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A linear map $\alpha$ is a nilpotent endomorphism if ${\alpha }^q=0$ for some positive integer $q$, and a semisimple endomorphism if it is diagonalizable over $\mathbb{C}$. Decompose $V$ into its generalized eigenspaces $V_{\lambda }=\ker (\alpha -\lambda I{)}^d$, where $d=\dim V$. Define ${\alpha }_s$ on $V_{\lambda }$ to be $\lambda I$, and set ${\alpha }_n=\alpha -{\alpha }_s$. Then ${\alpha }_s$ is diagonalizable, ${\alpha }_n$ is nilpotent, and both preserve these vector subspaces and commute. ThusFor uniqueness, suppose $\alpha =S+N$ with $S$ semisimple, $N$ nilpotent and $SN=NS$. Both commute with $\alpha$, so preserve each $V_{\lambda }$. On an eigenspace of $S$ of eigenvalue $\mu$ inside $V_{\lambda }$, the map $\alpha$ is $\mu I+N$ and has only the eigenvalue $\mu$. Since the same subspace lies in $V_{\lambda }$, $\mu =\lambda$. Hence $S=\lambda I$ on $V_{\lambda }$, proving $S={\alpha }_s$ and $N={\alpha }_n$. This is the additive Jordan–Chevalley decomposition.

We need a polynomial consequence of this decomposition. Hermite interpolation supplies a polynomial $p$ with $p(\alpha )={\alpha }_s$, by prescribing $p(t)\equiv \lambda (\mathrm{mod}(t-\lambda {)}^d)$ for each eigenvalue. For any endomorphism $T$, its semisimple part can likewise be expressed as a polynomial in $T$ with zero constant term: if zero is an eigenvalue, its interpolation condition already forces this; if not, add the independent condition $p(0)=0$.

On $\mathrm{End}(V)$, the maps $\mathrm{ad}{\alpha }_s$ and $\mathrm{ad}{\alpha }_n$ commute. The first is diagonalizable, with eigenvalues $\lambda -\mu$ on $\mathrm{Hom}(V_{\mu },V_{\lambda })$. The second is nilpotent, sincewhich vanishes for $k\ge 2q-1$ when $N^q=0$. Uniqueness therefore proves adjoint compatibility of additive Jordan decomposition: $\mathrm{ad}{\alpha }_s=(\mathrm{ad}\alpha {)}_s$.

Now assume $\alpha \in M$. The condition $[\alpha ,W]\subseteq U\subseteq W$ implies that both $W$ and $U$ are invariant under $\mathrm{ad}\alpha$, and every polynomial in $\mathrm{ad}\alpha$ with zero constant term maps $W$ into $U$. Define $\beta$ to be multiplication by $\overline{\lambda }$ on $V_{\lambda }$. It commutes with $\alpha$. On $\mathrm{Hom}(V_{\mu },V_{\lambda })$, $\mathrm{ad}\beta$ acts by $\overline{\lambda -\mu }$. Polynomial interpolation on the finite set of differences gives $\mathrm{ad}\beta =q(\mathrm{ad}{\alpha }_s)$ with $q(0)=0$. The preceding paragraph then expresses $\mathrm{ad}\beta$ as a polynomial in $\mathrm{ad}\alpha$ with zero constant term. Consequently $[\beta ,W]\subseteq U$, so $\beta \in M$.

The assumed trace orthogonality nilpotence lemma now follows directly. On $V_{\lambda }$, $\alpha =\lambda I+{\alpha }_n$ and $\beta =\overline{\lambda }I$, while the matrix trace of the nilpotent restriction of ${\alpha }_n$ is zero. HenceEvery summand is nonnegative, so every eigenvalue of $\alpha$ is zero. Its Jordan–Chevalley decomposition therefore has ${\alpha }_s=0$, and $\alpha$ is nilpotent. Notice that neither $U$ nor $W$ was required to be a Lie subalgebra.

Reading the coefficients of the linear map givesUnder the inverse identification, the four standard real basis vectors map respectively to $(1,0)$, $(i,0)$, $(0,1)$ and $(0,i)$ in ${\mathbb{C}}^2$. Applying $A$ and taking real and imaginary parts therefore gives the columnsThus the real matrix isFor verification, write its blocks as $\left(\begin{array}{cc}{R}_{\theta }& I\\ I& R_{-\varphi }\end{array}\right)$, where $R_{\psi }$ is the planar rotation matrix. Block elimination gives $\det B=\det (R_{\theta -\varphi }-I)=2-2\cos (\theta -\varphi )$. Therefore the determinant identity isThis is an instance of the realification determinant identity. The identification is real-linear; it is not an identification of complex vector-space dimensions.

