It must be proved that every vector space admits a basis; this will follow from the following more precise theorem:

Theorem 1 follows from this statement by taking L = ∅.

To prove Theorem 2, we note that the set Λ of free subsets of E contained in S, ordered by inclusion, is an inductive set (Set Theory, III, § 2, no. 4), by virtue of § 1, no. 11; so is the set M of free subsets containing L and contained in S. By Zorn's Lemma, M admits a maximal element B and it suffices to prove that the vector subspace of E generated by B is equal to E. This follows immediately from the definition of B and the following lemma:

Suppose that there were a relation $\mu b + \sum_i \lambda_i a_i = 0$ with $\mu \in K$ and $\lambda_i \in K$ for all $i \in I$, the family $(\lambda_i)$ having finite support; if $\mu \neq 0$, it would follow that $b = -\sum_i (\mu^{-1} \lambda_i) a_i$ and hence $b \in F$ contrary to the hypothesis; hence $\mu = 0$ and the relation becomes $\sum_i \lambda_i a_i = 0$, which implies $\lambda_i = 0$ for all $i \in I$ by hypothesis; whence the lemma.

We note first that if $E$ admits an infinite basis $B$, it follows from § 1, no. 12, Corollary 2 to Proposition 23 that every other basis of $E$ is equipotent to $B$. We may therefore confine our attention to the case where $E$ has a finite basis of $n$ elements. We note that every monogenous vector space over $K$, not reduced to 0, is a simple $K$-module (I, § 4, no. 4, Definition 7), for it is generated by each of its elements $\neq 0$, by virtue of the relation $\mu a = (\mu \lambda)(\lambda^{-1}a)$ for $\mu \in K$, $\lambda \in K$ and $\lambda \neq 0$. Hence if $(a_i)_{1 \leq i \leq n}$ is a basis of $E$, then $E = \bigoplus_{i=1}^n Ka_i$ up to isomorphism and the subspaces $E_k = \bigoplus_{i=1}^\kappa Ka_i$ for $0 \leq k \leq n$ form a Jordan-Hölder series of $E$, $E_k/E_{k-1}$ being isomorphic to $Ka_k$. Theorem 3 then follows in this case from the Jordan-Hölder Theorem (I, § 4, no. 7, Theorem 6).

A proof can be given independent of the Jordan-Hölder Theorem, by showing by induction on $n$ that, if $E$ admits a basis of $n$ elements, every other basis $B'$ has at most $n$ elements. The proposition is obvious for $n = 0$. If $n \geq 1$, $B'$ is non-empty; then let $a \in B'$. By Theorem 2 (no. 1) there exists a subset $C$ of $B$ such that $\{a\} \cup C$ is a basis of $E$ and $a \notin C$, since $\{a\} \cup B$ is obviously a generating system of $E$. As $B$ is a basis of $E$, $C = B$ is impossible (no. 1, Corollary to Theorem 2) and hence $C$ has at most $n - 1$ elements. Let $V$ be the subspace generated by $C$ and $V'$ the subspace generated by $B' - \{a\}$; $V$ and $V'$ are both supplementary to the subspace $Ka$ of $E$ and hence are isomorphic (§ 1, no. 10, Proposition 13). As $V$ admits a basis with at most $n - 1$ elements, $B' - \{a\}$ has at most $n - 1$ elements by the induction hypothesis and hence $B'$ has at most $n$ elements.

If V is the vector subspace of E generated by M, M contains a basis B' of V (no. 1, Theorem 2) and as B' is a free subset of E, it is contained in a basis B of E (no. 1, Theorem 2); then Card(B') $\leq$ Card(B), whence the corollary.

Theorems 2 and 3 immediately imply the following proposition:

(ii) For two vector spaces Ks^m and K_s^n to be isomorphic (m and n integers $\geq 0$), it is necessary and sufficient that m = n.*

(iii) In a vector space E of finite dimension n, every generating system has at least n elements; a generating system of E with n elements is a basis of E.

(iv) In a vector space E of finite dimension n, every free subset has at most n elements; a free subset with n elements is a basis of E.

