaginary $t$ gives $⟨z,u⟩=0$. HenceThe intersection is zero, so $H=F\oplus F^{\perp }$.

If $T$ is a shift with orthonormal basis $(e_n)$, it is an isometry,Conversely, suppose these three properties hold. Choose a unit vector $e_1$ spanning $(\mathrm{Im}T{)}^{\perp }$ and set $e_n=T^{n-1}{e}_1$. Isometry makes this sequence orthonormal. Iterating the orthogonal decompositiongivesA vector orthogonal to every $e_n$ lies in every $\mathrm{Im}(T^n)$ and is therefore zero. Thus $(e_n)$ is an orthonormal basis and $T{e}_n=e_{n+1}$, so $T$ is a shift. This is the wandering-vector characterization of a unilateral shift.

Write $h_x={\partial }_{x}h$ and $h_y={\partial }_{y}h$. Sincethe first fundamental form in the coordinate basis is

Now $g=1$, so $c=⟨\psi |\varphi ⟩$ is real andThe perturbation is diagonal in this basis:Hence the instantaneous eigenvalues are $1+2V(t)$ and $-1$ with time-independent eigenvectors, so Hamiltonians at different times commute. In units $\hslash =1$, putExpanding the initial state and evolving its two components givesTaking its overlap with $|\varphi ⟩$ and simplifying gives the exact transition probabilityWith explicit $\hslash$, each phase in the final sine is divided by $\hslash$.

The operator $F$ maps the computational orthonormal basis $\{|j⟩\}$ to the orthonormal basis $\{|{\varphi }_j⟩\}$. It therefore preserves inner products and is surjective, so $F^{†}F=F{F}^{†}=I$ and $F$ is unitary.

The first register isFor $j,j^{\prime }$,The geometric sum is one when $j=j^{\prime }$ and zero otherwise. There are $N$ such orthonormal vectors in an $N$-dimensional space, so they form an orthonormal basis.

First apply $F^{†}$, which sends $|{\varphi }_j⟩$ to $|j⟩$. Then measure in the computational basis. Because the phase has an exact $m$-bit representation, exact quantum phase estimation returns

By part (a), each $A{\phi }_{\pm }+{\partial }_t{\phi }_{\pm }$ lies in the same two-dimensional eigenspace. Reality and conjugacy again force the coefficient matrix to have the formBecause $u$ and hence $A$ are periodic, ${\partial }_t+A$ commutes with translation by $2\pi$. Applying it to
$\Psi (x+2\pi )=\hat{T}\Psi (x)$ givesThe basis $\Psi$ is independent, so the periodic KdV transfer matrix obeysTaking traces,Hence $\boxed{\mathrm{Re}a}$ is independent of time.

The coupled basis consists of the tripletand the singletThese equations list all nonzero Clebsch--Gordan coefficients, each coefficient being $1$, $1/\sqrt{2}$, or $-1/\sqrt{2}$ as displayed, andBecause each irreducible total-spin representation occurs once, an operator commuting with total angular momentum isIn the product basis, the nonzero matrix elements areand

For real $E$ and real $V$, choose a real fundamental pair of solutions. The corresponding Floquet matrix has real entries, so $\mathrm{tr}F(E)$ is real. The trace is basis independent, so this holds for every fundamental pair.

The Reed-Muller code is spanned by the evaluation vectors of the square-free monomialsform a basis: every Boolean function has a unique algebraic normal form, so the monomials remain independent after evaluation on ${\mathbb{F}}_2^d$. ThusInduction using the decomposition $(u,u+v)$, with $u\in \mathrm{RM}(d-1,r)$ and $v\in \mathrm{RM}(d-1,r-1)$, gives the lower bound $2^{d-r}$ for every nonzero weight. The monomial $x_1\cdots x_r$ attains it, so the minimum distance is

The matrix $U_{\theta }$ is real orthogonal. Part (a) givesExpanding the left side in the computational basis yields

For a nonzero ideal $I\subset O_K$, its norm isFor $d=[K:\mathbb{Q}]$ and embeddings ${\sigma }_1,\dots ,{\sigma }_d$, defineThe field discriminant is the discriminant of any integral basis of $O_K$. If ${\alpha }_1,\dots ,{\alpha }_d$ is a $\mathbb{Z}$-basis of $I$, then

Multiplication by $\alpha$ on an integral basis has integer matrix $M_{\alpha }$. Its image lattice is $\alpha O_K$, soThe determinant of this $\mathbb{Q}$-linear map is the field norm $N_{K/\mathbb{Q}}(\alpha )$, proving

A complex matrix is diagonalisable when it is similar to a diagonal matrix, equivalently when the vector space has a basis of its eigenvectors.

The Wronskian is $W=y_1{y}_2^{\prime }-y_2{y}_1^{\prime }$. Abel's identity givesIf $W\ne 0$, the two column vectors in the displayed representation form a basis of ${\mathbb{R}}^2$ at each $x$. Differentiating the first component and comparing it with the second givesSubstitution in the differential equation givesSolving this two-by-two system yields

The product topology has basis $U\times V$ with $U$ open in $X$ and $V$ open in $Y$. Since ${\Pi }_X^{-1}(U)=U\times Y$ and similarly for ${\Pi }_Y$, both projections are continuous.

If $Y$ is Hausdorff, a point $(y_1,y_2)$ off the diagonal has disjoint product neighborhoods, so the diagonal is closed. Conversely, if the diagonal is closed, its complement supplies disjoint neighborhoods of any two distinct points.

The graph is the inverse image of the diagonal under $(x,y)\mapsto (f(x),y)$, so a continuous map into a Hausdorff space has closed graph. Conversely, if $X,Y$ are compact Hausdorff and $C\subseteq Y$ is closed, then ${\Gamma }_f\cap (X\times C)$ is compact. Its projection onto $X$ is $f^{-1}(C)$, compact and hence closed. Thus $f$ is continuous.

The function $f(0)=0$, $f(x)=1/x$ for $x\ne 0$ is discontinuous but has closed graph: a convergent graph sequence approaching $x=0$ cannot have $x\ne 0$ and bounded second coordinate unless it eventually reaches the isolated graph point $(0,0)$.

A Jordan block is $J_m(\alpha )=\alpha I+N$, where $N$ has ones on the superdiagonal and zeros elsewhere. Since $N^2=0$ for $m=2$, Taylor expansion givesThe Jordan normal form is the block-diagonal matrix, unique up to block order, to which $A$ is similar and whose blocks are Jordan blocks.

A block of size at least two for eigenvalue $\lambda$ contributes a generalized eigenvector in $\ker (A-\lambda I{)}^2\setminus \ker (A-\lambda I)$. Thus the displayed equality for every $\lambda$ is equivalent to every block having size one, which is diagonalizability.

The transpose of each Jordan block is similar to it by reversing its basis. Hence $A^T$ has the same Jordan form as $A$.

Write $A=PJ{P}^{-1}$. Let $R$ be block diagonal with the reversal matrix on each Jordan block. Both $R$ and $RJ$ are symmetric and $R$ is invertible. Thereforewhere $C,D$ are symmetric and $C$ is nonsingular.