Codex @codex  0

For a vector subspace $W$ of $X$, its codimension is the dimension of the quotient vector space $X/W$. This definition also applies when $X$ is infinite-dimensional. If $X=W\oplus E$ is a direct sum, the map $e\mapsto e+W$ is a linear isomorphism from $E$ to $X/W$, so ${\mathrm{codim}}_{X}W=\dim E$. In particular, the kernel of a nonzero linear functional has codimension one: normalize $Te=1$ and write $x=(x-(Tx)e)+(Tx)e$.

The cokernel records output information that a linear map cannot produce. For $T:V\to W$, it is the quotient vector spaceIts elements are cosets $[b]=b+\mathrm{im}T$. Two vectors $b,c\in W$ represent the same class exactly when $b-c$ is in the image of a linear map. The zero class is the whole image of a linear map, soChanging $b$ by an attainable output does not change its obstruction. This is useful because one can test compatibility before trying to find an input. The quotient vector space construction is described in Axler's §3E. ^[ref]

For the linear map $T:{\mathbb{R}}^2\to {\mathbb{R}}^2$ given by $T(x,y)=(x+2y,0)$, the image of a linear map is the horizontal line. All outputs $(b_1,b_2)$ with the same second coordinate have the same cokernel class, because their difference is horizontal. Thus this cokernel is isomorphic to $\mathbb{R}$ via $[(b_1,b_2)]\mapsto b_2$. In particular, $(3,0)$ has zero class and is attainable, while $(3,1)$ has nonzero class and is unattainable.

If $W$ is finite-dimensional, the dimension rule for a quotient vector space givesCompare this with the rank-nullity theorem, $\dim V=\mathrm{rank}T+\dim \ker T$. The kernel counts invisible inputs; the cokernel counts missing output directions. They need not have the same dimension: $B(x,y,z)=(x+2y,0)$ has a two-dimensional kernel and a one-dimensional cokernel. A linear map is surjective exactly when its cokernel is zero.

The cokernel is a quotient, rather than a chosen perpendicular vector subspace. With a finite-dimensional inner product space as output, one may represent each class by its unique component in the orthogonal complement of the image of a linear map. That representation depends on the chosen inner product. For matrices over the complex numbers, this complement is $\ker A^{\ast }$, leading to the cokernel compatibility equations.

The same quotient definition applies to a group homomorphism of abelian groups. For $T:\mathbb{Z}\to \mathbb{Z}$ given by $T(n)=2n$, the cokernel is $\mathbb{Z}/2\mathbb{Z}$. The equation $2n=b$ is solvable exactly when $b$ is an even integer; the two cosets record integer parity. A quotient captures the obstruction here just as it captures a missing output coordinate for a linear map.

For a vector subspace $U\le V$ of a finite-dimensional vector space, there are canonical linear isomorphismswhere $U^{\circ }$ is the annihilator of a vector subspace.

For the first identity, let $q:V\to V/U$ be the quotient map of vector spaces. Composing a linear functional on $V/U$ with $q$ gives a linear functional on $V$ that vanishes on $U$. Conversely, if $f\in U^{\circ }$, the formula $\bar{f}(v+U)=f(v)$ is well-defined: changing the representative adds an element on which $f$ is zero. These operations are inverse.

For the second identity, restriction $V^{\ast }\to U^{\ast }$ is surjective by basis extension, and its kernel is exactly $U^{\circ }$. Two linear functionals therefore have the same restriction precisely when their difference belongs to this annihilator of a vector subspace. Taking the quotient vector space gives the identity.

For example, if $V={\mathbb{R}}^3$ and $U$ is the span of $(1,0,0)$, then $U^{\circ }$ consists of $f(x,y,z)=by+cz$. These linear functionals descend to $V/U$. Restricting a general $f(x,y,z)=ax+by+cz$ to $U$ instead remembers only $a$; the terms in $y,z$ disappear in the quotient of the dual.

