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Every commutator $[A,B]=AB-BA$ has zero trace, by the cyclic identity. Hence its linear span $W$ is contained in $\ker \mathrm{Tr}$. For the reverse inclusion, the matrix units satisfyThe off-diagonal matrix units and the differences $E_{ii}-E_{nn}$ for $i<n$ span all traceless matrices: a traceless diagonal part is ${\sum }_{i<n}{a}_{ii}(E_{ii}-E_{nn})$. Thus every element of the kernel is a linear combination of commutators, provingThis establishes that traceless matrices are spanned by commutators. For $n=1$, both vector subspaces are zero. Notice that the argument needs a span of commutators; it does not assume that every such sum has already been written as a single commutator.

To prove that trace is the unique normalized cyclic linear functional, let $E_{ij}$ be the matrix units. For $i\ne j$, $E_{ii}{E}_{ij}=E_{ij}$ whereas $E_{ij}{E}_{ii}=0$, so the assumed identity for the linear functional gives $f(E_{ij})=0$. Also, $E_{ij}{E}_{ji}=E_{ii}$ and $E_{ji}{E}_{ij}=E_{jj}$, giving $f(E_{ii})=f(E_{jj})$. Write this common value as $c$. Since the matrix units form a basis,For $n=1$ the same conclusion follows immediately from the one-dimensional basis $E_{11}$. Finally, $f(I)=nc$; if this equals $n$, then $c=1$ over either specified field. The normalization makes $f$ exactly the trace.

The definition of matrix multiplication givesThe middle equality uses commutativity of scalar multiplication in $\mathbb{R}$ or $\mathbb{C}$, together with interchange of finite sums. Thus the trace is unchanged by cyclically moving the two factors.

For matrices $A,B$ and scalars $r,s\in \mathbb{F}$, the definition of the trace givesHence the trace is a linear functional on the vector space of $n\times n$ matrices.

Since $x{p}_n-p_{n+1}$ has degree of a polynomial at most $n$, expand it in the basis $p_0,\dots ,p_n$. Multiplication by $x$ is self-adjoint for this weighted inner product: $⟨xp,q⟩=⟨p,xq⟩$. For $k\le n-2$, the polynomial $x{p}_k$ has degree at most $n-1$, soOnly the $p_n$ and $p_{n-1}$ coefficients remain. Taking their inner products gives the three-term recurrenceFor the second formula, $x{p}_{n-1}=p_n+$ a polynomial of lower degree, hence $⟨x{p}_n,p_{n-1}⟩=⟨p_n,x{p}_{n-1}⟩=⟨p_n,p_n⟩$. At $n=0$, take $p_{-1}=0$, ${\beta }_0=0$ and the same formula for ${\alpha }_0$. Positivity of the weight ensures that all the squared norms in the denominators are nonzero.

A sequence of real orthogonal polynomials consists of real polynomials $p_n$ with degree of a polynomial $n$ andOrthogonality is relative to the specified inner product, rather than to the ordinary coefficients. A monic polynomial has leading coefficient one, so a monic orthogonal polynomial has the form $p_n(x)=x^n+\text{terms of lower degree}$. In particular, the monic degree-zero polynomial is $p_0=1$.

In cylindrical coordinates, the transverse velocity is $A{\mathbf{e}}_{\theta }/r$, so its contribution to the circulation is $Ad\theta$. The vertical contribution is $U(z)dz$, the differential of an antiderivative of $U$, and therefore has zero integral around any closed curve. If $k$ is the oriented winding number of the curve's projection about the origin, thenIn particular, a single positive winding gives $2\pi A$; reversing orientation reverses the sign. The nonzero circulation is compatible with irrotational flow because the domain is not simply connected. For $A\ne 0$, an enclosing curve cannot bound a surface contained entirely in the domain to which Stokes theorem would apply.

