Finite-dimensional spaces, quotients and duality

Unless otherwise stated, $F$ is a field and every space in a dimension formula is finite-dimensional.

1. Why vectors need not be columns

A vector space is a set with addition and scalar multiplication satisfying the usual abelian-group and distributive axioms over a field $F$. Columns are one model. Polynomials of degree at most $d$, denoted $P_d(F)$, and linear maps between spaces are others. The zero polynomial is included in $P_d(F)$ although its degree is not ordinarily defined.

A subspace contains zero and is closed under sums and scalar multiples. A list spans if every vector is a linear combination of its entries; it is independent if only the zero coefficients give zero. A basis is both. A finite spanning list can be reduced to a basis by deleting dependent entries; an independent list can be extended to a basis by adding vectors outside its span until the given finite spanning list is spanned.

Exchange lemma. An independent list of $r$ vectors in the span of $s$ vectors has $r\le s$. To see this, express the first independent vector in the spanning list, use a nonzero coefficient to replace one spanning vector by it, and repeat. At each stage independence ensures the next independent vector has a nonzero coefficient on an as-yet unreplaced spanning vector. There must be at least $r$ such slots. Comparing two bases in both directions proves that their lengths agree; this common length is $\dim V$.

Consequently, in dimension $n$, any $n$ independent vectors form a basis, as do any $n$ spanning vectors. Counting a list is useful only after its independence or spanning has been justified.

For subspaces $U,W$, their sum is $U+W=\{u+w\}$. It is a direct sum, written $U\oplus W$, precisely when $U\cap W=\{0\}$, equivalently each vector in the sum has a unique decomposition. Extend a basis of $U\cap W$ separately to bases of $U$ and $W$; the combined extended list gives

$$\dim(U+W)=\dim U+\dim W-\dim(U\cap W).$$

To check independence, move the $U$ and $W$ combinations to opposite sides: the common vector lies in the intersection, so the extra basis coefficients vanish, then the intersection coefficients vanish. Every subspace of a finite-dimensional space has a complement by basis extension, but the complement is usually not unique or canonical.

2. Linear maps and dimension bookkeeping

A map $T:V\to W$ is linear if $T(\alpha v+\beta u)=\alpha Tv+\beta Tu$. Its kernel and image are subspaces. Choose a basis $k_1,\ldots,k_r$ of $\ker T$ and extend to a basis with $v_1,\ldots,v_s$. Then $Tv_1,\ldots,Tv_s$ form a basis of $\operatorname{im}T$: spanning follows by expanding any input; a dependence among these images places the corresponding input combination in the kernel, and independence of the extended basis makes all coefficients zero. Thus

$$\dim V=\dim\ker T+\dim\operatorname{im}T.$$

This is rank–nullity. In equal finite dimensions, injectivity and surjectivity are equivalent, but the equivalence can fail in infinite-dimensional spaces.

A basis $B=(b_1,\ldots,b_n)$ identifies $V$ with $F^n$ through its coordinate isomorphism. For bases $B,C$, the matrix $[T]_{C\leftarrow B}$ has columns $[Tb_j]_C$. For an endomorphism, changing basis by $[v]_B=P[v]_{B'}$ gives

$$[T]_{B'}=P^{-1}[T]_BP.$$

This is similarity, with the same basis change in domain and codomain. Independent changes of domain and codomain for a general map produce $Q^{-1}AP$, a different equivalence problem.

3. Quotients make an ignored subspace into zero

For a subspace $U\subseteq V$, define $v\sim w$ when $v-w\in U$. This is an equivalence relation. Its classes are cosets $v+U$, and the quotient $V/U$ has operations

$$(v+U)+(w+U)=(v+w)+U,\qquad \alpha(v+U)=\alpha v+U.$$

These are well-defined because changing representatives adds elements of $U$. The projection $\pi:V\to V/U$ is linear and surjective, with kernel $U$. If $u_1,\ldots,u_r,v_1,\ldots,v_s$ is an extended basis as above, then $v_1+U,\ldots,v_s+U$ form a quotient basis, proving

$$\dim(V/U)=\dim V-\dim U.$$

A quotient is a space of equivalence classes, not a chosen complementary subspace, though any chosen complement maps isomorphically to it.

Factorization theorem. A linear map $T:V\to W$ descends to a map $\widetilde T:V/U\to W$ satisfying $T=\widetilde T\pi$ exactly when $U\subseteq\ker T$. In that case $\widetilde T(v+U)=Tv$ is well-defined and unique. For $U=\ker T$, this map is injective onto $\operatorname{im}T$, giving the first isomorphism theorem

$$V/\ker T\cong\operatorname{im}T.$$

The formula itself proves all assertions: representatives differ by a vector killed by $T$, and a quotient class maps to zero exactly when its representative lies in the kernel.