If the E_t are canonically identified with subspaces of $E = \bigoplus_{t \in I} E_t$ and B_t is a basis of E_t ($t \in I$), then $B = \bigcup_{t \in I} B_t$ is a basis of E ($§ 1$, no. 11, Proposition 19); whence relation (1) since the B_t are pairwise disjoint.

Consider the vector space $\rho^*(E) = D \otimes_A E$ over D obtained by extending the ring of scalars to D ($§ 5$, no. 1) and let $\phi : x \mapsto 1 \otimes x$ be the canonical mapping of E into $\rho^*(E)$; if $(a_\lambda)$ is a basis of E, $(\phi(a_\lambda))$ is a basis of $\rho^*(E)$ ($§ 5$, no. 1, Proposition 4); the proposition then follows from Theorem 3.

There exists in A at least one maximal ideal m (I, $§ 8$, no. 6, Theorem 1) and, as A/m is a field, the conditions of Proposition 3 are fulfilled.

As the quotient vector space $E/F$ is a free module, we know (§ 1, no. 11, Proposition 21) that $F$ is a direct factor of $E$; relation (3) is then a special case of formula (1) (no. 2).

This is another way of expressing Proposition 4 (§ 1, no. 9).

the relation

holds, or, if all the spaces are finite dimensional,

Let $I_k = \operatorname{Im}\ u_k = \operatorname{Ker}\ u_{k+1}$ for $0 \leq k \leq n-1$; $I_{k+1}$ is therefore isomorphic to $E_{k+1}/I_k$, hence (formula (3)) $\dim I_k + \dim I_{k+1} = \dim E_{k+1}$ for $0 \leq k \leq n-2$ and moreover $\dim I_0 = \dim E_0$ and $I_{n-1} = E_n$, hence $\dim I_{n-1} = \dim E_n$. Replacing $\dim E_i$ by its expression as a function of the $\dim I_k$ in the two sides of (5), we obtain on each side $\sum_{k=0}^{n-1} \dim I_k$, whence the corollary.

(7)

It suffices to apply Corollary 2 to the exact sequence

(§ 1, no. 8, Proposition 10) taking account of the fact that

(no. 2, Proposition 2).

The first assertion is obvious from (3); further, if $\dim E$ is finite, the relation $\dim F = \dim E$ implies $\dim(E/F) = 0$ by (3) and a vector space of dimension 0 reduces to 0.

(8)

If further $\dim E$ is finite, the two sides of (8) are equal if and only if E is the direct sum of the family $(F_i)$.

The inequality (8) follows from (3) and the fact that E is isomorphic to a quotient of $\bigoplus_i F_i$ (§ 1, no. 7, formula (28)). The second assertion is a particular case of § 1, no. 10, Corollary 5 to Proposition 16, for the equality of the two sides of (8) implies that $\dim F_i = 0$ except for a finite number of indices.

The inequality $\operatorname{codim}_E F \leq \dim E$ is obvious from (9) and if $\dim E$ is finite the relation $\operatorname{codim}_E F = \dim E$ implies $\dim F = 0$ and hence $F = \{0\}$. The remainder of the proposition follows from this, for

Note that if E is finite dimensional, (10) is a consequence of (7) and (9).

Vector subspaces of dimension 1 (resp. dimension 2) of a vector space E are often called lines passing through 0 (resp. planes passing through 0) (or simply lines (resp. planes) if no confusion arises (cf. § 9, no. 3)), by analogy with the language of Classical Geometry; a subspace of E is called a hyperplane passing through 0 (or simply a hyperplane) if it is of codimension 1. Hyperplanes can also be defined as the maximal elements of the set $\mathscr{S}$ of vector subspaces of E distinct from E, ordered by inclusion. There is a one-to-one correspondence between the subspaces of E containing a subspace H and the subspaces of $E/H$ (I, § 4, no. 7, Theorem 4); if E is of dimension $\geq 1$, $\mathscr{S}$ is non-empty and to say that H is maximal in $\mathscr{S}$ means that $E/H$ contains no subspace distinct from $\{0\}$ and $E/H$, which implies that $E/H$ is generated by any of its elements $\neq 0$, in other words it is of dimension 1.