For a linear map $T:V\to W$ with $V$ finite-dimensional, the rank-nullity theorem statesTo prove it, choose a basis $v_1,\dots ,v_k$ of the kernel of a linear map and use basis extension to obtain $v_1,\dots ,v_n$ as a basis of $V$. The vectors $T{v}_{k+1},\dots ,T{v}_n$ span the image of a linear map: applying $T$ to a basis expansion kills the first $k$ terms. They are also linearly independent. Indeed, if ${\sum }_{j>k}{a}_{j}T{v}_j=0$, then ${\sum }_{j>k}{a}_j{v}_j\in \ker T$, so it is a combination of $v_1,\dots ,v_k$. Independence of the full basis forces every $a_j=0$. Consequently $\dim \mathrm{im}T=n-k$, proving the theorem.

Here the image of a linear map has dimension $r\in \{0,1,2,3\}$, since it lies in ${\mathbb{R}}^3$. ThusEvery possibility occurs: for $0\le r\le 3$, map $(x_1,\dots ,x_5)$ to $(x_1,\dots ,x_r,0,\dots ,0)\in {\mathbb{R}}^3$. This map has rank $r$ and kernel dimension $5-r$.

The Fréchet derivative at $p=(x_0,y_0)$ is a linear map $L:{\mathbb{R}}^2\to \mathbb{R}$ for whichThis is differentiability; if it holds, $L(h_1,h_2)=D_{1}f(p)h_1+D_{2}f(p)h_2$.

To prove the sufficient condition, take a rectangle about $p$ contained in the open domain. Split the increment along its coordinate directions:The one-variable mean value theorem, applicable because the restrictions are differentiable, represents these two terms asThe zero-increment cases are interpreted directly. Subtract $D_{1}f(p)h+D_{2}f(p)k$. By continuity of both partial derivatives at $p$, the remainder is at most $\epsilon (|h|+|k|)\le \sqrt{2}\epsilon \sqrt{h^2+k^2}$ once the increment is small. This is the required little-oh remainder and proves differentiability.

The converse is false. Set $f(x,y)=x^2\sin (1/x)$ for $x\ne 0$, and $f(0,y)=0$. At each point with $x=0$, its Fréchet derivative is zero, since the function increment has magnitude at most $h^2=o(\sqrt{h^2+k^2})$. It is smooth away from this line. Butis not continuous there. Thus a function can be differentiable everywhere while its partial derivatives are discontinuous.

For $h$, every point with $x\ne 0$ is a smooth point. At the origin, $|h(x,y)|\le |xy|\le (x^2+y^2)/2$, so $h$ is differentiable with zero derivative. At $(0,y_0)$ with $y_0\ne 0$, the putative first partial derivative would require the limit $y_0\sin (1/x)$, which does not exist. Consequently the exact set of differentiability is

A fiber metric on a real vector bundle $E\to M$ is a smoothly varying positive-definite inner product on each fiber. Choose a trivializing open cover $\{U_i\}$ and a smooth partition of unity $\{{\rho }_i\}$ subordinate to it. The partition theorem gives nonnegative functions summing to one, a locally finite family of subsets of supports, and $\mathrm{supp}{\rho }_i\subset U_i$; the usual Hausdorff second-countable smooth manifold hypotheses ensure this theorem applies. Transfer the Euclidean inner product to each local vector bundle trivialization, obtaining $h_i$, and setEach weighted term extends smoothly by zero outside $U_i$, and local finiteness makes the sum smooth in every vector bundle trivialization. At each $x$ some weight is positive, so $h_x(v,v)>0$ for every nonzero $v$. This proves existence of a fiber metric, with no orientability or triviality assumption.

A vector bundle morphism covering the identity is a smooth map $F:E^{\prime }\to E^{\prime \prime }$ that preserves base points and is a linear map on each fiber. Its induced map on the module of smooth sections is $s\mapsto F\circ s$, and is $C^{\infty }(M)$-linear. We prove the converse by constructing bundle morphisms from maps of smooth sections.

First the given map $\alpha$ is local. If a global section $s$ vanishes on a neighborhood of $x$, take a smooth bump function $\chi$ supported there with $\chi =1$ near $x$. Then $\chi s=0$, so $\chi \alpha (s)=\alpha (\chi s)=0$, and hence $\alpha (s)(x)=0$. Thus sections agreeing near $x$ have images agreeing at $x$.