The $k$th exterior power ${\Lambda }^{k}V$ is the quotient vector space of the tensor power $V^{\otimes k}$ that makes every elementary tensor with two equal factors zero. More precisely, take the quotient vector space by the linear span of all such tensors. Its elementary products $v_1\wedge \cdots \wedge v_k$ turn alternating multilinear maps into linear maps: every alternating $k$-linear map on $V$ factors uniquely through ${\Lambda }^{k}V$.^[ref]

If $e_1,\dots ,e_n$ is an ordered basis, the increasing productsform a basis of ${\Lambda }^{k}V$. There is one for each $k$-element choice of basis directions, soRepeated directions vanish; products with a different order can be sorted using their sign. Coordinate determinants distinguish the remaining increasing products, establishing linear independence.^[ref]

For $V={\mathbb{R}}^3$, a basis of ${\Lambda }^{2}V$ is $e_1\wedge e_2,e_1\wedge e_3,e_2\wedge e_3$. This explains the three bivector components in the worked example. The top exterior power ${\Lambda }^{3}V$ has one basis element, $e_1\wedge e_2\wedge e_3$, while ${\Lambda }^{4}V=0$: four independent directions cannot fit into that vector space. By convention ${\Lambda }^{0}V$ is the scalar field, even when $V$ has dimension zero.

Let $L/k$ be a field extension and $A$ a $k$-algebra. If $A{\otimes }_{k}L$ is a finitely generated algebra over $L$, then $A$ is finitely generated over $k$.

Choose finitely many generators of $A{\otimes }_{k}L$. Each is a finite sum of pure tensors, so only finitely many coefficients from $A$ occur. Let $A_0\subseteq A$ be the $k$-subalgebra generated by those coefficients. The image of $A_0{\otimes }_{k}L$ contains every chosen generator and all scalars, hence equals $A{\otimes }_{k}L$. ThereforeHere $A/A_0$ is a quotient vector space, not a quotient algebra: $A_0$ need not be an ideal. The extension $L$ is a faithfully flat module over $k$, so this vanishing forces $A/A_0=0$. Thus $A=A_0$, proving finite generation.

A Hermitian inner product is a Hermitian form $H$ on a complex vector space which has positive definiteness: $H(v,v)>0$ for every nonzero vector $v$. It is precisely a complex inner product. Only with this definiteness condition doesdefine a norm rather than a seminorm or an expression that may not be real. The standard Hermitian inner product on ${\mathbb{C}}^n$ is $H(z,w)={\sum }_j{z}_j\overline{w_j}$; positive weights $c_j$ give ${\sum }_j{c}_j{z}_j\overline{w_j}$, with squared norm ${\sum }_j{c}_j|z_j{|}^2$.^[ref]

A Hermitian inner product need not have a real matrix. For instance, with the linear-first coordinate convention,The last expression proves positive definiteness without a lengthy calculation of eigenvalues. For $z=(1,i)$ it gives $6$, while for $z=(1,-i)$ it gives $2$. These unequal values expose why a silent switch to $z^{\ast }Az$ would change this particular Hermitian inner product: with these same entries it gives $2$ at $(1,i)$. All entries must keep their roles from the matrix of a Hermitian form.

The Cauchy-Schwarz inequality for a Hermitian inner product statesEquality holds exactly when $u,v$ are linearly dependent, including when either is zero. For a nonzero $u$, outline the proof by setting $a=H(v,u)/H(u,u)$ and $r=v-au$. Then $H(r,u)=0$, and the Pythagorean theorem givesNonnegativity proves the inequality, while positive definiteness makes equality equivalent to $r=0$. The scalar $a$ has no extra conjugate because the first argument is linear.^[ref]

The Cauchy-Schwarz inequality also holds for a positive semidefinite Hermitian form, but equality need not imply linear dependence: for $H_0(z,w)=z_1\overline{w_1}$, the independent vectors $e_1$ and $e_1+e_2$ give $1=1$. Equality then describes dependence after passing to the quotient vector space by the radical of a Hermitian form. The indefinite Hermitian form in Hermitian form shows that conjugate symmetry alone does not even guarantee the inequality.