In a vector space of finite dimension $n \geq 1$, the hyperplanes are the subspaces of dimension $n - 1$ by formula (3).

It suffices to show that for all $x \notin F$ there exists a hyperplane H containing F and not containing x. By hypothesis $F \cap Kx = \{0\}$ and hence the sum M of F and $Kx$ is direct. Let N be supplementary to M in E; E is then the direct sum of $H = F + N$ and $Kx$ and H is therefore a hyperplane with the desired property.

This follows immediately from relations (12) and (13).

If E is an infinite-dimensional vector space, there are injective (resp. surjective) endomorphisms of E which are not bijective (Exercise 9).

Let K, K' be two fields, $\sigma : K \to K'$ an isomorphism of K onto K', E a vector K-space, E' a vector K'-space and $u : K \to K'$ a semi-linear mapping relative to $\sigma$ (§ 1, no. 13); the dimension of the subspace $u(E)$ of E' is also called the rank of $u$. It is also the rank of $u$ considered as a linear mapping of E into $\sigma_*(E')$, for every basis of $u(E)$ is also a basis of $\sigma_*(u(E))$.

If K is the field of scalars of E, E is isomorphic to a space $K_s^{(I)}$ and therefore $E^*$ is isomorphic to $K_d^I$ (§ 2, no. 6, Proposition 10). As $K_s^{(I)}$ is a subspace of $K_d^I$, $\dim E = \mathrm{Card}(I) \leq \dim E^*$ (no. 3, Corollary 4 to Proposition 4); further, if I is finite, $K_d^I = K_d^{(I)}$ (cf. Exercise 3(d)).

This follows from the fact that every vector subspace is a direct factor (no. 3, Proposition 4) and from § 2, no. 1, Propositions 1 and 2.

It follows in particular that for every vector subspace M of E the canonical homomorphism $E^*/M' \to M^*$, where $M'$ is the subspace of $E^*$ orthogonal to M (§ 2, no. 4), is bijective.

The first assertion and the fact that if E is finite-dimensional $c_E$ is bijective are special cases of § 2, no. 7, Proposition 14. Suppose that E is infinite-dimensional, so that we may assume that $E = K_s^{(L)}$, where L is an infinite set and therefore $E^* = K_d^L$. Let $(e_\lambda)_{\lambda \in L}$ be the canonical basis of E and let $(e_\lambda^*)_{\lambda \in L}$ be the corresponding family of coordinate forms in $E^*$ (§ 2, no. 6); the vector subspace of $E^*$ generated by the $e_\lambda^*$ is just the direct sum $F' = K_d^{(L)}$ and the hypothesis that L is infinite implies that $F' \neq E^*$. Then there exists a hyperplane $H'$ of $E^*$ containing $F'$ (no. 3, Proposition 8) and, as $E^*/H'$ is non-zero, its dual is also non-zero (Corollary to Theorem 4), which is identified with the orthogonal $H''$ of $H'$ in $E^{**}$ (§ 2, no. 6, Corollary to Proposition 9). But $H'' \cap c_E(E)$ is contained in the image under $c_E$ of the orthogonal of $F'$ in E, which is by definition 0; hence $c_E(E) = E^{**}$ is impossible.

E will usually be identified with the subspace of $E^{**}$ the image of $c_E$.

Let E, F be two vector spaces over a field K and $u : E \to F$ a linear mapping. We shall define canonical isomorphisms:

(1) Of the dual of Im(u) = u(E) onto Im($^t u$) = $^t u(F^*)$.

(2) Of the dual of Ker(u) = $^{-1}u(0)$ onto Coker($^t u$) = $E^*/^t u(F^*)$.

(3) Of the dual of Coker(u) = F/u(E) onto Ker($^t u$) = $^t u^{-1}(0)$.