Worked example. On $P_2(F)$ let $T(p)=p'$ with the formal polynomial derivative, and first assume $F$ has characteristic zero. Its kernel is the constants and its image is $P_1(F)$. Thus $P_2(F)/P_0(F)\cong P_1(F)$, with quotient basis $x+P_0,x^2+P_0$ mapping to $1,2x$. In characteristic two, however, $(x^2)'=0$: the kernel is $\operatorname{span}(1,x^2)$ and the image consists of constants. The general theorem remains true; the characteristic-zero computation does not.

4. The dual space and coordinate-free transpose

The dual $V^*$ consists of linear functions $\varphi:V\to F$. For a basis $b_1,\ldots,b_n$, its dual basis $b^1,\ldots,b^n$ is defined by $b^i(b_j)=\delta_{ij}$. Every functional is uniquely $\sum_i\varphi(b_i)b^i$, so $\dim V^*=n$.

The dual map of $T:V\to W$ is $T^*:W^*\to V^*$, defined by $T^*\psi=\psi\circ T$. In dual bases its matrix is the ordinary transpose $A^{\mathsf T}$, including over $\mathbb C$. There is no complex conjugation: this is algebraic duality, not a Hermitian adjoint. Composition reverses:

$$(ST)^*=T^*S^*.$$

This follows by evaluating either side on an arbitrary functional and vector.

For $U\subseteq V$, its annihilator is

$$U^\circ=\{\varphi\in V^*: \varphi(u)=0\text{ for all }u\in U\}.$$

An adapted basis shows $\dim U^\circ=\dim V-\dim U$. The quotient's dual identifies naturally with the annihilator: $\eta\mapsto\eta\circ\pi$ is an isomorphism $(V/U)^*\to U^\circ$, because precisely the functionals killing $U$ descend through the quotient.

For a linear map,

$$\ker T^*=(\operatorname{im}T)^\circ,\qquad \operatorname{im}T^*=(\ker T)^\circ.$$

The first equality is immediate by evaluation. For the second, inclusion follows because $\psi(Tv)=0$ whenever $Tv=0$. To prove the reverse inclusion, if $\varphi$ kills $\ker T$, define a functional on $\operatorname{im}T$ by $Tv\mapsto\varphi(v)$; it is well-defined, and basis extension extends it to all of $W$. This extended $\psi$ satisfies $T^*\psi=\varphi$. Taking dimensions gives equality of row rank and column rank without choosing a row-reduction algorithm.

Finally the evaluation map $J:V\to V^{**}$, $J(v)(\varphi)=\varphi(v)$, is injective because a nonzero vector can be included in a basis and detected by its first dual functional. Equal finite dimensions make it an isomorphism. This identification is canonical. An isomorphism $V\cong V^*$ generally requires a chosen basis or a nondegenerate bilinear form and is not canonical in the same way.

5. Exercises

P01 — Foundation. In $F^3$, find the dimension and a complement of $U=\{(x,y,z):x+y+z=0\}$.

P02 — Core. For $T:F^3\to F^2$, $T(x,y,z)=(x+y,y+z)$, find its kernel, image and quotient-space isomorphism.

P03 — Core. For $U=\operatorname{span}((1,1,0),(0,1,1))\subseteq\mathbb R^3$, find $U^\circ$.

P04 — Core. Given $B=((1,0),(1,1))$ in $\mathbb R^2$, write its dual basis as row functionals in standard coordinates.

P05 — Diagnosis. Let $U=\operatorname{span}(e_1)$ and $T(x,y)=x+y$. Does $T$ descend to $F^2/U$? Explain using representatives.

P06 — Proof. Prove that $T$ is surjective exactly when $T^*$ is injective, for finite-dimensional spaces.

P07 — Proof. Prove that $U=\{v:\varphi(v)=0\text{ for every }\varphi\in U^\circ\}$.

P08 — Extension. Let $S:F^{\mathbb N_0}\to F^{\mathbb N_0}$ shift a sequence right by inserting zero at its first coordinate. Show that $S$ is injective but not surjective. What hypothesis of the finite-dimensional equivalence is absent?

P09 — Core. For formal differentiation on $P_2(F)$ in characteristic two, give a quotient basis for $P_2/\ker T$ and its image under the first isomorphism theorem.

6. Full solutions

P01. The linear condition solves $z=-x-y$, so $(1,0,-1),(0,1,-1)$ is a basis and $\dim U=2$. The vector $e_3$ does not satisfy the condition, so $F^3=U\oplus\operatorname{span}(e_3)$. This works over any field, since $1\ne0$.

P02. The equations give $x=-y$, $z=-y$, so the kernel is $\operatorname{span}((-1,1,-1))$. The images of $e_1,e_3$ are the two standard vectors in $F^2$, so the image is all of $F^2$. The induced isomorphism is $(x,y,z)+\ker T\mapsto(x+y,y+z)$; it is well-defined and bijective by §3.