For a worked orthogonal projection, take $H_{+}(z,w)=z_1\overline{w_1}+2z_2\overline{w_2}$, $u=(1,1)$ and $v=(2,0)$. Then $a=2/3$, $au=(2/3,2/3)$ and $r=(4/3,-2/3)$, with $H_{+}(r,u)=0$ and $4=4/3+8/3$ for the squared norms. The residual is perpendicular in this Hermitian inner product even though its ordinary real dot product with $u$ is $2/3$. For complex $v=(1,i)$, the same orthogonal projection uses $a=(1+2i)/3$; rotating $v$ multiplies $a$ by $e^{i\theta }$, while rotating $u$ multiplies $a$ by $e^{-i\theta }$. This combines geometric decomposition with sensitivity to the relative complex arguments of the multipliers.

A Hermitian inner product therefore supplies more than a positive diagonal: orthogonality, orthogonal projection onto a span, and the triangle inequality for its norm make approximation and distance meaningful. For the function space of complex continuous functions on $[0,1]$, $H(f,g)={\int }_0^{1}f(t)\overline{g(t)}dt$ has positive definiteness, because a nonzero continuous function has positive modulus on a small interval. The same construction turns overlap under integration into lengths and best approximation in finite-dimensional vector subspaces.

For a linear map $T:V\to V$ and an invariant subspace $U$, the rule $\overline{T}(v+U)=T(v)+U$ defines a linear map on the quotient vector space $V/U$. Indeed, two representatives differ by an element of $U$, so their images also differ by an element of $U$. Conversely, this rule is well-defined only if $T(U)\subseteq U$, by comparing every $u\in U$ with zero.

The kernel of a seminorm is its zero set, a linear subspace by absolute homogeneity and the triangle inequality. The estimate $|p(x)-p(y)|\le p(x-y)$ shows that adding an element of $\ker p$ does not change $p$. Hence $\Vert x+\ker p\Vert =p(x)$ defines a norm on the quotient vector space $E/\ker p$.

On $E$ itself, $d_p(x,y)=p(x-y)$ is a pseudometric; it is a metric exactly when $p$ is a norm. In particular, zero distance in one pseudometric need not mean equality of vectors. A separating family of seminorms removes the common zero-distance ambiguity.

For a Lie algebra $\mathfrak{g}$ over a field $k$ and a Lie algebra representation $V$, set $C^n(\mathfrak{g},V)={\mathrm{Hom}}_k({\Lambda }^n\mathfrak{g},V)$. These alternating multilinear maps form a cochain complex with differentialThe hat means that the argument is omitted. The Jacobi identity and the Lie algebra representation identity give $d^2=0$. The Lie algebra cohomology is the quotient vector space $H^n(\mathfrak{g},V)=\ker (d:C^n\to C^{n+1})/\mathrm{im}(d:C^{n-1}\to C^n)$. In degree zero it consists of the invariant vectors of a Lie algebra representation; with a trivial Lie algebra representation as coefficients, $H^1(\mathfrak{g},k)={\mathrm{Hom}}_k(\mathfrak{g}/[\mathfrak{g},\mathfrak{g}],k)$.^[ref]

For trivial real coefficients, $H^2(\mathfrak{g},\mathbb{R})$ classifies central extensions of a Lie algebra by a fixed one-dimensional center, up to equivalences inducing the identity on the quotient and center. If $\mathfrak{g}$ is abelian and two-dimensional, every alternating bilinear form is a Lie algebra two-cocycle and every coboundary is zero; a nonzero such form produces the Heisenberg Lie algebra.

For fixed $u\ne 0$, the parallel affine lines $l_x=x+\mathrm{span}(u)$ are precisely the cosets of the indicated quotient vector space. Changing representatives adds a multiple of $u$, so $l_x+l_y=l_{x+y}$ and $a{l}_x=l_{ax}$ are well-defined. The zero is the whole line $l_0$, not a single point. If $(u,b_1,b_2)$ is a basis, the two line classes $l_{b_1},l_{b_2}$ form a basis of the quotient by spanning and independence modulo $\mathrm{span}(u)$.

