Level 1B: Linear Algebra Done Wrong

The Defect. The standard linear algebra course starts with matrices and row reduction, a computational approach that treats vectors as column arrays and linear maps as grids of numbers. This works for computation but obscures the geometry: a vector is not a column of numbers; a linear map is not a grid. The coordinate-free approach of Sergei Treil’s Linear Algebra Done Wrong treats vectors as abstract elements of a vector space and linear maps as structure-preserving functions, introducing coordinates only at the very end as a computational convenience. This page follows that approach.

The One-Line Reveal. The entire page follows from one definition: a vector space is a set with addition and scalar multiplication satisfying eight axioms. Everything else (dual spaces, quotient spaces, exterior algebra, determinants) is constructed from this single definition by universal properties and dimension counting.

Scope. All vector spaces in this course are finitely generated: spanned by finitely many vectors. For spaces not spanned by any finite list, even the existence of a basis is equivalent to the axiom of choice; that set-theoretic territory is never needed downstream and is avoided here. A field is a system of scalars with addition, multiplication, and division, of characteristic different from $2$, such as $\mathbb{Q}$, $\mathbb{R}$, or $\mathbb{C}$; nothing below uses more than the field axioms together with $2 \neq 0$, which the polarisation step of Lemma 4.4 needs.

Definition. A vector space $V$ over a field $k$ is a set with addition $+: V \times V \to V$ and scalar multiplication $\cdot: k \times V \to V$ satisfying eight axioms: associativity and commutativity of addition, additive identity, additive inverses, scalar identity $1v = v$, compatibility $(\alpha\beta)v = \alpha(\beta v)$, and distributivity of scalar multiplication over vector addition and over scalar addition.

Definition. A function $T: V \to W$ between vector spaces is a linear map if $T(\alpha u + \beta v) = \alpha T(u) + \beta T(v)$ for all $u, v \in V$ and $\alpha, \beta \in k$. The set of all linear maps from $V$ to $W$ is denoted $\operatorname{Hom}(V, W)$ or $L(V, W)$.

Theorem 1.1. $\operatorname{Hom}(V, W)$ is itself a vector space under pointwise addition and scalar multiplication.

Proof. Each axiom is inherited from $W$. The zero map $0(v) = 0_W$ is the additive identity. For $T_1, T_2 \in \operatorname{Hom}(V, W)$, the pointwise sum $(T_1 + T_2)(v) = T_1(v) + T_2(v)$ is linear: it distributes over arguments because each $T_i$ does and $W$‘s addition does. Likewise $(\alpha T)(v) = \alpha T(v)$ is linear. The remaining axioms (associativity, commutativity, distributivity, identities, inverses $-T$) hold pointwise because they hold in $W$ at every argument. $\blacksquare$

Definition. A list $e_1, \dots, e_n$ of vectors of $V$ is linearly independent if the only scalars with $\sum_i a_i e_i = 0$ are $a_1 = \dots = a_n = 0$; it spans $V$ if every $v \in V$ can be written $\sum_i a_i e_i$. A basis is a list that is both.

Coordinates and their uniqueness. If $e_1, \dots, e_n$ is a basis, every $v$ has an expression $v = \sum_i v_i e_i$, and the scalars $v_i$ (the coordinates of $v$) are unique: if $\sum_i v_i e_i = \sum_i v'_i e_i$, then $\sum_i (v_i - v'_i) e_i = 0$, so independence forces $v_i = v'_i$ for every $i$. $\blacksquare$

Theorem 1.2 (bases exist). If finitely many vectors span $V$, then $V$ has a basis.

Proof. Among all lists spanning $V$, choose one of minimal length, $w_1, \dots, w_m$. If it were dependent, some nontrivial combination $\sum_i a_i w_i = 0$ would have $a_p \neq 0$ for some $p$; solving gives $w_p = -a_p^{-1} \sum_{i \neq p} a_i w_i$. Deleting $w_p$ leaves a shorter list that still spans: any expression over the full list can be rewritten with $w_p$ replaced by that combination. This contradicts minimality, so the minimal spanning list is independent, hence a basis. $\blacksquare$

Lemma 1.3 (exchange). If $v_1, \dots, v_k$ are linearly independent and $w_1, \dots, w_m$ span $V$, then $k \leq m$.