Put $r^2=x^2+y^2$. On the specified domain, $r>0$, and direct differentiation gives the divergence and vorticityFor example, the third component of the curl is $A(r^2-2x^2)/r^4+A(r^2-2y^2)/r^4=0$; the other two components vanish because $U$ depends only on $z$ and the transverse velocity is independent of $z$. Hence the flow is both incompressible and irrotational exactly whenThe exclusion of the axis matters: these are ordinary derivatives on the domain, with no assertion of regularity on the removed axis.

Completing the square in the optimization constraint givesThe left side is nonnegative, so every feasible point has $x_1^2\ge 4$. Equality is feasible precisely when $x_2=3x_1/2$ and $x_1=\pm 2$. Thus the global minimum isThere is no maximum: for every $|x_1|\ge 2$, the values $x_2=3x_1/2\pm \sqrt{x_1^2/4-1}$ satisfy the optimization constraint, allowing $x_1^2$ to be arbitrarily large. This direct argument also establishes that the two stationary points are the only local minima on the two branches of the constraint curve.

For a twice differentiable function, convexity on ${\mathbb{R}}^2$ is equivalent to its Hessian matrix being positive semidefinite everywhere. Here the Hessian matrix is constant:Its diagonal entries are positive, so positive semidefiniteness is equivalent to $\det H=8-{\alpha }^2\ge 0$. The required parameter range isThe endpoints give convex functions with one flat direction; strict inequalities give a Hessian matrix defining a positive-definite quadratic form and a strictly convex function.

Use an isometry of the Poincare disc model to move the centre to zero. A hyperbolic disc of radius $\rho$ then has Euclidean radius $R=\tanh (\rho /2)$ by the hyperbolic distance formula. The area element of a surface of the Riemannian metric is $4rdrd\theta /(1-r^2{)}^2$, soTherefore the area is independent of the centre, and

The radial segment is a geodesic of the Poincare disc model. Its hyperbolic length isOne can also verify that it minimizes length of a curve: along any piecewise smooth curve from zero to radius $r$, the Euclidean speed is at least the absolute radial speed. Integrating $2|dr|/(1-r^2)$ is at least the absolute change in $\log ((1+r)/(1-r))$, with equality on the radial segment. Thus the answer is independent of $\theta$.

A suitable Poincare disc automorphism is the Möbius transformationIt sends $a$ to zero and has inverse $(w+a)/(1+\overline{a}w)$. To verify the isometry directly, note thatThese identities show both that $f$ maps the unit disc onto itself and that $2|df|/(1-|f{|}^2)=2|dz|/(1-|z{|}^2)$, preserving the Riemannian metric.

The Poincare disc model is the open unit disc $D=\{z=x+iy:|z|<1\}$ equipped with the Riemannian metricThis normalization gives Gaussian curvature $-1$. In particular, a Euclidean curve has hyperbolic length $\int 2|dz|/(1-|z{|}^2)$; the boundary circle is infinitely far away in hyperbolic distance.

Let $u_n={\mathbb{P}}_0(X_n=0)$ and $f_n={\mathbb{P}}_0(T=n)$. Decomposing at the first return and using the Markov property gives the renewal equationFor $0<s<1$, define the probability generating functions $U(s)={\sum }_{n\ge 0}{u}_n{s}^n$ and $F(s)={\sum }_{n\ge 1}{f}_n{s}^n={\mathbb{E}}_0[s^T]$. Absolute convergence, or nonnegative summation, justifies the convolution identity $U=1+FU$. The binomial series givesSolving the renewal identity gives the first-return generating function of the simple symmetric random walkAs consistency checks, $F(s)\to 1$ as $s\uparrow 1$, confirming recurrence, while $F^{\prime }(s)=s/\sqrt{1-s^2}\to \infty$. By monotone convergence of ${\sum }_{n}n{f}_n{s}^{n-1}$, this last limit is ${\mathbb{E}}_{0}T=\infty$, confirming null recurrent states.

An irreducible Markov chain has positive recurrent states if the expected first return time to a state is finite; this then holds for every state. A recurrent chain with infinite mean return time has null recurrent states.