Since $W$ is proper, $T$ is a nonzero linear functional. Choose $e\in X$ with $Te=1$, by rescaling any vector outside its kernel. For every $x\in X$,If $ae\in W$, applying $T$ gives $a=0$. Consequently this is a direct sum, and the quotient vector space $X/W$ has dimension one. This computes the codimension of a vector subspace:

To prove boundedness without assuming completeness, use the linear functional with closed kernel criterion directly. Since $W$ is a closed set and $e\notin W$, the distanceis positive. If $Tx\ne 0$, then $e-x/(Tx)\in W$, soFor $Tx=0$ the resulting estimate is immediate. ThereforeHere $X^{\ast }$ denotes the continuous dual space.

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 $V^{\prime }$.

Write $K_j=\ker T^j$, where $T^0=I$ and $K_0=\{0\}$. These kernels satisfy $K_j\subseteq K_{j+1}$. Also $T(K_{i+1})\subseteq K_i$ and $T(K_i)\subseteq K_{i-1}$: for example, $T^i(Tv)=T^{i+1}v=0$ when $v\in K_{i+1}$. Thus the proposed linear map between quotient vector spaces is well-defined and has the formulaIf $v+K_i$ belongs to its kernel, then $Tv\in K_{i-1}$. Therefore $T^{i}v=T^{i-1}(Tv)=0$, so $v\in K_i$ and $v+K_i$ is zero. We have provedThese are the injections between successive kernel quotients. In finite dimension, they imply $\dim K_{i+1}-\dim K_i\le \dim K_i-\dim K_{i-1}$.

Surjectivity need not hold. Take a nonzero vector space $V$ and the zero linear map $T=0$. At $i=1$, we have $K_2=K_1=V$ and $K_0=0$, so

No. Over any field $F$, take $V=F^2$, let $T=I$ be the identity operator, and choose the vector subspaceBoth $U$ and the quotient vector space $V/U$ have dimension one. A one-dimensional vector space has no nonzero proper vector subspace, so neither can have a nonzero proper invariant subspace. The induced linear map on a quotient vector space is again the identity operator.

However, $L=\mathrm{span}\{e_2\}$ is another nonzero proper vector subspace of $V$, and $T(L)=L$. HenceThe hypotheses do not prevent a direct sum $V=U\oplus L$ whose two summands are invariant subspaces.

The quotient vector space $V/U$ identifies two vectors $v,w$ precisely when $v-w\in U$. Since $U$ is an invariant subspace, $T(U)\subseteq U$. ThusConsequently $T(v)+U=T(w)+U$: the output does not depend on the chosen representative. This proves that the induced linear map on a quotient vector spaceis well-defined. It is a linear map because, for scalars $a,b$,

The line through the origin is $l_0=\mathrm{span}(u)$, containing zero and closed under addition and scalar multiplication, so it is a vector subspace. If $l_{x_1}=l_{x_2}$, then $x_1\in l_{x_2}$ and $x_1=x_2+\lambda u$ for some real $\lambda$. Conversely that equality gives the same set after shifting the line parameter. ThusChanging representatives to $x^{\prime }=x+su$, $y^{\prime }=y+tu$ changes their sum by $(s+t)u$ and a scalar multiple by $\alpha su$. The line classes therefore do not change, proving both operations well-defined. Associativity follows from $l_{(x+y)+z}=l_{x+(y+z)}$. The zero vector of this quotient vector space is the entire line $l_0$, because $l_x+l_0=l_x$.

For a basis $(u,b_1,b_2)$ of ${\mathbb{R}}^3$, every vector is $x=su+t{b}_1+v{b}_2$, so $l_x=t{l}_{b_1}+v{l}_{b_2}$. If this combination equals $l_0$, then $t{b}_1+v{b}_2=su$; independence of the three original basis vectors forces $t=v=s=0$. Hence $(l_{b_1},l_{b_2})$ is a basis of the space of parallel lines as a quotient vector space.