We write I = Im(u), N = Ker(u), C = Coker(u); from the exact sequences

(14) $0 \to N \to E \xrightarrow{p} I \to 0,\quad 0 \to I \xrightarrow{j} F \to C \to 0$

we derive, by transposition (Corollary to Theorem 5), the exact sequences

(15)

Moreover, as $u = j \circ p,\ ^t u = ^t p \circ ^t j$; the exact sequences (15) thus define canonical isomorphisms of $C^*$ onto Ker($^t u$), of $I^*$ onto Im($^t u$) and of $N^*$ onto Coker($^t u$), since $^t p$ is injective and $^t j$ surjective. To be precise, let $y \in \mathrm{Im}(u)$, $z \in \mathrm{Ker}(u)$, $t \in \mathrm{Coker}(u)$, $y' \subset \mathrm{Im}(^t u)$, $z' \in \mathrm{Coker}(^t u)$, $t' \in \mathrm{Ker}(^t u)$; when $y'$, $z'$, $t'$ are canonically identified with linear forms on $\mathrm{Im}(u)$, $\mathrm{Ker}(u)$ and $\mathrm{Coker}(u)$ respectively, then

(16) $\langle y, y' \rangle = \langle x, y' \rangle$ for all $x \in E$ such that $u(x) = y$;

(17) $\langle z, z' \rangle = \langle z, x^* \rangle$ for all $x^* \in E^*$ whose class mod. $^t u(F)$ is equal to $z'$;

(18) $\langle t, t' \rangle = \langle s, t' \rangle$ for all $s \in F$ whose class mod. $u(E)$ is equal to $t$.

In particular we derive from these results:

(i) For u to be injective (resp. surjective), it is necessary and sufficient that $^t u$ be surjective (resp. injective).

(ii) $\operatorname{rg}(u) \leq \operatorname{rg}(^t u)$ and $\operatorname{rg}(u) = \operatorname{rg}(^t u)$ if $\operatorname{rg}(u)$ is finite.

The second assertion follows from the above and Theorem 4.

(i) $\dim F' \geq \operatorname{codim}_E F;$ for $\dim F'$ to be finite, it is necessary and sufficient that $\operatorname{codim}_E F$ be finite, and then $\dim F' = \operatorname{codim}_E F$.

(ii) The orthogonal of F' in E is equal to F.

(iii) Every finite-dimensional subspace G' of E is the orthogonal of some subspace of E, necessarily equal to the orthogonal of G' in E and of finite codimension.*

(i) We know that F' is isomorphic to the dual (E/F)* (§ 2, no. 6, Corollary to Proposition 9) and hence the assertion follows from Theorem 4, since $\dim(E/F) = \operatorname{codim}_E F$ by definition.

(ii) Let $F_1$ be the orthogonal of F' in E; clearly $F \subset F_1$ and the orthogonal $F'_1$ of $F_1$ is equal to F' (§ 2, no. 4); the canonical linear mapping $(E/F_1)^* \to (E/F)^*$, the transpose of $E/F \to E/F_1$ is therefore bijective (§ 2, no. 6, Corollary to Proposition 10); it then follows from Proposition 10 that the canonical mapping $E/F \to E/F_1$ is bijective, which implies $F_1 = F$.

(iii) Let G' be a subspace of E* of finite dimension $p$ and let F be its orthogonal in E; then $\operatorname{codim}_E F \leq \dim G'$. For, if $(a_i^*)_{1 \leq i \leq p}$ is a basis of G', F is the kernel of the linear mapping $x \mapsto (\langle x, a_i^* \rangle)$ from E to $K_s^p$ whose rank is at most $p$ (no. 4, Proposition 9), whence the conclusion (no. 4). Then let F' be the orthogonal of F in E*; it follows from (i) that $\dim F' \leq \dim G'$; but on the other hand obviously $G' \subset F'$, whence $F' = G'$ (§ 2, no. 3, Corollary 4 to Proposition 4).

Then $\operatorname{codim}_E F$ is equal to the rank of the set of the $x_i^*$ and every linear form on E which is zero on F is a linear combination of the $x_i^*$. Then $\operatorname{codim}_E F \leq p$ and in order that $\operatorname{codim}_E F = p$, it is necessary and sufficient that the $x_i^*$ be linearly independent.