Choose a local frame $e_1,\dots ,e_r$ on $U$, and a bump function equal to one on a smaller neighborhood $V$ of $x$ and supported in $U$. Multiplying the frame by that bump and extending by zero gives global smooth sections $t_j$ whose restrictions to $V$ are the frame. If $s(x)=0$, write $s=\sum a_j{e}_j$ on $V$. A second bump extends each $a_j$ to a global smooth function ${\tilde{a}}_j$ agreeing near $x$. By locality and $C^{\infty }(M)$-linearity,Every fiber vector $v\in E_x^{\prime }$ is the value of a global smooth section, by the same bumped-frame construction. Define $F_x(v)=\alpha (s)(x)$ for any such section. The just-proved vanishing statement makes this well-defined. The maps $F_x$ are linear maps, and locally their matrix columns are the smooth sections $\alpha (t_j)$ in a frame of $E^{\prime \prime }$. Thus $F$ is smooth, is a vector bundle morphism, and satisfies $\alpha (s)=F\circ s$. Fiberwise evaluation also proves uniqueness.

Finally apply the fiber metric construction to the tangent bundle. A Riemannian metric $g$ defines the musical isomorphismPositive definiteness makes it a fiberwise bijection; its inverse is smooth because inverse metric matrices vary smoothly. Hence $TM$ and $T^{\ast }M$ are isomorphic as real smooth vector bundles on every such manifold. The isomorphism depends on the chosen metric and is not canonical.

Use the three left-coset representatives $I,S,ST$ for ${\Gamma }_0(2)\backslash {\mathrm{SL}}_2(\mathbb{Z})$. If a nonzero weight-four cusp form $h$ existed, its coset norm of a modular formwould be a nonzero weight-twelve modular form for the full group: right multiplication permutes its factors. At infinity $h$ has order at least one in $q$. The other two factors correspond to the zero modular cusp of cusp width two and each have order at least $1/2$ in $q$. Thus $H$ has order at least two there.

The ratio $H/\Delta$ is weight zero, holomorphic on the half-plane because $\Delta$ has no zeros there, and holomorphic at the modular cusp with value zero. It descends to a holomorphic function on the compact full modular curve. Such a function is constant, hence zero, contradicting $H\ne 0$. ThereforeConstant terms at the two modular cusps define a linear map $M_4({\Gamma }_0(2))\to {\mathbb{C}}^2$ whose kernel is this modular cusp space. It is injective, so the dimension is at most two. The forms $E_4(\tau )$ and $E_4(2\tau )$ are holomorphic weight-four forms for this group. For the second, the same conjugation used in 1(c) proves transformation, and $E_4(2S\tau )|$ after the weight factor is $2^{-4}{E}_4(\tau /2)$, proving holomorphy at zero. Their constant terms are both one but their $q$ coefficients are respectively $240$ and zero, so they are independent. We obtainThis is the weight-four Eisenstein basis at level two.

Write a finite quiver as $Q=(Q_0,Q_1,s,t)$, with vertex and arrow sets and source/target maps. A representation of a quiver assigns a vector space $X_i$ to each vertex and a linear map $f_{\rho }:X_{s(\rho )}\to X_{t(\rho )}$ to each arrow. A quiver representation morphism $h:X\to Y$ is a family satisfying $h_{t(\rho )}{f}_{\rho }=g_{\rho }{h}_{s(\rho )}$.

The path algebra $A=kQ$ has every directed path, including each length-zero path $e_i$, as a basis. Multiplication is composition when endpoints match, and zero otherwise; in $pq$, the path $q$ is traversed first. The orthogonal idempotents satisfy $1={\sum }_i{e}_i$.

The path-algebra module equivalence is explicit. From a representation, form $M={⨁}_i{X}_i$, let $e_i$ project onto $X_i$, and let each path act by the composite of its arrow maps. Conversely, an $A$-module gives $X_i=e_{i}M$ and $f_{\rho }(x)=\rho x$. An $A$-module homomorphism restricts to the required vertex maps, and a compatible family extends by direct sum. These constructions are mutually inverse up to their evident natural identifications.

$kQ$ is finite-dimensional exactly when $Q$ is finite and has no oriented cycle. For a finite acyclic quiver, paths have length at most $|Q_0|-1$. An oriented cycle has arbitrarily many distinct powers, giving infinitely many basis paths. If arbitrary infinite quivers are allowed, finiteness of both vertices and arrows is also necessary; the unital module correspondence above uses finite $Q_0$.

