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The Jordan normal form theorem says that every complex square matrix is similar to a direct sum of Jordan blocks, uniquely up to their order. For an eigenvalue $\lambda$, if its blocks have sizes $n_1,\dots ,n_r$, then its algebraic multiplicity and geometric multiplicity areand its contribution to the minimal polynomial is $(t-\lambda {)}^{{\max }_j{n}_j}$. Thuswhere $s_{\lambda }$ is the largest $\lambda$-block size.

For the given matrix,Moreover,HenceThe eigenvalue $-1$ therefore has one Jordan block of size two, so

With states ordered as $(A,B,C)$ and rows representing the current state, the Markov chain has transition matrix

Two matrices are equivalent when $A^{\prime }=PAQ$ for invertible matrices $P$ and $Q$ of the appropriate sizes. A change of basis in the codomain multiplies a matrix on the left, while a change of basis in the domain multiplies it on the right. Hence two matrices of the same linear map in two pairs of bases are equivalent.

Conversely, start with the map $\alpha :F^n\to F^m$ having matrix $A$ in the standard bases. Given $A^{\prime }=PAQ$, choose domain and codomain bases whose change-of-coordinate matrices produce $Q$ and $P$; then $A^{\prime }$ represents the same map in those bases. This proves the equivalence.

The column rank of $A$ is the dimension of the span of its columns, and its row rank is the dimension of the span of its rows. If $A$ represents $\alpha :F^n\to F^m$, its column rank is $\mathrm{rank}\alpha$. The transpose represents the dual mapso the row rank is $\mathrm{rank}{\alpha }^{\ast }$.

If $\mathrm{rank}\alpha =r$, thenwhose dimension is $m-r$: extend a basis of $\mathrm{im}\alpha$ to one of $F^m$ and use the dual basis. Rank-nullity now gives $\mathrm{rank}{\alpha }^{\ast }=r$. This proves the equality of row rank and column rank.

For $I\subseteq [m]$, the independent vectors $\{v_i:i\in I\}$ lie in the coordinate subspace supported onwhich has dimension $|S_I|$. Hence $|I|\le |S_I|$, and the stated form of Hall marriage theorem gives an injection $f$ with $f(i)\in \mathrm{supp}(v_i)$.

Let $V$ be the $n\times m$ matrix whose $i$th column is $v_i$. Its column rank is $m$. By equality of row and column rank, it has $m$ linearly independent rows. Let $B$ be their indices. The $m\times m$ submatrix $V_B$ is invertible, so its columns are independent. Those columns are exactly the nonzero coordinates of $B{v}_1,\dots ,B{v}_m$, and therefore these truncated vectors are linearly independent.

Expanding $\det V_B$ gives a permutation $f:[m]\to B$ for whichThus $f(i)\in \mathrm{supp}(v_i)$. Finally, order the coordinates with $B$ first. The matrix whose columns are the $v_i$ together with the $e_j$ for $j\notin B$ is block triangular, with diagonal blocks $V_B$ and an identity matrix. Its determinant is nonzero. Consequentlyis a basis of ${\mathbb{C}}^n$. This is the simultaneous basis exchange from a nonzero minor.

Let $X$ be the $173\times 3$ design matrix and write $p=3$. The displayed estimate in row $j$ isWithits standard error isThe $t$-value and two-sided $p$-value arerespectively.

The inverse first-fundamental-form matrix isThus $2H=\mathrm{tr}((I^S{)}^{-1}I{I}^S)$ vanishes exactly whenThis is the minimal surface equation for a graph.

The Markov property at deterministic times givesThus $(Z_n)$ is a discrete-time Markov chain with transition matrixwhen the minimal chain is nonexplosive, $P(h)=e^{hQ}$.

For an irreducible chain, state $i$ is recurrent in continuous time exactly whenand it is recurrent for the skeleton exactly when ${\sum }_{n\ge 0}{p}_{ii}(nh)=\infty$. If $t\in [nh,(n+1)h]$, the event of staying at $i$ suppliesIntegrating over each interval shows that the integral diverges exactly when the series does. Irreducibility then makes recurrence equivalent for the two chains. This is recurrence equivalence for a CTMC and its fixed-time skeleton.

On the finite state space, $P_t=e^{tQ}$ andThereforeThe matrix semigroup satisfies $\frac{d}{dt}{P}_{t}f=P_{t}Qf$. Integrating from $0$ to $t$ gives the backward integral equation

Let $P$ be the state-by-asset matrix of discounted gains and identify $Y$ with its state vector. Fix $q^0\in \mathcal{Q}$. If $q\ge 0$, $P^{T}q=0$, and $q\ne 0$, thenbelongs to $\mathcal{Q}$. The assumption and passage to the limit imply $q^{T}Y\ge 0$ for every nonnegative $q\in \ker P^T$.

The finite-state superhedging alternative now gives $\theta$ withThis residual cannot vanish identically, since then every $Q\in \mathcal{Q}$ would have $E_{Q}Y=q^{T}P\theta =0$. Hence it is strictly positive in at least one state, which has positive physical probability. This is the required inequality.

For the Fibonacci transition matrix above,where $L_n$ is the $n$th Lucas number. It counts closed symbolic itineraries of length $n$. Since $7$ is prime, the itineraries fixed by the seventh shift have least period either $1$ or $7$. Hence the number of cyclic classes of primitive length-seven itineraries isDistinct primitive cyclic itineraries give distinct cycles, so $F$ has at least four distinct $7$-cycles. This is the fibonacci transition-graph cycle count.

