Opposite vertices differ in all three bits. Thus the four added edges areThey form a perfect matching and raise the degree of every vertex by one.

Label vertices by triples of bits and join pairs differing in exactly one bit. Draw the four vertices with first bit zero as an outer square, and the four with first bit one as an inner square, in the same cyclic order $00,01,11,10$ for the last two bits. Connect corresponding corners by four nonintersecting spokes. Every cube edge is present and there are no crossings, giving the required planar embedding.

At step $k$, choose a real shift ${\mu }_k$, factor $A_k-{\mu }_{k}I=Q_k{R}_k$, and setThe last equality proves orthogonal similarity at every step. Thus all iterates have the same eigenvalues; if $A_k$ is symmetric, the last expression is symmetric, proving symmetry by induction.

For the specified matrix, let $e=(1,\dots ,1{)}^{\mathsf{T}}$. Then $A=I-2e{e}^{\mathsf{T}}/n$ and $e^{\mathsf{T}}e=n$, so $A^{\mathsf{T}}=A$ and $A^2=I$. It is a Householder reflection, hence orthogonal. With the standard positive-diagonal QR convention, part (a) gives $Q=A$, $R=I$ at the first unshifted step, and therefore $A_1=RQ=A$. The same repeats forever: the unshifted algorithm is stationary.

If arbitrary diagonal signs in QR are allowed, $R=D$, $Q=AD$ for a sign matrix $D$, and the next iterate is $DAD$. Its off-diagonal magnitudes are unchanged, so even then only sign changes occur, not progress toward a diagonal form. This explains the role of the usual convention in the literal stationary-iterate statement.

The first column is five times $(4,0,3,0{)}^{\mathsf{T}}/5$. Its orthogonal companion in the first-and-third-coordinate plane is $(3,0,-4,0{)}^{\mathsf{T}}/5$, and the second column has positive residual seven times that vector. The remaining coordinate directions can therefore be read off immediately rather than performing a general four-dimensional orthogonalization. One obtainsThe columns of $Q$ are visibly orthonormal. Taking their inner products with each column of the given matrix gives the displayed $R$, which also verifies $QR=A$ and positive diagonal entries.

A real square QR factorization is $A=QR$ with $Q^{\mathsf{T}}Q=I$ and $R$ upper triangular. For an invertible matrix it can be made unique by requiring positive diagonal entries of $R$.

If $A$ is orthogonal, then $R=Q^{\mathsf{T}}A$ is orthogonal as well. An orthogonal upper triangular matrix is diagonal with entries $\pm 1$: its first column has just its first entry, which must have modulus one, and orthogonality then zeros the rest of the first row; repeat inductively on the lower block. Thusand under the positive-diagonal convention $\boxed{R=I}$, $Q=A$.

The best-approximation characterization gives $⟨f-q,q⟩=0$, since $q\in {\Pi }_k$. ConsequentlyTaking nonnegative square roots proves $\boxed{\Vert q\Vert \le \Vert f\Vert }$. Thus an orthogonal projection cannot increase the norm.

As explained by admissible polynomials for a singular inverse-linear weight, there is a domain issue: the constant polynomial does not belong to the weighted space, because ${\int }_0^1{x}^{-1}dx=\infty$. Hence the full space ${\Pi }_{n-1}$ is not a subspace of $V$, and the preceding inner-product projection theorem cannot be applied to it as stated.

Nevertheless a meaningful finite-error best approximation does exist after restricting to admissible polynomials. The target $x^n$ lies in $V$ for every $n\ge 1$, since its squared integrand near zero is $x^{2n-1}$. A polynomial $p\in {\Pi }_{n-1}$ has finite error exactly when $p(0)=0$: otherwise the residual tends to a nonzero constant and its weighted squared integral diverges; if $p(0)=0$, it has at least a simple zero and is integrable in the required norm. Thus the admissible approximation space isFor $n=1$ it is $\{0\}$, giving best approximation zero. For $n\ge 2$ it is a finite-dimensional inner-product subspace with positive definite Gram matrix, so it has a unique orthogonal projection of $x^n$. Therefore the full polynomial-space formulation is not defined inside $V$, but the finite-error interpretation has a unique best approximation. Declaring that no best approximation can exist under any interpretation would miss this admissible subspace.

Write $q(x)=a+bx$. Orthogonality of $x^2-q$ to $1,x$ givesThe entries are obtained by integrating $1/x,1,x,x^2$ over the given interval. Subtracting the second equation divided by two from the first gives $(\log 3-1)a=-1/3$. ThusThe Gram matrix is positive definite, with determinant $4(\log 3-1)>0$, so this orthogonal projection is the unique best approximation.

Interpret the weighted function space as measurable functions modulo equality almost everywhere, so the integral form is a genuine inner product. Assume $q\in {\Pi }_k\subseteq V$. If the residual is orthogonal to ${\Pi }_k$, then for $p\in {\Pi }_k$,Thus $q$ is a best approximation. Conversely, if it is best, take any $r\in {\Pi }_k$ and vary through $q+tr$. The polynomialhas a minimum at $t=0$, so its derivative there vanishes. Hence $\boxed{⟨f-q,r⟩=0}$ for every $r\in {\Pi }_k$. The same Pythagorean identity proves uniqueness of the best approximation.

Define $\boxed{\Vert v\Vert =\sqrt{⟨v,v⟩}}$. Positive definiteness gives nonnegativity and separation of zero, and bilinearity gives absolute homogeneity. Cauchy–Schwarz follows by minimizing the nonnegative quadratic $\Vert u-tv{\Vert }^2$ in real $t$. It then gives $\Vert u+v{\Vert }^2\le (\Vert u\Vert +\Vert v\Vert {)}^2$, establishing the triangle inequality. Thus this is a norm.

The variables $(m-1)S_X^2/{\sigma }^2$ and $(n-1)S_Y^2/{\sigma }^2$ are independent chi-squared variables with $m-1$ and $n-1$ degrees of freedom. Independence follows because the two whole samples are independent. Their sum is chi-squared with the sum of the degrees of freedom, soWithin each normal sample the sample mean is independent of its sample variance, and independence of the samples makes the pair of means jointly independent of the pair of variances. Hence this pooled chi-squared variable is independent of the standardized mean difference, which is the independence required for the Student pivot.

Independence makes a difference of the two normal sample means normal, with means subtracting and variances adding. Therefore

A state is recurrent if, starting there, the probability of returning at some strictly positive time is one. It is transient if that return probability is strictly less than one. Suppose $i,j$ communicate. Choose $r,s$ with $p_{ji}(r)>0$ and $p_{ij}(s)>0$. Chapman–Kolmogorov givesIf $i$ is recurrent, summing over $n$ and using the supplied return-series criterion makes the return series for $j$ diverge, so $j$ is recurrent. Interchanging the states proves equivalence; therefore transience is also constant on a communicating class.

For the nearest-neighbor walk, return in an odd number of steps is impossible. A return in $2n$ steps requires exactly $n$ steps each way, soAt $p=1/2$, the return series diverges like $\sum n^{-1/2}$. If $0<p<1$ and $p\ne 1/2$, then $4p(1-p)<1$, so it converges by exponential decay. At $p=0$ or one the walk never returns. Hence $\boxed{\text{recurrence occurs exactly at }p=1/2}$.