Choose only the orientation $1\to 2\to 3$. The interval representations of an equioriented three-vertex quiver $I[a,b]$ have $k$ at vertices $a,\dots ,b$, zero elsewhere, and identity arrows within that interval. The complete list isHere is an elementary proof, without the Gabriel theorem. For $X_1\stackrel{f}{\to }{X}_2\stackrel{g}{\to }{X}_3$, set $K=\mathrm{im}f\cap \ker g$. Choose $F$ complementing $K$ in $\mathrm{im}f$, $G$ complementing $K$ in $\ker g$, and $H$ complementing $\mathrm{im}f+\ker g$ in $X_2$. Then $X_2=K\oplus F\oplus G\oplus H$, and $g$ is injective on $F\oplus H$. Lift bases of $K,F$ to a complement of $\ker f$ in $X_1$, and extend the bases of $gF,gH$ to $X_3$. These bases split $X$ into precisely the six kinds of interval block. Every block has endomorphism ring $k$, hence is indecomposable, and their different supports make them pairwise nonisomorphic. The same basis argument handles arbitrary vertex dimensions; each indecomposable block itself is finite-dimensional.

Apply the polynomial method in combinatorics to the finite-field Kakeya set $A$. We will actually obtain the stronger estimateLet $\mathcal{P}$ be the vector space over the prime field ${\mathbb{F}}_p$ of multivariate polynomials in $n$ variables of total degree of a polynomial at most $p-1$. By the dimension of a bounded-total-degree polynomial space, its monomial basis consists of $x_1^{a_1}\cdots x_n^{a_n}$ with $a_i\ge 0$ and ${\sum }_i{a}_i\le p-1$. Adding a slack exponent gives $n+1$ nonnegative exponents summing to $p-1$; the stars and bars count shows thatIf $|A|<\dim \mathcal{P}$, the evaluation linear map $\mathcal{P}\to {\mathbb{F}}_p^A$ has a nonzero element in its kernel of a linear map. Thus there is a nonzero polynomial $P$ of total degree of a polynomial $d\le p-1$ vanishing at every point of $A$. Its degree cannot be zero, since $A$ contains a line and is nonempty. Write $P_d$ for its nonzero top homogeneous polynomial part.

For every nonzero $v\in {\mathbb{F}}_p^n$, the directional hypothesis supplies an affine line in a vector spaceDirections are one-dimensional vector subspaces; rescaling a representative does not change this line. The univariate polynomial $P(a_v+tv)$ has degree at most $d<p$ and vanishes for all $p$ values of $t$. The root bound for a polynomial makes it the zero polynomial. Its coefficient of $t^d$ is exactly $P_d(v)$, so $P_d(v)=0$ for every nonzero $v$. Because $d>0$, $P_d(0)=0$ too.

To finish, we prove the relevant polynomial nonvanishing below the field size. A polynomial over ${\mathbb{F}}_p$ of degree at most $p-1$ in each variable cannot vanish on all of ${\mathbb{F}}_p^n$ unless it is zero. Induct on $n$. The one-variable case is the root bound for a polynomial. For more variables, writeFixing the first $n-1$ coordinates gives a univariate polynomial with $p$ roots of a polynomial, so all its coefficients are zero. Hence every $Q_j$ vanishes on ${\mathbb{F}}_p^{n-1}$, and the induction hypothesis makes every $Q_j$ zero. Applying this to $P_d$, whose total degree of a polynomial is less than $p$, contradicts its choice as nonzero.

Therefore $|A|\ge \dim \mathcal{P}$. Finally,which proves the requested lower bound. The crucial observation in the finite-field Kakeya polynomial bound is that complete lines force the highest homogeneous polynomial part to vanish in every direction.

A nilpotent endomorphism $x$ satisfies $x^N=0$ for some positive integer $N$. A nilpotent Lie algebra $\mathfrak{g}$ has a terminating lower central series of a Lie algebra: with ${\gamma }_1=\mathfrak{g}$ and ${\gamma }_{r+1}=[\mathfrak{g},{\gamma }_r]$, one has ${\gamma }_r=0$ for some $r$. These are different conditions: the first concerns a particular linear map, whereas the second concerns repeated Lie brackets.

The representation form of Engel theorem says that if a finite-dimensional Lie subalgebra $L\subseteq {\mathrm{End}}_k(V)$ consists entirely of nilpotent endomorphisms, with $0<\dim V<\infty$, then $V$ contains a nonzero vector annihilated by every element of $L$. Consequently there is a basis in which every element of $L$ is strictly upper triangular. No assumption on the characteristic of a field or algebraic closure is required.