Proof. We show by induction on $j$ that for each $j \leq \min(k, m)$ there is a subset $S_j \subseteq \{1, \dots, m\}$ with $|S_j| = m - j$ such that the list $v_1, \dots, v_j$ together with $\{w_i : i \in S_j\}$ spans $V$. For $j = 0$, take $S_0 = \{1, \dots, m\}$. Assume the claim for $j$, with $j < k$. Since $v_{j+1}$ lies in the span of the list, $v_{j+1} = \sum_{i \leq j} c_i v_i + \sum_{i \in S_j} d_i w_i$. If all $d_i$ were zero, then $\sum_{i \leq j} c_i v_i - v_{j+1} = 0$ would be a nontrivial dependence among $v_1, \dots, v_{j+1}$ (its $v_{j+1}$ coefficient is $-1$), contradicting independence. So some $d_p \neq 0$ with $p \in S_j$; solving for $w_p$ shows the list with $w_p$ replaced by $v_{j+1}$ still spans, which is the claim for $j + 1$ with $S_{j+1} = S_j \setminus \{p\}$. If $k > m$, the induction runs to $j = m$, giving $S_m = \emptyset$: the list $v_1, \dots, v_m$ alone spans $V$. Applying the induction step once more would express $v_{m+1}$ as a combination of $v_1, \dots, v_m$ (there are no $w$‘s left to carry any $d_i$), contradicting independence. Hence $k \leq m$. $\blacksquare$

Corollary 1.4 (dimension is well defined). Any two bases of $V$ have the same length. This common length is the dimension $\dim V$.

Proof. Apply Lemma 1.3 twice: with basis one playing the independent list and basis two the spanning list, and vice versa. $\blacksquare$

Lemma 1.5 (basis extension). If $u_1, \dots, u_r$ are linearly independent in $V$ and $w_1, \dots, w_m$ span $V$, then the $u$‘s can be extended, using some of the $w$‘s, to a basis of $V$.

Proof. Choose $S \subseteq \{1, \dots, m\}$ maximal such that $u_1, \dots, u_r$ together with $\{w_i : i \in S\}$ remains independent. Such an $S$ exists (the empty set works) and there are finitely many subsets, so a maximal one can be chosen. We claim $u_1, \dots, u_r$ and $\{w_i : i \in S\}$ together span $V$. If not, some $w_l$ lies outside their span; then adjoining $w_l$ keeps the list independent. Indeed, suppose $\sum_i a_i u_i + \sum_{i \in S} b_i w_i + c\, w_l = 0$ with some coefficient nonzero. If $c \neq 0$, then $w_l$ is a combination of the others, contrary to its choice; so $c = 0$, and then all other coefficients vanish because the shorter list is already independent. This contradicts the maximality of $S$. So the list is independent and spans $V$, i.e. it is a basis extending the $u$‘s. $\blacksquare$

Theorem 1.6 (subspaces are finitely spanned). Every subspace $U$ of a finitely spanned space $V$ is finitely spanned.

Proof. Let $w_1, \dots, w_m$ span $V$. By the exchange lemma (Lemma 1.3), every linearly independent list in $V$ has at most $m$ members (apply the lemma with the list in question playing the independent part and the $w$‘s the spanning part). So the independent lists in $U$ have lengths bounded by $m$; among them, choose one of maximal length, $u_1, \dots, u_r$ (the empty list qualifies, so a maximal one exists). We claim the $u$‘s span $U$. If some $u \in U$ lay outside their span, then $u_1, \dots, u_r, u$ would be a longer independent list in $U$: a dependence $\sum_i a_i u_i + b u = 0$ with $b \neq 0$ would solve for $u$ as a combination of the $u$‘s, forcing $b = 0$, and then all $a_i = 0$ by independence; so the longer list is independent, contradicting maximality. Hence the $u$‘s span $U$. $\blacksquare$