The set $G'$ of linear combinations of the $x_i^*$ is a subspace of $E^*$ and F is the orthogonal of $G'$ in E, hence $\operatorname{codim}_E F = \dim G'$ by Theorem 7; further $\dim G' \leq p$ and the relation $\dim G' = p$ means that $(x_i^*)$ is a free system (no. 2, Proposition 1); hence the corollary.

(ii) Let $(x_i)_{1 \leq i \leq p}$ be a finite sequence of elements of E. For $(x_i)$ to be a free system, it is necessary and sufficient that there exist a sequence $(x_i^*)_{1 \leq i \leq p}$ of linear forms on E such that $\langle x_i, x_j^* \rangle = \delta_{ij}$.

Clearly (ii) follows from (i) by considering E as identified with a subspace of $E^{**}$ by means of $c_E$ (Theorem 6). Let $G'$ be the subspace of $E^*$ generated by the $x_i^*$ and F its orthogonal in E; E/F and $G'$ can each be canonically identified with the dual of the other; if the family $(x_i^*)$ is free, there is in E/F a basis $(\dot{x}_i)$ the dual of $(x_i^*)$ and every representative system $(x_i)$ of the classes $\dot{x}_i$ has the required properties. Conversely, the existence of the system $(x_i)$ such that $\langle x_i, x_j^* \rangle = \delta_{ij}$ implies that for all $i$ the subspace of $E^*$ orthogonal to K.$x_i$ contains the $x_j^*$ of index $j \neq i$ but does not contain $x_i^*$, hence the system $(x_i^*)_{1 \leq i \leq p}$ is free.

The space $K_d^S$ is canonically identified with the dual of $E = K_d^{(S)}$ and $f(s) = \langle e_s, f \rangle$ for $s \in S$ and $f \in K_s^S$, $(e_s)_{s \in S}$ being the canonical basis of E. Corollary 2 then shows that the condition is sufficient. Conversely, suppose that $\dim V \geq p$, so that there exists a subspace $G'$ of V of dimension $p$; let F be the orthogonal of $G'$ in E, so that $\dim(E/F) = p$. It follows from no. 1, Theorem 2 that there exist $p$ elements $s_i \in S$ such that the $e_{s_i}$ $(1 \leq i \leq p)$ form a basis of a supplement F of E (applying Theorem 2 (no. 1) to a free subset consisting of a basis of E and the generating system the union of this free subset and the canonical basis of E); we then take the $f_i$ to be the elements of a basis of G' dual to the basis of E/F consisting of the classes of the $e_{s_i}$ mod. F.

As M (resp. N) is the orthogonal of M' (resp. N') in E (Theorem 7), M ∩ N is the orthogonal of M' + N' in E and hence M' + N' is the orthogonal of M ∩ N in E* (Theorem 7 (iii)).

Theorem 7 gives another characterization of hyperplanes in E:

This statement merely expresses Theorem 7 for subspaces of E of codimension 1 and subspaces of E* of dimension 1.

If H is a hyperplane and $x_0^*$ a linear form such that H = $x^{*-1}(0)$, the relation

which characterizes the elements $x \in H$, is called an equation of H.

More generally, if $(x_i^*)$ is a family of linear forms on E and F denotes the vector subspace the intersection of the hyperplanes $x_i^{*-1}(0)$, the relation "for all $i, \langle x, x_i^* \rangle = 0$" characterizes the elements of F; the relations

form a system of equations of the subspace F. Theorem 7 (ii) expresses the fact that every vector subspace of E can be defined by a system of equations.

Theorem 7 (i) and (ii) proves moreover that a subspace F of finite codimension $p$ can be defined by a system of $p$ equations

(19)

where the forms $x_i^*$ are linearly independent. Conversely, Corollary 1 to Theorem 7 shows that a subspace F defined by a system of $p$ equations (19) is of co-dimension $\leq p$ and that it is of codimension $p$ if and only if the $x_i^*$ are linearly independent; it amounts to the same to say that F cannot be defined by a system consisting of at most $p - 1$ of the equations (19).