We prove the common-vector assertion by induction on $\dim L$, simultaneously for every nonzero finite-dimensional representation space. The zero algebra is immediate. First, nilpotence of commutation by a nilpotent endomorphism follows fromIf $x^N=0$, the right-hand side vanishes for $r\ge 2N-1$, in every characteristic of a field. For any proper Lie subalgebra $M⊊L$, its Adjoint representation on $L/M$ therefore consists of nilpotent endomorphisms. Its image has dimension at most $\dim M<\dim L$, so the inductive assertion gives a nonzero class $y+M$ annihilated by $M$. Equivalently $y\notin M$ and $[M,y]\subseteq M$. Thus the Engel normalizer lemma gives $M⊊N_L(M)$.

Choose a maximal proper Lie subalgebra $M$. Its normalizer of a Lie subalgebra must be all of $L$, so $M$ is an ideal of a Lie algebra. Moreover $L/M$ has dimension one: otherwise a one-dimensional Lie subalgebra of this quotient would have a proper inverse image strictly between $M$ and $L$. By induction the space $W=\{v\in V:Mv=0\}$ is nonzero. It is invariant under $L$, since $m(xw)=x(mw)+[m,x]w=0$. Write $L=M+kx$. The restriction of the nilpotent endomorphism $x$ to $W$ has a nonzero kernel, and any vector in that kernel is annihilated by all of $L$. This completes the induction. Apply the assertion repeatedly to $V/kv$ to obtain a complete invariant flag $0=V_0\subset V_1\subset \cdots \subset V_d=V$ with $L{V}_j\subseteq V_{j-1}$. A basis adapted to that flag makes every matrix strictly upper triangular.

A nilpotent abstract Lie algebra need not act by nilpotent matrices. For $V\ne 0$, the scalar algebra $kI\subseteq \mathfrak{gl}(V)$ is an abelian Lie algebra, hence a nilpotent Lie algebra, but $I^N=I\ne 0$. No change of basis makes $I$ strictly upper triangular. The failed implication is precisely the distinction described in nilpotent Lie algebras need not act nilpotently.

If $\mathfrak{g}$ is a nilpotent Lie algebra, then $(\mathrm{ad}x{)}^r\mathfrak{g}\subseteq {\gamma }_{r+1}$, so every $\mathrm{ad}x$ is a nilpotent endomorphism. Conversely, apply Engel theorem to $\mathrm{ad}\mathfrak{g}$ acting on $\mathfrak{g}$. Its invariant flag is lowered by every Adjoint representation matrix, so every product of $d=\dim \mathfrak{g}$ such matrices vanishes. Since ${\gamma }_{d+1}$ is spanned by expressions $\mathrm{ad}{x}_1\cdots \mathrm{ad}{x}_d(y)$, the lower central series of a Lie algebra terminates. ThereforeFinally suppose $k$ is an algebraically closed field. If some $\mathrm{ad}x$ is not a nilpotent endomorphism, it has a nonzero eigenvalue $\lambda$ and an eigenvector $y$, giving $[x,y]=\lambda y$. The vectors $x,y$ are linearly independent and span a nonabelian two-dimensional Lie subalgebra. Conversely, a Lie subalgebra of a nilpotent Lie algebra is nilpotent because its lower central series of a Lie algebra lies termwise in the ambient series. A nonabelian two-dimensional Lie algebra has a basis $u,v$ with $[u,v]=v$: choose $v$ spanning the nonzero derived algebra and rescale a complementary vector. Its Adjoint representation has the nonzero eigenvalue $1$, so it is not nilpotent. This proves the two-dimensional subalgebra criterion for Lie algebra nilpotence:

A representation of a quiver assigns a $k$-vector space $X_i$ to each vertex $i\in Q_0$ and a linear map $f_{\rho }:X_{s(\rho )}\to X_{t(\rho )}$ to each arrow $\rho \in Q_1$. Here $s(\rho )$ and $t(\rho )$ are its source and target. A quiver representation morphism $\theta :X\to Y$ is a family of linear maps ${\theta }_i:X_i\to Y_i$ such that every arrow square commutes:Composition and identities are defined vertex by vertex. A quiver representation morphism is an isomorphism precisely when every ${\theta }_i$ is invertible.