An irreducible positive recurrent chain admits a stationary distribution. If such a distribution $\pi$ existed for this walk, its stationarity equations would giveThus ${\pi }_{j+1}-{\pi }_j={\pi }_j-{\pi }_{j-1}$, so ${\pi }_j=A+Bj$. Nonnegativity for all positive and negative integers forces $B=0$. A constant sequence cannot sum to one over $\mathbb{Z}$, so no stationary probability distribution exists. Hence the walk is null recurrent, not positive recurrent: all its states are null recurrent states. The first-return generating function in part (c) also verifies directly that its mean return time is infinite.

A Markov chain is irreducible if every state can reach every other state in finitely many steps with positive probability. It is a recurrent Markov chain if, starting at any state, it returns to that state with probability one; for an irreducible chain this property is shared by all states.

For the simple symmetric random walk on $\mathbb{Z}$, taking $|j-i|$ steps all in the appropriate direction reaches $j$ from $i$ with probability $2^{-|j-i|}>0$. Thus the chain is irreducible. Its return probabilities areBy Stirling formula, $p_{00}^{(2k)}\sim 1/\sqrt{\pi k}$, so their sum diverges. The standard recurrence criterion ${\sum }_{n\ge 0}{p}_{00}^{(n)}=\infty$ therefore proves the walk is recurrent.

The printed hint's constant is erroneous: the actual limit of $\sqrt{k}{p}_{00}^{(2k)}$ is $1/\sqrt{\pi }$. Its positivity is what the recurrence argument requires, but the reciprocal is necessary for the correct numerical value.

Write $J_r(\lambda )$ for a size-$r$ Jordan block. The characteristic polynomial fixes the total dimensions at $2$ and $-7$, while the minimal polynomial fixes the largest Jordan block at each eigenvalue. Consequently, the blocks for $2$ have total size four, largest size two, and at least one size-two block. The blocks for $-7$ both have size one. The complete list, up to similarity, isThese Jordan normal forms are inequivalent because their $2$-eigenspaces have different dimensions, two and three.

The characteristic polynomial of the given matrix is $(t-1)(t-2{)}^2$. Solving the two eigenvalue equations givesFor example, multiplying the matrix by these three vectors gives respectively $1$, $2$ and $2$ times the original vector. The eigenvectors form a basis: the matrix with these columns has nonzero determinant $-1$. Thus an eigenbasis and its diagonal matrix are

Write $L={\min }_i{X}_i$, $R={\max }_i{X}_i$ and $M={\max }_i|X_i|=\max \{-L,R\}$. The joint probability density factors asBy the Fisher-Neyman factorization theorem, $T=(L,R)$ is a sufficient statistic. The same factorization shows that the smaller statistic $M$ is also a sufficient statistic. This gives the sufficient statistic for a symmetric uniform sample.

In fact, the likelihood-ratio criterion for minimal sufficiency identifies $M$ as a minimal sufficient statistic: two sample points with the same $M$ have identical likelihoods for all $\theta$; with different $M$, their likelihoods have different parameter supports, and so are not proportional as functions of $\theta$. Equivalently, the condition that $f_{\theta }(x)=c{f}_{\theta }(y)$ for every $\theta$ and some positive constant $c$ holds exactly when their absolute maxima agree.

To see rigorously that $T$ cannot be recovered from $M$ almost surely, fix any $\theta >0$. The sample has a distribution invariant under simultaneous sign reversal. If $T=h(M)$ almost surely, the same identity would hold for the sign-reversed sample on another probability-one set. Since $M$ is unchanged by sign reversal, it would follow thatand hence $L=-R$ almost surely. But this event has probability zero for a finite sample from a continuous uniform distribution: it entails $X_i=-X_j$ for some pair of indices, including the possible case $X_i=0$. Therefore $T$ is not a function of the sufficient statistic $M$. The pair of extrema is not minimally sufficient, for every $n\ge 1$, even though it is sufficient.