If $A$ is an invertible matrix, thenThus $AB$ and $BA$ are similar matrices, and similar matrices have the same characteristic polynomial.

Taking the matrix trace of the adjugate expansion and using part (b) givesHence $\mathrm{Tr}(B_i)=(n-i)c_i$. Taking traces in $B_i=A{B}_{i-1}+c_{i}I$ gives, for $1\le i<n$,For $i=n$, the same formula follows by taking the trace of $-A{B}_{n-1}=c_{n}I$. Iterating the recursion from part (a) also givesConsequently the Faddeev–LeVerrier algorithm isStarting with $c_0=1$, this recursively expresses every $c_i$ using only $\mathrm{Tr}(A),\dots ,\mathrm{Tr}(A^i)$; these are the Newton identities for the characteristic polynomial.

Let $M(t_1,\dots ,t_n)=\mathrm{diag}(t_1,\dots ,t_n)-A$. Expanding the determinant along row $i$ shows thatwhere $C_{ii}$ is the corresponding cofactor. Since a diagonal entry of the adjugate matrix is that same cofactor, the multivariable chain rule along the diagonal $t_1=\cdots =t_n=t$ yieldsTherefore

The adjugate matrix is the transpose of the cofactor matrix, and it satisfiesPut $M=tI-A$ and substitute the two given expansions into this identity:Comparing coefficients of powers of $t$ givesHere $c_0=1$ because the characteristic polynomial is monic.

An unbiased estimator $T$ of a parameter $\theta$ satisfies ${\mathbb{E}}_{\theta }[T]=\theta$ for every permissible value of $\theta$. Since the sample mean is linear,so $\hat{\mu }$ is unbiased.

Use the identityThe two expectations on the right are $n{\sigma }^2$ and $n\mathrm{Var}(\overline{X})={\sigma }^2$, respectively. HenceThusThe unbiased sample variance instead divides the same sum of squares by $n-1$.

For observations $x_1,\dots ,x_n$, the log-likelihood of the normal distribution model isDifferentiating first with respect to $\mu$ gives $\hat{\mu }=\overline{x}$. Substituting this value and differentiating with respect to ${\sigma }^2$ gives the normal mean and variance maximum-likelihood estimators

Let the common initial state be $x$ and the robber's final state be $y$. Part (a) gives their joint distribution asLet ${\tau }_y=\min \{n\ge 0:X_n=y\}$ be the cop's hitting time of $y$ when starting from $x$. For a chain started from $y$, condition on its first step to obtainHence the expected catch time, averaged over the displayed joint law, isUsing Kac's lemma once more givesThe convention ${\tau }_y=0$ correctly counts an immediate catch when the robber's reverse step leaves it at the common starting state.

Conditional on $X_0=x$, the random variable $T$ is the first return time $T_x^{+}$. The Kac's lemma mean-recurrence formula for a finite irreducible Markov chain givesAveraging over $X_0\sim \pi$ therefore yields

By the definition of conditional probability and stationarity,This is the transition matrix of the time-reversed Markov chain.

A level-$\alpha$ test is uniformly most powerful for a composite alternative when its power is at least that of every other level-$\alpha$ test at every parameter value in that alternative.

For every fixed ${\lambda }_1<{\lambda }_0$, the likelihood ratio is increasing in the same statistic $S$, and the size-$\alpha$ threshold is always the null quantile $n{c}_{\alpha }$, independent of ${\lambda }_1$. The Neyman-Pearson lemma therefore makes the test from part (a) most powerful against every individual ${\lambda }_1<{\lambda }_0$. Consequently it is

The power function of a test is the probability of rejection as a function of the true parameter. Since $S=n\overline{X}\sim \Gamma (n,\lambda )$,At the null value, $\beta ({\lambda }_0)=\alpha$.

The Neyman-Pearson lemma says that for simple hypotheses with densities $f_0,f_1$, a size-$\alpha$ test that rejects where $f_1/f_0>k$ and accepts where $f_1/f_0<k$, with possible randomization on equality to attain size $\alpha$, is most powerful among all tests of size at most $\alpha$. To prove it, let ${\phi }^{\ast }$ be this test and $\phi$ any competing test. Pointwise,After integration,which proves maximal power.

For the exponential sample, with $S={\sum }_i{X}_i=n\overline{X}$,Since ${\lambda }_1<{\lambda }_0$, this ratio is increasing in $S$. The likelihood-ratio test therefore rejects for large $\overline{X}$:where $G_{n,\lambda }$ is the distribution function of the rate-parameterized Gamma distribution $\Gamma (n,\lambda )$.