A linear map $T:X\to Y$ between normed spaces is a bounded linear operator if $\Vert Tx\Vert \le C\Vert x\Vert$ for one finite $C$ and every $x$. This implies $\Vert Tx-Ty\Vert \le C\Vert x-y\Vert$, hence continuity. Conversely, continuity at zero supplies $\delta >0$ such that $\Vert x\Vert <\delta$ implies $\Vert Tx\Vert <1$. Apply this to $\delta x/(2\Vert x\Vert )$ to obtain $\Vert Tx\Vert \le 2\Vert x\Vert /\delta$ for nonzero $x$. Thus boundedness and continuity are equivalent for linear maps.

The operator norm is $\Vert T\Vert ={\sup }_{\Vert x\Vert \le 1}\Vert Tx\Vert$, equivalently ${\sup }_{x\ne 0}\Vert Tx\Vert /\Vert x\Vert$. The bounded operators form a vector space, since $\Vert (S+T)x\Vert \le (\Vert S\Vert +\Vert T\Vert )\Vert x\Vert$ and scalar multiples remain bounded. Taking suprema proves the triangle inequality and absolute homogeneity for the operator norm; $\Vert T\Vert =0$ forces $Tx=0$ for every $x$. Thus $B(X,Y)$ is a normed space, whether or not $X,Y$ are complete.

Group the matrix entries by columns. Each column of $BA$ is $B$ times the corresponding column of $A$. In the column-ordered basis of matrix units, the linear map $L_B$ consequently has a block diagonal matrix with $n$ copies of $B$. The determinant of a block diagonal matrix is the product of its block determinants, givingThis argument does not require $B$ to be an invertible matrix.

Right multiplication by $B^{\ast }$ acts separately on the rows. After transposing each row its matrix is $(B^{\ast }{)}^T=\overline{B}$, so its determinant is $(\overline{\det B}{)}^n$. Composing this linear map with left multiplication and using part (a) givesHere the determinant is taken on the $n^2$-dimensional complex vector space of matrices.

If $A$ is a Hermitian matrix, then $(BA{B}^{\ast }{)}^{\ast }=BA{B}^{\ast }$, so $M_B$ preserves $W$. Addition and multiplication by real scalars commute with this operation, making $T_B$ a real linear map.

A real basis of $W$ consists of $E_{ii}$ and, for $i<j$, $E_{ij}+E_{ji}$ and $i(E_{ij}-E_{ji})$. It has $n^2$ elements and is also a complex basis of all matrices. Indeed every matrix is uniquely $H+iK$ with $H,K$ Hermitian matrices, by takingReal linear independence of the displayed basis therefore implies complex linear independence. Since its images under $M_B$ are Hermitian matrices, the matrix of $M_B$ in this complex basis has exactly the same real entries as the matrix of $T_B$ in the real basis. Their determinants are equal. There is no further squaring when passing to this real subspace:This is the determinant of Hermitian congruence, including the case of singular $B$.

For a linear map $A:{\mathbb{R}}^m\to {\mathbb{R}}^n$, define the operator norm byIt is finite. To see this without assuming continuity of $A$, write $x=\sum x_j{e}_j$: the triangle inequality bounds $\Vert Ax\Vert$ by $\sum |x_j|\Vert A{e}_j\Vert$. Every norm in finite dimensions is equivalent to the Euclidean norm. One proof first uses this same coordinate bound to establish continuity of the norm on the Euclidean unit sphere; compactness then gives a strictly positive minimum there. Thus the coordinate bound above is at most a fixed multiple of $\Vert x\Vert$.

Absolute homogeneity and the triangle inequality for this operator norm follow from those for $\Vert Ax\Vert$ and taking suprema. If $\Vert A\Vert =0$, then $Ax=0$ for every $x$, hence $A=0$, and the converse is immediate. For any $B:{\mathbb{R}}^{\ell }\to {\mathbb{R}}^m$, the defining bound $\Vert Ax\Vert \le \Vert A\Vert \Vert x\Vert$ givesThis is submultiplicativity of the operator norm.

Differentiability here is understood in the sense of the Fréchet derivative: there is a linear map $L:{\mathbb{R}}^n\to {\mathbb{R}}^m$ such thatThe map $L=Df(x)$ is the Fréchet derivative. In finite dimensions a linear map is bounded, so $\Vert Lh\Vert \le C\Vert h\Vert$ for some $C$. The defining estimate therefore impliesThis is precisely continuity at $x$: differentiability implies continuity.

