A complete excursion from $B$ along any one arm and back to $B$ has mean length $L=2n$. This follows from the same path recurrence, or from Kac's lemma: the graph has $3n$ edges, the stationary distribution of simple random walk assigns mass $\mathrm{deg}(B)/(2|E|)=3/(6n)=1/(2n)$ to $B$, so its mean return time is $L=2n$.

Let $x$ be the mean time to hit $E$ starting at $B$. On each visit to $B$, the first step reaches $E$ with probability $1/3$; with probability $2/3$ it begins a wrong-arm excursion of mean duration $L$, after which the problem restarts. ThusThe walk first takes mean time $n^2$ to reach $B$, so the requested time from $A$ is

Until the walk first reaches $B$, it is the simple random walk on a path of length $n$ with reflection at the endpoint $A$. If $h_i$ is the expected hitting time of $B$ from the vertex at distance $i$ from $A$, thenSolving this second-difference equation gives $h_i=n^2-i^2$. Therefore

Put $q=p^n$. Every extension of $k={\mathbb{F}}_p$ has characteristic $p$, so the iterated Frobenius endomorphism satisfiesThe root set is $K=\{a\in F:a^q=a\}$. It contains $0$ and $1$. If $a,b\in K$, thenso it is closed under subtraction and multiplication. If $a\in K$ is nonzero, thenThus $K$ is closed under multiplicative inverses as well, and is therefore a field. This is the polynomial characterization of a finite field; in fact it is the finite field with $q$ elements, since $X^q-X$ has degree $q$ and no repeated roots.

Proceed by induction on $d=\mathrm{deg}f$. A nonconstant polynomial over a field has an irreducible factor $h$. Part (a) gives an extension $k_1/k$ containing a root ${\alpha }_1$ of $h$, hence of $f$. The factor theorem giveswhere $f_1$ is monic of degree $d-1$. Apply the induction hypothesis over $k_1$ and compose the resulting field extensions. The final field $F$ contains roots ${\alpha }_1,\dots ,{\alpha }_d$ andEquivalently, every polynomial has a splitting field.

Because $f$ is irreducible, the ideal $(f)$ is maximal in the polynomial ring $k[X]$. Henceis a field extension of $k$. Let $\alpha =X+(f)\in F$. Evaluation in the quotient gives $f(\alpha )=0$, and the factor theorem in $F[X]$ givesfor some $g\in F[X]$. This is the standard adjoining a root of an irreducible polynomial construction.

For every nonnegative integer $k$,Consequently, if $m_{AB}(AB)=0$, thenBy the divisibility characterization of the minimal polynomial,Interchanging $A$ and $B$ also gives $m_{AB}(t)\mid t,m_{BA}(t)$. Therefore the Minimal polynomials of AB and BA can differ only in the exponent of the factor $t$, and that exponent differs by at most one.

If $AB$ is diagonalizable, its minimal polynomial has no repeated roots. The preceding divisibility says that every nonzero Jordan block of $BA$ has size one and every zero-eigenvalue block has size at most two. Squaring $BA$ preserves the one-dimensional nonzero blocks and sends every size-at-most-two nilpotent block to zero. Thus the Jordan normal form of $(BA{)}^2$ is diagonal, so

TakeThen $AB=B\ne 0$ and $(AB{)}^2=0$, whereas $BA=0$. Hence their minimal polynomials are

For all but finitely many $s\in \mathbb{C}$, the matrix $A-sI$ is invertible. Part (a), applied to $A-sI$ and $B$, then givesBoth sides are polynomials in the two variables $s,t$. Since they agree for infinitely many values of $s$, their coefficients as polynomials in $s$ agree. Setting $s=0$ yieldseven when $A$ is singular.

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