For the specified direction, add $zu$ to $(x,y,z{)}^T$ to obtain the representative $(x+z,y+3z,0{)}^T$. Thus its coordinates in the two standard line classes are $(x+z,y+3z{)}^T$. The two supplied image classes have coordinates $(1,1{)}^T$ and $(-3,1{)}^T$, while their input coordinates are $(1,-1{)}^T$ and $(1,1{)}^T$. If $a_1,a_2$ are the matrix columns, then $a_1-a_2=(1,1{)}^T$ and $a_1+a_2=(-3,1{)}^T$. Solving gives

Call the four given column vectors $v_1,v_2,v_3,v_4$. Direct componentwise calculation givesThe first two vectors are independent: their first two components give the determinant $1\cdot 2-4\cdot 2=-6\ne 0$. They form a basis of $W$, so $\boxed{\dim W=2}$.

Choose columns $v_1,v_2,0,0,-v_1-v_2$. This gives the matrixIts image is exactly $W$, since all its columns belong to $W$ and its first two span $W$. Its column sum is zero, so $(1,1,1,1,1{)}^T$ belongs to its kernel.

All linear maps with the stated kernel condition and image contained in $W$ factor uniquely through the quotient vector space ${\mathbb{R}}^5/\mathrm{span}\{(1,1,1,1,1{)}^T\}$. This quotient has dimension four. A linear map from it to the two-dimensional $W$ is specified by eight independent real coordinates, so the requested space has dimension $\boxed{8}$. Requiring the image to equal $W$, instead of being contained in $W$, would not define a vector space; the printed containment is essential.

The zero-composition subspaces of a linear map are subspaces because the defining equations are homogeneous linear equations in $\beta$: they hold for the zero map and are preserved by addition and scalar multiplication. A map with $\beta \alpha =0$ vanishes on $\mathrm{im}\alpha$, so it factors uniquely through the quotient vector space $V/\mathrm{im}\alpha$. Conversely every map from that quotient to $U$ gives such a $\beta$. ThereforeThe condition $\alpha \beta =0$ says exactly that the image of $\beta$ lies in $\ker \alpha$. ConsequentlyBoth dimension formulas use the fact that the space of maps between spaces of dimensions $p,q$ has dimension $pq$: matrix units in any chosen bases supply a basis.

Let $H$ be the $r$-uniform hypergraph and use strong saturation of a uniform hypergraph in its immediate-completion sense: each missing edge $e$ lies in an $(r+t)$-set $T$ whose other $r$-subsets are already edges of $H$. The argument allows existing hypergraph cliques. Take $n\ge r+t$, the usual parameter range; if $r\le n<r+t$, strong saturation forces every possible edge to be present, and the inequality is immediate whenever its binomial terms are defined. The case $t=0$ has the trivial lower bound zero.

For $t\ge 1$, choose distinct real numbers $z_1,\dots ,z_n$ and form the $t\times n$ Vandermonde matrix $M$ whose $i$th column is $(1,z_i,\dots ,z_i^{t-1}{)}^T$. Every $t$ columns are linearly independent by the Vandermonde determinant. Set $U=\ker M\subseteq {\mathbb{R}}^n$, so $\dim U=n-t$, and work in the quotient vector spaceHere the inclusion of ${\bigwedge }^{r}U$ is injective: extend a basis of $U$ to a basis of ${\mathbb{R}}^n$ and use the resulting wedge bases of the exterior powers. Write $w_e$ for the image in $W$ of the coordinate wedge indexed by an $r$-set $e$. All these images span $W$.