First suppose each $g_j$ is differentiable at $x_j$. Write $g_j(x_j+t)=g_j(x_j)+v_{j}t+r_j(t)$, where $v_j\in {\mathbb{R}}^m$ and $r_j(t)=o(|t|)$. Define the linear map $Lh={\sum }_j{v}_j{h}_j$. Given $\epsilon >0$, choose a common sufficiently small neighbourhood so that $\Vert r_j(h_j)\Vert \le \epsilon |h_j|$ for all $j$, with $r_j(0)=0$. ThenThis proves differentiability, withConversely, suppose $f$ is differentiable with Fréchet derivative $L$. Along the $j$th coordinate direction $e_j$, the other summands cancel exactly:Thus $g_j$ is differentiable at $x_j$, with $g_j^{\prime }(x_j)=L{e}_j$. The equivalence works because each summand varies in only one coordinate; it is stronger than merely having partial derivatives.

Two norms $N_1,N_2$ on the same vector space are equivalent norms, or Lipschitz equivalent, if there are constants $0<c\le C<\infty$ such that $c{N}_1(v)\le N_2(v)\le C{N}_1(v)$ for every $v$. The constants are uniform over the whole space.

Let $N$ be any norm on ${\mathbb{R}}^n$. The triangle inequality and the Cauchy-Schwarz inequality giveThe reverse triangle inequality then gives $|N(x)-N(y)|\le C\Vert x-y{\Vert }_2$, so $N$ is Euclidean-continuous; this continuity was derived, rather than assumed. On the compact Euclidean unit sphere, $N$ attains a strictly positive minimum $c$, because it vanishes only at zero. Homogeneity yieldsFinally a linear map $A:{\mathbb{R}}^n\to {\mathbb{R}}^m$ satisfies $\Vert Ax{\Vert }_2\le ({\sum }_{i,j}|a_{ij}{|}^2{)}^{1/2}\Vert x{\Vert }_2$, again by Cauchy-Schwarz inequality. The equivalence just proved transfers this bound to any norms on the source and target. Hence every such linear map is continuous.

First $w\in {\ell }^2$, since ${\sum }_{n\ge 1}{4}^{-n}=1/3$. Any finite linear dependence among $w$ and the $e_m$ has the form $aw+{\sum }_{m\in F}{b}_m{e}_m=0$ for a finite index set $F$. At an index outside $F$, its coordinate equals $a{2}^{-n}$, so $a=0$. Each remaining coordinate then gives $b_m=0$. Therefore the set is linearly independent, and the prescribed values define a unique linear map on its algebraic span.

Take the tailsThey belong to $V$, satisfy $f(v_N)=1$, and haveThus $v_N\to 0$ while $f(v_N)$ does not approach $f(0)=0$. The map is not continuous, and part (b) also shows it is unbounded. This is a discontinuous functional detected by vanishing sequence tails, illustrating why the finite-dimensional conclusion in part (a) cannot be extended to arbitrary normed spaces.

The Weak Hilbert Nullstellensatz states that if $k$ is an algebraically closed field, every maximal ideal of $k[Y_1,\dots ,Y_s]$ isEquivalently, every proper ideal has a common zero. We prove the more general field statement, the Zariski lemma: a field finitely generated as a $k$-algebra is a finite field extension of $k$.

Write such a field as $L=k[b_1,\dots ,b_s]$. Choose a maximal set of algebraically independent elements $t_1,\dots ,t_r$ among these generators and denote the remaining generators by $c_1,\dots ,c_h$. Each $c_j$ is an algebraic element over $K=k(t_1,\dots ,t_r)$, so $L/K$ is a finite field extension. Clearing the denominators in their monic equations, choose a nonzero $d\in k[t_1,\dots ,t_r]$ such that each $c_j$ is an integral element overBecause $L$ is a field and $d^{-1}\in L$, we have $L=D[c_1,\dots ,c_h]$. Thus $L$ is an integral extension of $D$.

A subring over which a field is integral must itself be a field. Indeed, for $0\ne a\in D$, an integral equation for $a^{-1}$ multiplied by $a^{m-1}$ expresses $a^{-1}$ as an element of $D$. But if $r>0$, the ring $D$ is not a field: choose an irreducible polynomial $q(t_1)\in k[t_1]$ which does not divide $d$. There are infinitely many monic irreducible polynomials in $k[t_1]$, by the usual product-plus-one argument, whereas only finitely many can divide the nonzero $d$. The prime ideal $(q)$ stays proper after inverting $d$, so $q$ has no inverse in $D$. Therefore $r=0$, proving that $L/k$ is a finite field extension, and in particular an algebraic field extension.

