The annihilator of a vector subspace isChoose a basis $e_1,\dots ,e_r$ of $U$ and extend it to a basis $e_1,\dots ,e_d$ of $V$. In the dual basis, precisely $f_{r+1},\dots ,f_d$ span $U^{\circ }$. ThereforeThe canonical restriction linear transformationhas kernel of a linear map $U^{\circ }$. It is surjective: extend any linear functional on $U$ by assigning arbitrary values, for instance zero, to the additional basis vectors. The First isomorphism theorem for vector spaces gives the explicit linear isomorphismIt is natural because the displayed map itself makes no basis choice; a basis was used only to prove surjectivity. In particular, $U^{\prime }$ is canonically a quotient vector space of $V^{\prime }$, rather than canonically a subspace of $V^{\prime }$.

For $\varphi \in V^{\prime }$ and $v\in V$,so $I^{\prime }$ is the identity map on the dual space. If $S:U\to V$ and $T:V\to W$, then, for $\psi \in W^{\prime }$ and $u\in U$,Equality on all vectors provesApply this to $R{R}^{-1}=R^{-1}R=I$. It gives both $R^{\prime }(R^{-1}{)}^{\prime }=I^{\prime }$ and $(R^{-1}{)}^{\prime }{R}^{\prime }=I^{\prime }$. Consequently the dual map is invertible and

The dual map acts by function composition:The composition is a linear functional on $V$, and pointwise operations give $T^{\prime }(a\varphi +b\psi )=a{T}^{\prime }\varphi +b{T}^{\prime }\psi$. Thus this really is a linear transformation; its direction reverses that of $T$.

The dual space isthe vector space of linear functionals, with pointwise operationsGiven the basis $e_1,\dots ,e_n$, its dual basis consists of the coordinate linear functionalsIn particular, $\dim V^{\prime }=\dim V=n$.

Write $f=m_T$. Part (ii) identifies the first superdiagonal block exactly:where $f^{\prime }$ is the formal derivative. Thus the candidate is $g=f^{\prime }$. If the irreducible polynomial $f$ has $f^{\prime }\ne 0$, its polynomial degree makes $f$ and $f^{\prime }$ coprime. Equivalently, $f$ is a separable polynomial. This includes every irreducible polynomial over a field of characteristic zero, and every irreducible polynomial over a perfect field.

Under this separability hypothesis, Bézout's identity for polynomials gives $a(x)f(x)+b(x)f^{\prime }(x)=1$. Evaluating at $A$ proves that $C=f^{\prime }(A)$ is an invertible matrix. Moreover, $f(B)$ has zero diagonal blocks, because $f(A)=0$. It is therefore a nilpotent matrix with $f(B{)}^N=0$. The only product of $N-1$ strictly upward block steps that reaches block $(1,N)$ uses the first superdiagonal every time, soFor $N=1$, the corresponding assertion is simply $f(B{)}^0=I\ne 0$. The block upper triangular matrix $B$ hasThe minimal polynomial divides this characteristic polynomial. Since $f$ is an irreducible polynomial, it must be a power of $f$; the exact nilpotency index just found forces

The arbitrary-field qualification is necessary. In the rational function field $F={\mathbb{F}}_2(t)$, take $f(x)=x^2-t$. This is an irreducible polynomial: a square of a rational function has even order of vanishing at $t=0$, while $t$ has odd order. Its formal derivative is zero. With $N=2$,The off-diagonal block in $f(B)$ is zero, and $m_U=f$, rather than $f^2$. No alternative polynomial $g$ coprime to $f$ could represent that block: $g(A)=0$ would imply $m_T=f\mid g$. Thus the printed coprimality claim fails as stated over a general field.

There is also a precise general replacement. In characteristic $p>0$, let $q=p^r$ be the largest power for which $f(x)=h(x^q)$, so that $h^{\prime }\ne 0$; set $q=1$ in characteristic zero. Irreducibility of $f$ implies that $h$ is an irreducible polynomial, so $h$ and $h^{\prime }$ are coprime. The binomial theorem in characteristic $p$ gives $B^q=D^q+K^q$. Expanding $h(D^q+K^q)$, with the coefficients equivalently supplied by the Hasse derivatives, giveswhere $H$ commutes with $K$ and is a block upper triangular matrix whose diagonal blocks are $h^{\prime }(A^q)$. These blocks are invertible by Bézout's identity for polynomials, since $h(A^q)=f(A)=0$. Thus $H$ is invertible, andThe same characteristic polynomial and divisibility argument now prove the corrected arbitrary-field answer:This is the minimal polynomial of a companion block chain; $q=1$ recovers the intended separable case.

Write $B=D+K$, where $D$ has $A$ in each diagonal block and zero elsewhere, while $K$ has $I$ immediately above the diagonal and zero elsewhere. These block matrices satisfy $DK=KD$: either product has $A$ immediately above the diagonal. Also, $K^i$ has $I$ on its $i$th superdiagonal and is zero elsewhere.