P03. A functional has form $\alpha x+\beta y+\gamma z$. Vanishing on the two spanning vectors requires $\alpha+\beta=0$ and $\beta+\gamma=0$. Thus $U^\circ=\operatorname{span}(x-y+z)$ as a space of functionals.

P04. If $(x,y)=c_1(1,0)+c_2(1,1)$, then $c_2=y$ and $c_1=x-y$. Thus $b^1=(1,-1)$ and $b^2=(0,1)$. They evaluate to the required Kronecker deltas on $B$.

P05. No. Zero and $e_1$ represent the same quotient class, but $T(0)=0$ and $T(e_1)=1$. Equivalently $U$ is not contained in $\ker T$.

P06. If $T$ is onto, $\ker T^*=(\operatorname{im}T)^\circ=W^\circ=0$. Conversely if its image is a proper subspace, extend a basis of that image to a basis of $W$; a nonzero dual functional supported on an added basis vector kills the image. It belongs to $\ker T^*$, contradicting injectivity.

P07. Inclusion of $U$ follows by definition. If $v\notin U$, extend a basis of $U$ by $v$ and then to a basis of $V$. The functional taking value one at $v$ and zero on all other adapted basis vectors lies in $U^\circ$ but does not kill $v$. Thus no vector outside $U$ belongs to the displayed set.

P08. If $S(a_0,a_1,\ldots)=(0,a_0,a_1,\ldots)$ is zero, every $a_j$ is zero, so it is injective. No sequence whose first coordinate is one is in its image, so it is not surjective. The space is infinite-dimensional; equal domain and codomain alone do not restore the finite-dimensional implication.

P09. The kernel is $\operatorname{span}(1,x^2)$, so a quotient basis is $x+\ker T$. Its image is the constant polynomial one, a basis of the one-dimensional image. Rank one plus nullity two equals dimension three.

7. Summary

Bases justify dimension; rank–nullity tracks information lost by a map. Quotients forget precisely a specified subspace, and their well-defined maps are controlled by kernels. Duality turns subspaces into annihilators and reverses arrows, with ordinary transpose rather than Hermitian adjoint. Polynomial operators in the next chapter add a second algebraic structure to this picture.

Product-space supplement

Let $F$ be a field. All vector spaces in one product use this same field. A dimension formula below assumes the displayed factors are finite-dimensional; the definition of a product does not require this assumption. Write $F^{\mathbb N}$ for the space of all sequences indexed by the positive integers, not just the finitely supported ones.

Finite products: structure, coordinates and a universal property

Definition. For an integer $r\ge1$ and vector spaces $V_1,\ldots,V_r$ over $F$, their external product is

$$V_1\times\cdots\times V_r=\{(v_1,\ldots,v_r):v_i\in V_i\},$$

with componentwise addition and scalar multiplication. The zero vector is the tuple of zeros. Each vector-space axiom holds in every coordinate because it holds in the corresponding factor, so the product is a vector space over $F$.

The projection $\pi_i$ sends a tuple to its $i$th coordinate. The injection $j_i:V_i\to V_1\times\cdots\times V_r$ puts its input in coordinate $i$ and zero elsewhere. Both maps are linear, as can be checked in each coordinate. The projection is surjective; the injection is injective. Moreover,

$$\pi_i j_i=I_{V_i},\qquad \pi_i j_k=0\quad(i\ne k),\qquad v=\sum_{i=1}^rj_i(\pi_i v).$$

Consequently the product is the internal direct sum of the subspaces $j_i(V_i)$. For finite families the external product and external direct sum have the same underlying tuples and operations. This finite-family statement must not be transferred to an infinite family without a finite-support condition.

Universal property. Given a vector space $W$ and linear maps $T_i:W\to V_i$, there is exactly one linear map $T:W\to V_1\times\cdots\times V_r$ such that $\pi_iT=T_i$ for every $i$.

Proof. Define $T(w)=(T_1w,\ldots,T_rw)$. Each coordinate of $T(aw+bz)$ equals the corresponding coordinate of $aT(w)+bT(z)$, so $T$ is linear. Its projections have the required values. Any other map with these projections has exactly the same coordinates at each $w$, hence is equal to $T$. No choice of bases or finite-dimensionality is needed. ∎

Finite-dimensional basis theorem. If $B_i=(b_{i1},\ldots,b_{in_i})$ is a basis of $V_i$, the concatenated list of all $j_i(b_{ik})$, in any fixed factor order and basis order, is a basis of the product. In particular,

$$\dim(V_1\times\cdots\times V_r)=\sum_{i=1}^r\dim V_i.$$

Proof. Expand every coordinate of a tuple in its factor basis and use the displayed finite decomposition to obtain spanning. For independence, apply $\pi_i$ to a zero linear combination of the concatenated list. This leaves a zero linear combination of $B_i$, whose coefficients are all zero. Repeating for every $i$ makes every coefficient zero. Thus the list is a basis, of the stated length. A zero-dimensional factor contributes an empty list and causes no exception. ∎

