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The characteristic polynomial isTo prove triangularizability, use induction on $n$. The result is immediate for $n=1$. Over $\mathbb{C}$, ${\chi }_A$ has a root $\lambda$, so $A$ has an eigenvector $v_1$. Extend it to a basis. In this basis,By induction, a change among the remaining $n-1$ basis vectors makes $B$ upper triangular. Thus the triangularization over an algebraically closed field gives

The minimal polynomial $m_A$ is the monic polynomial of least degree satisfying $m_A(A)=0$. To establish existence without quoting the Cayley-Hamilton theorem, let $T$ be an upper-triangular matrix similar to $A$, with diagonal entries ${\lambda }_1,\dots ,{\lambda }_n$, and putThenThe factors $T-{\lambda }_{k}I$ commute, so applying all of them successively lowers the invariant flag to zero:Similarity gives the same polynomial identity for $A$. Hence a nonzero monic annihilating polynomial of degree $n$ exists, and a least-degree one exists.

If $m$ and $\tilde{m}$ were two monic annihilating polynomials of the same least degree, then $m-\tilde{m}$ would be an annihilating polynomial of smaller degree unless it were zero. Thus the minimal polynomial is unique, and the minimal polynomial bound from a triangular invariant flag gives

Integer row and column operations giveThus the Smith normal form isThe two relation vectors are the columns of the displayed matrix, so its cokernel is the presented module. Consequently

The Smith normal form of an integer $m\times n$ matrix $A$ is a diagonal matrixfor whichwith $U\in G{L}_m(\mathbb{Z})$ and $V\in G{L}_n(\mathbb{Z})$. It exists and is unique.

The structure theorem for finitely generated modules over a principal ideal domain, specialized to $\mathbb{Z}$, says that every finitely generated abelian group has a unique invariant-factor decompositionEquivalently, its finite part is a direct sum of cyclic groups of prime-power order.

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.