The binomial theorem for commuting matrices therefore yieldsOnly $r=i$ can contribute to block $(k,k+i)$, provided that block lies in the matrix. HenceThus ${\beta }_{i,j}=\left(\genfrac{}{}{0ex}{}{j}{i}\right)$, interpreted as an element of the field. The formula remains valid in positive field characteristic: some of these coefficients can then vanish. This possibility matters in the next part.

The displayed matrix is the transpose convention for a companion matrix. Expanding $\det (xI-A)$ along its last row givesIndeed, the last entry contributes $(x+{\alpha }_{n-1})x^{n-1}$ and the entry in column $j<n$ contributes ${\alpha }_{j-1}{x}^{j-1}$; the cofactor signs cancel those from the superdiagonal entries $-1$. ConsequentlyThe Cayley-Hamilton theorem gives $f(A)=0$, so $m_T$ divides $f$. To exclude a smaller minimal polynomial, let $e_1,\dots ,e_n$ be the standard basis. The first $n-1$ rows of $A$ showIf ${\sum }_{j=0}^{n-1}{c}_j{A}^j=0$, multiplying on the left by $e_1^{\mathsf{T}}$ gives ${\sum }_j{c}_j{e}_{j+1}^{\mathsf{T}}=0$, so every coefficient is zero. No nonzero polynomial of degree below $n$ vanishes at $A$. Since $f$ and $m_T$ are monic polynomials,

The Cayley-Hamilton theorem statesLet $d=\dim V$. Extend the field to an algebraically closed field and choose a basis in which the matrix of $T$ is an upper triangular matrix, as permitted. Write its diagonal entries as ${\lambda }_1,\dots ,{\lambda }_d$ and put $V_i=\mathrm{span}(e_1,\dots ,e_i)$. Each $V_i$ is an invariant subspace, andApply $T-{\lambda }_{d}I$ first, then $T-{\lambda }_{d-1}I$, and continue. The resulting product sends $V_d$ into $V_0=\{0\}$. These factors are commuting operators, soThe determinant of an upper triangular matrix gives ${\chi }_T(x)=(-1{)}^d{\prod }_i(x-{\lambda }_i)$. Thus ${\chi }_T(T)=0$ after the field extension. Its entries were in the original field, whose embedding is injective, so they were already zero there. The zero-dimensional case is immediate as well.

For a linear transformation $T$ of a finite-dimensional vector space over a field $F$, its minimal polynomial $m_T$ is the unique monic polynomial of least degree satisfying $m_T(T)=0$. There is at least one nonzero polynomial with this property because the powers $I,T,T^2,\dots$ lie in the finite-dimensional vector space of linear maps $V\to V$. Dividing any other such polynomial by $m_T$ leaves a remainder of smaller degree that also vanishes at $T$, so that remainder must be zero. This also proves uniqueness.

If $C$ represents $T$ in any basis, the paper's characteristic polynomial isIt is independent of the basis, since changing the basis replaces $C-xI$ by a similar matrix, preserving its determinant. Its degree is $\dim V$ and its leading coefficient is $(-1{)}^{\dim V}$; $m_T$ is always monic.

Let the two oriented surfaces carry positive area forms $\alpha$ and $\beta$, obtainable from oriented Riemannian metrics. A top-degree form on a surface is automatically closed. On their product setIt is closed, andis a nowhere-vanishing four-form; the two squared terms vanish because each comes from a two-dimensional factor. Thus the product is symplectic, with no compactness assumption needed.

Use the standard convention that the compact manifold is nonempty and has no boundary. Orient it by ${\omega }^m$. Since $d\omega =0$, every power ${\omega }^k$ is closed. If ${\omega }^k=d\eta$ for some $1\le k\le m$, thenagain by Leibniz and closedness. Stokes theorem would give ${\int }_M{\omega }^m=0$. But this integral is strictly positive in the symplectic orientation, a contradiction. ThereforeThe boundaryless hypothesis matters: a compact disk has a symplectic area form but no nonzero ordinary second cohomology class.

The nonvanishing top-degree form ${\omega }^m$ is a volume form. Declare an ordered tangent basis positive when its value under ${\omega }^m$ is positive. Smooth nonvanishing ensures these choices agree on overlapping oriented charts and vary continuously. Hence a symplectic manifold is orientable; this is its symplectic orientation.

The usual Weyl character formula requires a compact connected Lie group. Choose a maximal torus $T$, a positive root system ${\Phi }^{+}$, Weyl group $W$, and $\rho =\frac{1}{2}{\sum }_{\alpha \in {\Phi }^{+}}\alpha$. Irreducible representations are indexed by dominant integral highest weights $\lambda$ belonging to the character lattice of the actual group $T$. For $X\in L(T)$ such that $\exp X$ is regular,Weights are written as linear functionals on the complexified torus algebra; their values on the compact real torus algebra are imaginary. The denominator is $e^{\rho (X)}{\prod }_{\alpha >0}(1-e^{-\alpha (X)})$. If $\rho$ is not itself a character of $T$, the individual expressions are interpreted on its covering algebra; their quotient is single-valued and descends to $T$. The apparent singularities at nonregular elements are removable, and conjugacy into a maximal torus determines the character on the whole connected group.