Expanding the numerical propagation matrix,Its difference from $e^{kA}$ isTherefore the one-step local truncation error isand the method is first-order accurate globally.

The matrix exponential is the absolutely convergent seriesMultiplying the two expansions givesOn the other hand,Subtracting yieldsThis is the leading Lie-Trotter splitting commutator error.

Put $A_j=\left(\begin{array}{cc}{a}_j& b_j\\ 1& 0\end{array}\right)$. The identityfollows by induction: multiplying the formula for $n-1$ by $A_n$ gives the two stated recurrences in the first column and $b_n(p_{n-1},q_{n-1}{)}^T$ in the second. The associated Möbius transformations $z\mapsto a_j+b_j/z$, applied from the right to $a_n$, show that the first-column ratio is the displayed positive generalized continued fraction.

Now set every $b_j=1$. Taking determinants in the matrix identity givesConsequently adjacent convergents differ byThe recurrence and $a_j\ge 1$ imply $q_{n+1}\ge q_n+q_{n-1}$, so $q_n\to \infty$. The determinant signs show that the even convergents increase, the odd convergents decrease, and every even one is below every odd one. Their adjacent separation tends to zero, so both subsequences have a common limit $x$. Since $x$ lies between each adjacent pair, the convergence of a positive simple continued fraction gives exactly

If all $a_j=1$ as well, the recurrence and initial values giveThe characteristic roots of $F_{n+2}=F_{n+1}+F_n$ are
$\varphi =(1+\sqrt{5})/2$ and $\psi =(1-\sqrt{5})/2=-{\varphi }^{-1}$, henceTherefore $F_{n+2}/F_{n+1}\to \varphi$. Moreover,Using $F_n{\varphi }^{-n}\to 1/\sqrt{5}$ yieldsandThis is the all-one continued fraction and Fibonacci ratios case.

Let ${\hat{\theta }}_n$ be the MLE and let $I(\theta )$ be the one-observation Fisher information matrix. One standard form of the Wald statistic isAny consistent information estimator gives the same asymptotics. Under $H_0$,
$W_n({\theta }_0)\Rightarrow {\chi }_p^2$. If
$c_{p,\alpha }$ is the $(1-\alpha )$ quantile of ${\chi }_p^2$, reject whenThe asymptotic type-I error is $\alpha$.

By part (a), each $A{\phi }_{\pm }+{\partial }_t{\phi }_{\pm }$ lies in the same two-dimensional eigenspace. Reality and conjugacy again force the coefficient matrix to have the formBecause $u$ and hence $A$ are periodic, ${\partial }_t+A$ commutes with translation by $2\pi$. Applying it to
$\Psi (x+2\pi )=\hat{T}\Psi (x)$ givesThe basis $\Psi$ is independent, so the periodic KdV transfer matrix obeysTaking traces,Hence $\boxed{\mathrm{Re}a}$ is independent of time.

Translation by $2\pi$ preserves the scattering equation, so it acts linearly on its two-dimensional solution space. Because $u$ is real and the initial conditions are conjugate,
${\phi }_{-}=\overline{{\phi }_{+}}$, forcing the transfer matrix to have the formThe Wronskian of two solutions of
$-{\varphi }^{\prime \prime }+u\varphi =k^2\varphi$ is independent of $x$. Translation therefore preserves it, soConsequently

A density operator satisfiesAn $N\times N$ Hermitian matrix has $N^2$ real parameters, and the trace condition removes one, leavingreal free parameters.

If $A_H(t)=U(t{)}^{†}{A}_{S}U(t)$, equality of expectation values requiresTherefore

The kinetic matrix is diagonal:For the linear potential,andThus, in units ${\hslash }^2/(m{a}^2)$,The lower eigenvalue gives the two-mode variational bound for a linearly tilted square well

If the Schmidt rank is one, the decomposition is $|\psi ⟩=|\alpha ⟩|\beta ⟩$ after absorbing its sole coefficient, so the state is a product. Conversely, a product state has coefficient matrix $X=a{b}^T$, which has rank one. Therefore a bipartite pure state is entangled exactly when its Schmidt rank is at least two.

The reduction map sends $H_k$ onto $H_{k-1}$ with kernel of order $p$, so $|H_k|=p^{k-1}$. More explicitly, the lifting-the-exponent calculationshows that $1+p$ has order $p^{k-1}$ modulo $p^k$. Hence $H_k=⟨1+p⟩$ is cyclic of that order.

For the matrix claim, suppose $A\ne I$ and choose $m$ maximal such that $A=I+p^{m}B$, so some entry of $B$ is nonzero modulo $p$. Replacing $A$ by a suitable power, we may assume its order is a prime $q$. If $q\ne p$, binomial expansion modulo $p^{m+1}$ givesIf $q=p$, expansion modulo $p^{m+2}$ givesbecause $p$ is odd. Both contradict $A^q=I$, so $A=I$.

For $p=2$, $r=1$ fails because $-I=I+2(-I)$ has order two. If $r=2$, the same argument works; in the order-two case, for maximal $m\ge 2$,since the parenthesis is nonzero modulo two. Thus the smallest value is $r=2$.