Worked example. For $V=F^2\times P_1(F)$, let $P_1(F)$ denote the vector space of formal polynomials of degree at most one together with the zero polynomial, not a space of polynomial functions. A basis is

$$((1,0),0),\quad((0,1),0),\quad((0,0),1),\quad((0,0),x).$$

Hence $\dim V=4$. The second projection has image $P_1(F)$ and kernel $F^2\times\{0\}$, of dimensions two and two respectively. The ordered pair in each displayed vector separates the $F^2$ coordinate from the polynomial coordinate; a polynomial is not being identified with a single field scalar.

Infinite products are not finite-support direct sums

The product $F^{\mathbb N}$ contains every sequence $(a_1,a_2,\ldots)$ with entries in $F$, using componentwise operations. Its finite-support subspace is

$$F^{(\mathbb N)}=\{a\in F^{\mathbb N}:a_i=0\text{ for all but finitely many }i\}.$$

It is a subspace: the support of a sum is contained in the union of two finite supports, and scalar multiplication does not create a new nonzero coordinate outside the old support.

Let $e_i$ be the sequence with a one in coordinate $i$ and zeros elsewhere. The list $e_1,e_2,\ldots$ is independent: in any finite zero linear combination, coordinate projection recovers each coefficient as zero. Its span is exactly $F^{(\mathbb N)}$, because a linear combination is finite by definition, while every finitely supported sequence is such a combination. The sequence $(1,1,1,\ldots)$ belongs to the product but not that span. Thus the $e_i$ form a basis of the direct sum, not a basis of the full product. This conclusion does not assert a dimension for the full product or permit an infinite linear sum without a separate meaning of convergence.

The product universal property still holds for infinitely many coordinate maps: define the output tuple coordinatewise and check linearity and uniqueness coordinatewise. The finite formula expressing each tuple as a sum of its coordinate injections no longer holds algebraically for a tuple of infinite support.

Exercises

P01. For $F^2\times P_1(F)$, verify the displayed basis directly. Find the kernel and image of the first projection, and verify rank–nullity.

P02. Let $A:W\to U$ and $B:W\to Z$ be linear maps, and put $T(w)=(Aw,Bw)$. Prove $\ker T=\ker A\cap\ker B$. If $W$ is finite-dimensional, deduce a formula for $\operatorname{rank}T$. Explain, with an example, why $\operatorname{rank}T$ need not equal $\operatorname{rank}A+\operatorname{rank}B$.

P03. In $F^{\mathbb N}$, prove that the constant-one sequence is outside the span of the coordinate vectors $e_i$. Prove that deleting the first coordinate is a surjective but noninjective linear map on $F^{\mathbb N}$. Which finite-dimensional hypothesis prevents a contradiction with the injectivity–surjectivity equivalence?

Full solutions

P01. Every element has form $((a,b),c+dx)$ and has the unique expansion with coefficients $a,b,c,d$ in the displayed four vectors. Uniqueness follows first by the two coordinates in $F^2$, then by equality of polynomial coefficients. The first projection has image $F^2$ and kernel $\{(0,0)\}\times P_1(F)$. Both have dimension two. Thus rank plus nullity is $2+2=4$, the dimension of the domain.

P02. A product tuple is zero exactly when both its coordinates are zero. Therefore $Tw=0$ exactly when $Aw=0$ and $Bw=0$, proving the kernel identity. Rank–nullity gives

$$\operatorname{rank}T=\dim W-\dim(\ker A\cap\ker B).$$

Take $W=U=Z=F$ and $A=B=I_F$. Each component map has rank one, but $T(w)=(w,w)$ has the one-dimensional diagonal as its image. Its rank is one, not two. The two recorded outputs can carry dependent information.

P03. Any linear combination of coordinate vectors uses finitely many indices. Choose a positive integer outside those indices. That coordinate of the combination is zero, but the same coordinate of the constant-one sequence is one, which is nonzero in a field. Thus the sequence is outside the span. The deletion map is $D(a_1,a_2,\ldots)=(a_2,a_3,\ldots)$. It is linear coordinatewise. Any target $(b_1,b_2,\ldots)$ is the image of $(0,b_1,b_2,\ldots)$, so it is surjective; $D(e_1)=0$ with $e_1\ne0$, so it is not injective. The domain and codomain here are infinite-dimensional: the infinite independent coordinate list shows this. The injectivity–surjectivity equivalence requires equal finite dimensions, not merely the same domain and codomain.