The formula explicitly determines all irreducible characters from their highest weights, and its limit at the identity gives the dimension formulaCharacter multiplication and orthogonality then compute tensor-product multiplicities. For $S{U}_2$, highest weights are integers $m\ge 0$, and at $\mathrm{diag}(z,z^{-1})$,Connectedness cannot simply be omitted from this version. For a nontrivial finite group the maximal torus is trivial, yet characters can differ away from the identity. The root data on that torus alone cannot supply the characters of all disconnected compact groups.

Schur orthogonality for irreducible complex characters of a compact group, with normalized Haar measure, isEquivalent representations clearly have equal characters, since trace is invariant under conjugation. Conversely complete reducibility gives ${\chi }_V={\sum }_{\alpha }{m}_{\alpha }{\chi }_{\alpha }$, where the nonnegative integer multiplicity is recovered by $m_{\alpha }=⟨{\chi }_V,{\chi }_{\alpha }⟩$. Equal characters thus give equal multiplicities of every irreducible constituent, hence equivalent representations.

As usual here, representations are finite-dimensional and continuous. Start with any positive definite Hermitian form $h_0$ and normalized Haar measure $\mu$ on the compact group. DefineThis is Hermitian, sesquilinear, and positive definite: for $v\ne 0$ its continuous integrand $h_0(gv,gv)$ is strictly positive. Bi-invariance of Haar measure shows $h(av,aw)=h(v,w)$ for every $a\in G$.

If $U\subset V$ is invariant, then so is its orthogonal complement: for $v\perp U$ and $u\in U$, $h(av,u)=h(v,a^{-1}u)=0$. Hence $V=U\oplus U^{\perp }$ as a representation. Induct on dimension, splitting whenever a nontrivial proper invariant subspace exists. This proves every such representation is a direct sum of irreducible representations.

With the convention applicable also to disconnected compact groups, the Weyl group is $W=N_G(T)/T$, where the normalizer consists of elements taking $T$ to itself by conjugation. Put $\mathfrak{t}=L(T)$ and $\Lambda =\ker (\exp :\mathfrak{t}\to T)$. This is a full lattice with $T\cong \mathfrak{t}/\Lambda$.

If $\psi$ is a torus automorphism, exponential compatibility gives $\psi (\exp X)=\exp (D{\psi }_{I}X)$. Therefore $D{\psi }_I$ sends $\Lambda$ into itself; applying the inverse proves it sends the lattice onto itself. In a lattice basis this derivative is an element of $G{L}_r(\mathbb{Z})$.

Conjugation by a connected subgroup $Q\subset N_G(T)$ induces a continuous family of these integer matrices. Since $G{L}_r(\mathbb{Z})$ is discrete, this family is constant and equals its value at the identity. Its action on $\mathfrak{t}$ is therefore trivial, and surjectivity of the torus exponential shows that $Q$ centralizes $T$.

Apply this to the identity component $N_G(T{)}^0$. It lies in $G^0$ and centralizes $T$. The allowed maximal-abelian property of $T$ in $G^0$ gives $N_G(T{)}^0\subseteq T$, while the connected group $T$ lies in that identity component. Thus $N_G(T{)}^0=T$. The normalizer is a closed compact Lie subgroup, so its component quotient is discrete and compact, hence finite. Consequently $W$ is finite and discrete. For disconnected $G$, its action on the lattice need not be faithful; the component argument is essential.

The distinction is recorded in Weyl group of a disconnected compact Lie group.

Write ${\mathfrak{g}}_{\mathbb{C}}=L(G_{\mathbb{C}})$ and $\mathfrak{h}=L(H)$. A root is a nonzero weight $\alpha \in {\mathfrak{h}}^{\ast }$ of the adjoint action. Its root space isAt the group level these weights come from nontrivial characters of the Cartan subgroup acting by conjugation. The Cartan root-space decomposition isThe zero-weight space is $\mathfrak{h}$; for the complexification of a compact group the algebra is reductive, with any centre contained in this zero-weight space.

For the product in question,Take the diagonal Cartan subalgebra generated by $H_1=(\mathrm{diag}(1,-1),0)$ and $H_2=(0,\mathrm{diag}(1,-1))$. On $Z=x{H}_1+y{H}_2$, define ${\alpha }_1(Z)=2x$ and ${\alpha }_2(Z)=2y$. The roots are exactlyIf $E=E_{12}$ and $F=E_{21}$, their four root spaces are respectively $\mathbb{C}(E,0)$, $\mathbb{C}(F,0)$, $\mathbb{C}(0,E)$, and $\mathbb{C}(0,F)$. These four lines and the two-dimensional diagonal subalgebra form the full six-dimensional decomposition. There are no mixed roots: the two direct summands commute. This is the root system $A_1\times A_1$.