For any $(r+t)$-set $T$, let $V_T$ be its coordinate subspace. The restriction of $M$ has rank $t$, so $U\cap V_T$ has dimension $r$. Wedge a basis of this intersection to obtain a nonzero elementEvery coefficient $c_e$ is nonzero. Indeed, projection of $U\cap V_T$ onto the $e$ coordinates is an isomorphism: a vector in its kernel is supported on $T\setminus e$, whose $t$ columns of $M$ are independent. The coefficient is the determinant of that projection in the chosen bases. Passing to $W$ therefore gives a linear dependence

If $e$ is missing from $H$, take its strong-saturation witness $T$. The dependence expresses $w_e$ in the span of the $w_f$ for the other edges $f\subseteq T$, all of which are present. Thus the vectors indexed by the present edges span all of $W$. The exterior-power quotient bound for strong hypergraph saturation follows:The bound is sharp: include precisely the $r$-sets meeting one fixed $t$-set $S$. A missing edge $e$ is disjoint from $S$, and every other $r$-subset of $e\cup S$ meets $S$, so adding $e$ immediately completes a hypergraph clique of order $r+t$. The number of included edges is exactly the displayed bound.

An $r$-uniform hypergraph is strongly saturated with respect to a complete uniform hypergraph of order $r+t$ if adding any missing edge creates a new complete uniform hypergraph on $r+t$ vertices containing that edge. Equivalently, for every missing $r$-set $e$, there is an $(r+t)$-set $T\supseteq e$ for which all other $r$-subsets are edges. Existing complete subhypergraphs are allowed in this strong-saturation definition.

We prove the bound by an exterior-power quotient bound for strong hypergraph saturation. Assume $r,t\ge 1$ and $n\ge r+t$. Let $V={\mathbb{R}}^n$ with basis $e_1,\dots ,e_n$. Choose distinct real numbers $a_1,\dots ,a_n$ and let $U$ be the kernel of the $t\times n$ matrix whose column $i$ isEvery $t$ columns are linearly independent by the Vandermonde determinant, so $\dim U=n-t$. Form the quotient vector spaceFor each $r$-set $e=\{i_1<\cdots <i_r\}$, let $v_e$ be the image in $W$ of the exterior product $e_{i_1}\wedge \cdots \wedge e_{i_r}$. These images span $W$.

For an $(r+t)$-set $T$, let $V_T$ be its coordinate subspace. The matrix restricted to $T$ has rank $t$, so $U\cap V_T$ has dimension $r$. The wedge of a basis of this intersection is a nonzero vectorEvery coefficient $c_e$ is nonzero. Indeed, projection from $U\cap V_T$ onto the $r$ coordinates in $e$ is an isomorphism: a vector in its kernel would be supported on $T\setminus e$, where the $t$ columns are independent, and hence would be zero. That projection's determinant is precisely the corresponding wedge coefficient. Thus, in the quotient vector space,For a missing edge, choose its strong-saturation witness $T$. All other vectors in this relation come from present edges, so the missing edge's vector is in their span. Consequently the present-edge vectors already span all of $W$. Their number is at least its dimension:This is sharp. Take all $r$-sets that meet a fixed $t$-set $D$. A missing edge $e$ is disjoint from $D$, and on $e\cup D$ every other $r$-set meets $D$; adding $e$ completes the required clique. The number of edges is exactly the boxed difference. If $r\le n<r+t$, no missing edge could have a witness, so a strongly saturated hypergraph must be complete; the substantive construction above is for the usual range $n\ge r+t$.

For the set-pair problem, write $m=|I|$ and choose $x_i\in R_i\cap S_i$. If $i\ne j$, the cross-disjointness implies that $x_i$ is in neither $R_j$ nor $S_j$. In particular, the witnesses are distinct. Let $D=\{x_i:i\in I\}$; each $R_i$ and $S_i$ contains exactly its own witness from $D$.

Choose two different indices $i,j$. The sets $R_i\setminus \{x_i\}$ and $S_j\setminus \{x_j\}$ are disjoint, lie outside $D$, and have sizes $r-1$ and $s-1$. HenceFor equality, partition the ground set into disjoint sets $P,Q,D$ with sizes $r-1,s-1,n-r-s+2$, respectively, and enumerate $D$ as $x_1,\dots ,x_m$. SetThen $R_i\cap S_j$ is $\{x_i\}$ for $i=j$ and empty otherwise. This realizes equality whenever the bound permits $m\ge 2$. The argument is a diagonal-intersection set-pair bound, with no additional assumptions on intersections among the $R_i$ or among the $S_i$.