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First recall from the least-upper-bound property that every real interval is connected. For completeness, if $[a,b]=U\cup V$ were a separation with $a\in U$, let $s$ be the supremum of the points $t$ for which $[a,t]$ remains in $U$. Relative openness rules out both $s\in U$ and $s\in V$: in the first case one can move slightly to the right while staying in $U$, and in the second one finds points of $V$ slightly to the left of $s$, contrary to the definition of $s$. Thus no separation exists.

Part (a) now makes $g([0,1])$ a connected subset of the real line. Such a set is an interval: if it contained $u<v$ but omitted $y\in (u,v)$, its intersections with $(-\infty ,y)$ and $(y,\infty )$ would separate it. Since $g([0,1])$ contains $g(0)$ and $g(1)$, it consequently contains every number between them. Therefore for each such $y$ there is an $x\in [0,1]$ with $\boxed{g(x)=y}$. This derives the intermediate value theorem from connectedness.

Suppose for contradiction that the continuous image of a connected space $Y=f(X)$ is disconnected. Then there are disjoint nonempty sets $U,V$, open in the subspace topology on $Y$, with $Y=U\cup V$. By continuity, $f^{-1}(U)$ and $f^{-1}(V)$ are disjoint open subsets of $X$; by surjectivity both are nonempty, and their union is $X$. This is a separation of a topological space of $X$, contradicting that $X$ is a connected space. Hence $Y$ is connected.

At a point with $xy\ne 0$, the determinant of the derivative isThe inverse function theorem therefore supplies neighbourhoods on which $f$ has a continuously differentiable, hence differentiable, inverse. Concretely, if $(u,v)=f(x,y)$, the local inverse is the restriction ofwhich is continuously differentiable near the image whenever $y\ne 0$.

For $x\ne 0$, all component functions are continuously differentiable and the Jacobian matrix isAt a point $(0,y_0)$, the first component cannot be differentiable: along the first coordinate axis its difference quotient contains $h^{1/3}/h=h^{-2/3}$, which is unbounded as $h\to 0$. Therefore $f$ is continuously differentiable exactly at the points with $x\ne 0$.

Let $v_i$ be the vertex $i$ steps from $B$ toward $C$, and let $x_i$ be the mean time to reach $v_{i+1}$ from $v_i$. Part (b) gives $x_0=4n+1$. For $1\le i<n$, first-step analysis at $v_i$ givesbecause a step to $v_{i-1}$ must be followed by a fresh passage of mean $x_{i-1}$ back to $v_i$ before trying again. HenceBy the Strong Markov property, the expected time from $B$ to $C$ is the sum of these successive passage times:Adding the initial mean time $n^2$ from $A$ to $B$ gives

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