Apply the Zariski lemma to the residue field $k[Y_1,\dots ,Y_s]/\mathfrak{m}$ of a maximal ideal. When $k$ is an algebraically closed field, this field is $k$ itself. The images $a_i$ of $Y_i$ then give $\mathfrak{m}=(Y_1-a_1,\dots ,Y_s-a_s)$, proving the Weak Hilbert Nullstellensatz.

For the real Laurent polynomial ring, the same argument applies because it is the finitely generated algebraEvery residue field at a maximal ideal is consequently a finite field extension of $\mathbb{R}$, and so is $\mathbb{R}$ or $\mathbb{C}$. Here we use the fundamental theorem of algebra, which makes $\mathbb{C}$ the algebraic closure of $\mathbb{R}$ of degree two. The images of $X_1,X_2$ must be nonzero since these elements are units. It follows that the answer iswith $(\alpha ,\beta )$ and $(\overline{\alpha },\overline{\beta })$ defining the same maximal ideal. If both coordinates are real, evaluation has image $\mathbb{R}$; otherwise its image is $\mathbb{C}$. Thus every displayed kernel is maximal, and the Zariski lemma shows that this list is exhaustive.

For explicit generating sets of an ideal, there are three cases. For real nonzero $a,b$,If $\alpha =a\in {\mathbb{R}}^{\times }$ and $\beta =s+it$ with $t\ne 0$,If $\alpha =u+iv$ with $v\ne 0$ and $\beta =s+it$, thenThe last two quotients are $\mathbb{C}$; the nonzero coordinate condition ensures that the Laurent inverses exist in them. There are no further coincidences in the classification: equal kernels induce an $\mathbb{R}$-isomorphism between the evaluation image fields. The only $\mathbb{R}$-automorphisms of $\mathbb{C}$ are the identity and complex conjugation, so the two pairs must agree or be conjugate.

To prove the intersection assertion, fix a prime ideal $P$ and an element $f\notin P$. The integral domain $B=R/P$ has nonzero image $\bar{f}$. Its localization of a ring $B[{\bar{f}}^{-1}]$ is nonzero and still a finitely generated algebra over $\mathbb{R}$, so it has a maximal ideal $\mathfrak{n}$. By the Zariski lemma, the field $L=B[{\bar{f}}^{-1}]/\mathfrak{n}$ is finite over $\mathbb{R}$.

The image of $B$ in $L$ is a subring containing $\mathbb{R}$ and is a finite-dimensional $\mathbb{R}$-vector space. It is an integral domain, hence a field: multiplication by any nonzero element is an injective linear map of a finite-dimensional vector space and therefore surjective. The kernel of $B\to L$ is thus a maximal ideal which avoids $\bar{f}$, since $\bar{f}$ is invertible in $L$. Its inverse image in $R$ is a maximal ideal containing $P$ but avoiding $f$. As this works for every $f\notin P$,This is the Jacobson ring property for this Laurent polynomial ring.

For a nonzero finitely generated module $M$, its set of associated primes isHere ${\mathrm{Ann}}_R(m)=\{a\in R:am=0\}$ is the annihilator of the element. The Laurent polynomial ring $R$ is a Noetherian ring, so the nonempty collection of annihilators of nonzero elements of $M$ has a maximal member, say $P={\mathrm{Ann}}_R(m)$. It is proper because $m\ne 0$.

If $ab\in P$ and $b\notin P$, then $bm\ne 0$ andMaximality forces ${\mathrm{Ann}}_R(bm)=P$, hence $a\in P$. Thus $P$ is a prime ideal, proving $\boxed{{\mathrm{Ass}}_R(M)\ne \varnothing }$. This argument in fact only needs that $R$ is Noetherian, not the finite generation of $M$.

A proper ideal $Q$ is $P$-primary when its radical of an ideal is $P$ andEquivalently, $ab\in Q$ and $b\notin P$ imply $a\in Q$; equivalently, every zero divisor of $R/Q$ is nilpotent. These formulations distinguish the actual primary ideal property from the weaker assertion that there is only one minimal prime ideal.

For the final example, set $u=X_1-1$ and $v=X_2-1$. ThenSince $R/P\cong \mathbb{R}[X_2,X_2^{-1}]$ is an integral domain, $P$ is a prime ideal. Moreover $I\subseteq P$ and $u^2\in I$, so $\sqrt{I}=P$. Consequently $P$ is the unique minimal prime ideal over $I$.