The orthogonal of $u(E)$ in $F^*$ is ${}^t u^{-1}(0)$ (§ 2, no. 5, Corollary to Proposition 8) and the orthogonal of ${}^t u^{-1}(0)$ in F is therefore $u(E)$ (no. 5, Theorem 7 (ii)).

We shall obtain a more convenient criterion for systems of scalar linear equations

where the unknown $x$ takes its values in a vector space E over a field K, the $x_i^*$ are linear forms on E and the right hand sides $\eta_i$ elements of K.

If we consider a basis $(a_\lambda)_{\lambda \in L}$ of E, the system (21) is equivalent to the system of equations

with $x = \sum_{\lambda \in L} \xi_\lambda a_\lambda$, the solutions of (22) being of necessity families $(\xi_\lambda)$ of elements of K of finite support.

If F' is the subspace of $E^*$ generated by the $x_i^*$, the kernel of u is the orthogonal F of F' in E; if F' is of dimension r, F is of codimension r and conversely (no. 5, Theorem 7) and $\mathrm{rg}(u) = \mathrm{codim}_E F$ (no. 4, formula (11)).

be a system of scalar linear equations on a vector space E over a field K. For this system to have at least one solution, it is necessary that, for every family $(\rho_i)$ of scalars of finite support such that $\sum_i x_i^* \rho_i = 0,\ \sum_i \eta_i \rho_i = 0.$ If the rank of the system (21) is finite, this condition is also sufficient.

The condition is obviously necessary. It says that, if $F'$ is the subspace of $E^*$ generated by the family $(x_i^*)$, there exists a linear mapping $f : F' \to K_d$ such that $f(x_i^*) = \eta_i$ for all $i \in I$. If $F'$ is of finite dimension $r$, $F'$ is the orthogonal of a subspace $F$ of $E$ of codimension $r$ (no. 5, Theorem 7) and $F'$ is identified with the dual of $E/F$ (§ 2, no. 6, Corollary to Proposition 9); $f$ is therefore an element of the bidual $(E/F)**$. As $E/F$ is finite-dimensional there exists one and only one element $y \in E/F$ such that $f(x^*) = \langle y, x^* \rangle$ for all $x^* \in F'$ (no. 5, Theorem 6). The solutions of (21) are then the $x \in E$ whose canonical image in $E/F$ is $y$.

This will always be the case if the number of equations is finite and < n.

(23)

As the sequence (23) splits, this is a particular case of § 3, no. 7, Corollary 5 to Proposition 7 and § 3, no. 6, Proposition 5.

Because of Proposition 14, when $E'$ is a vector subspace of $E$, $j : E' \to E$ the canonical injection, $E' \otimes_K F$ is usually identified with a sub-$\mathbf{Z}$-module of $E \otimes_K F$ by means of the injection $j \otimes 1_F$. With this convention:

It is obviously sufficient to prove the particular case

Clearly the left hand side of (25) is contained in the right hand side. To prove the converse, we consider a basis $(f_\lambda)_{\lambda \in L}$ of F. Every element of $E \otimes_K F$ can then be expressed uniquely in the form $\sum_{\lambda \in L} x_\lambda \otimes f_\lambda$, where $x_\lambda \in E$ (§ 3, no. 7, Corollary 1 to Proposition 7); if $E'$ is a vector subspace of E, the relation $\sum_{\lambda \in L} x_\lambda \otimes f_\lambda \in E' \otimes_K F$ is equivalent, by Proposition 14, to $x_\lambda \in E'$ for all $\lambda \in L$. To say that $\sum_{\lambda \in L} x_\lambda \otimes f_\lambda$ belongs to each of the $M_\alpha \otimes_K F$ thus means that for all $\lambda \in L$ and all $\alpha \in A$, $x_\lambda \in M_\alpha$, that is $x_\lambda \in \bigcap_{\alpha \in A} M_\alpha$ for all $\lambda \in L$, which proves that the right hand side of (25) is contained in the left hand side.