For this Adams–Moulton method, the characteristic polynomials of a linear multistep method areExpansion at zero givesPart (a) therefore shows that the method has order exactly three.

It is consistent because its order is at least one. The roots of $\rho (w)=w(w-1)$ are $0$ and $1$; both lie in the closed unit disk, and the only unit-modulus root is simple. Thus the method is zero-stable. By the Dahlquist equivalence theorem, consistency and zero stability imply convergence. Hence the method is

Insert the exact solution into the linear multistep method and expand about $t_n$. Its defect isThe coefficient of $h^k{y}^{(k)}(t_n)$ isfor $k\ge 1$, while the constant coefficient is ${\sum }_l{\rho }_l$. The method has order at least $p$ exactly when these coefficients vanish for $0\le k\le p$.

On the other hand,The coefficient of $z^k$ in this expression is precisely the preceding order-condition coefficient. Therefore all coefficients through degree $p$ vanish exactly whenwhich proves the equivalence.

For each $k$, the coordinate projectionis continuous when $\mathbb{Z}$ has the discrete topology: if $d(x,y)<2^{-k}$, then $x_k=y_k$.

If $\gamma :[0,1]\to X$ were a continuous path, then ${\pi }_k\circ \gamma$ would be a continuous map from the connected interval $[0,1]$ to a discrete space, so it would be constant. This holds for every $k$, and therefore $\gamma$ itself must be constant. Since $X$ contains distinct points, no path joins every pair. Henceindeed, every path component is a singleton.

A metric space is complete when every Cauchy sequence converges to a point of the space.

Let $x^{(1)},x^{(2)},\dots$ be a Cauchy sequence in $X$, where $x^{(j)}=(x_1^{(j)},x_2^{(j)},\dots )$. For each positive integer $N$, choose $J_N$ such thatThis inequality says that the first $N$ coordinates of $x^{(j)}$ and $x^{(k)}$ agree. Consequently each coordinate $x_n^{(j)}$ is eventually constant. Let $x_n$ be its eventual value and put $x=(x_1,x_2,\dots )\in X$.

For $j\ge J_N$, the first $N$ coordinates of $x^{(j)}$ agree with those of $x$, soAs $N$ is arbitrary, $x^{(j)}\to x$. Therefore

Nonnegativity and symmetry are immediate. Moreover, $d(x,y)=0$ exactly when $n_{\min }(x,y)=\infty$, which means $x_n=y_n$ for every $n$.

For three sequences $x,y,z$, if $x$ and $y$ agree through coordinate $r-1$ and $y$ and $z$ agree through coordinate $s-1$, then $x$ and $z$ agree through coordinate $\min (r,s)-1$. Henceand thereforeThus $d$ is a metric; in fact, the stronger first inequality makes it an ultrametric.

For every invertible matrix $A$,The maps $l_C$ and $r_C$ are linear. Differentiate this identity at $A=I$ and use the chain rule and part (a):Thus $Df(C)(Ch)C=-h$. Given an arbitrary increment $k$, set $h=C^{-1}k$ to obtain the derivative of matrix inversionThis also proves differentiability at every $C\in V$.

A map $f:A\to {\mathbb{R}}^m$ is differentiable at $p\in A$ when there is a linear map $Df(p):{\mathbb{R}}^n\to {\mathbb{R}}^m$ such thatThe linear map $Df(p)$ is the Fréchet derivative of $f$ at $p$.

The multivariable chain rule states that if $f$ is differentiable at $p$ and $g$ is differentiable at $f(p)$, thenIndeed, write $f(p+h)=f(p)+Ah+r(h)$ and $g(f(p)+k)=g(f(p))+Bk+s(k)$, where $r(h)=o(|h|)$ and $s(k)=o(|k|)$. Since $k=Ah+r(h)=O(|h|)$,and the last two terms are $o(|h|)$. This proves the formula.

For matrix inversion near the identity matrix,The last term is $O(\Vert h{\Vert }^2)$ because inversion is bounded near $I$. Therefore the inversion map is differentiable at $I$ and

Interchange the two primes if necessary so that $p<q$. By the Sylow theorems, the number $n_q$ of Sylow $q$-subgroups satisfiesIf $n_q\ne 1$, then $n_q$ is $p$ or $p^2$. The first is impossible because $1<p<q$. The second would implySince $q>p$ is prime, it cannot divide $p-1$, so it must divide $p+1$. But $q>p$ then forces $q=p+1$, impossible because $p$ and $q$ are both odd. Therefore $n_q=1$. The unique Sylow $q$-subgroup is a nontrivial proper normal subgroup, and hence