Theorem 1.7 (rank–nullity). For every linear map $T: V \to W$,

Proof. The kernel $\ker T$ is a subspace of the finitely spanned $V$, so by Theorem 1.6 it is finitely spanned and by Theorem 1.2 it has a basis $u_1, \dots, u_r$ (empty, with $r = 0$, if $\ker T = \{0\}$). Extend it to a basis $u_1, \dots, u_r, v_1, \dots, v_s$ of $V$ by Lemma 1.5. The images $T(v_1), \dots, T(v_s)$ span $\operatorname{im} T$: any $v$ decomposes as $\sum_i b_i u_i + \sum_j c_j v_j$, and $T(v) = \sum_j c_j T(v_j)$ because the $u$-terms map to zero. They are independent: if $\sum_j a_j T(v_j) = 0$, then $T(\sum_j a_j v_j) = 0$, so $\sum_j a_j v_j \in \ker T$ equals some $\sum_i b_i u_i$; moving everything to one side gives a vanishing combination of the full basis $u_1, \dots, u_r, v_1, \dots, v_s$, so every coefficient is zero, in particular every $a_j$. Hence $\{T(v_1), \dots, T(v_s)\}$ is a basis of $\operatorname{im} T$, so $\dim \operatorname{im} T = s$, while $\dim V = r + s$ because $u_1, \dots, u_r, v_1, \dots, v_s$ is a basis of $V$. $\blacksquare$

The One-Line Reveal. The proof splits $V$ into two pieces: the part that maps to zero (the kernel) and the part that maps injectively (the extension). The splitting is possible by basis extension, and then the theorem is just counting.

Quick check.

1. Recite the eight vector-space axioms in one breath.
2. Why must a minimal spanning list be independent?
3. State the exchange lemma.
4. Why is dimension well defined?
5. What does basis extension give?

Answers: 1. Associative commutative addition with identity and inverses; $1v = v$; $(\alpha\beta)v = \alpha(\beta v)$; two distributivities. 2. A dependent member can be deleted while still spanning, contradicting minimality (Theorem 1.2). 3. An independent list in a finitely spanned space has length at most the spanning list (Lemma 1.3). 4. Lemma 1.3 applied in both directions between two bases (Corollary 1.4). 5. Any independent list extends, using pieces of a spanning list, to a basis (Lemma 1.5).

Definition. The dual space $V^* = \operatorname{Hom}(V, k)$ is the space of all linear functionals on $V$: linear maps $V \to k$.

Theorem 2.1. $\dim V^* = \dim V$.

Proof. Let $\{e_1, \dots, e_n\}$ be a basis of $V$. Define $e^j \in V^*$ (the dual basis) by $e^j(e_i) = \delta_{ij}$ (Kronecker delta). These are linearly independent: if $\sum_j a_j e^j = 0$, evaluate on $e_i$ to get $a_i e^i(e_i) = a_i = 0$. They span: for any $\varphi \in V^*$ and any $v = \sum_i v_i e_i$, linearity gives $\varphi(v) = \sum_i v_i \varphi(e_i)$, and since $v_i = e^i(v)$ ($e^i$ is linear and kills every $e_j$ with $j \neq i$), we get

So $\varphi = \sum_i \varphi(e_i)\, e^i$, and the dual basis spans $V^*$. Hence $\{e^1, \dots, e^n\}$ is a basis of $V^*$ and $\dim V^* = n = \dim V$. $\blacksquare$

The One-Line Reveal. The dual basis is defined by where it sends the basis vectors, and the delta $\delta_{ij}$ ensures it picks out exactly one coordinate. This is the coordinate-free version of “the transpose.”

Theorem 2.2. The map $\operatorname{ev}: V \to V^{**}$ defined by $(\operatorname{ev}(v))(\varphi) = \varphi(v)$ is an isomorphism.

Proof. $\operatorname{ev}$ is linear in $v$: for fixed $\varphi$, the assignment $v \mapsto \varphi(v)$ is linear, so $\operatorname{ev}(\alpha v_1 + \beta v_2)$ agrees with $\alpha\operatorname{ev}(v_1) + \beta\operatorname{ev}(v_2)$ on every $\varphi$, and two functionals agreeing on all arguments are equal. It is injective: if $\operatorname{ev}(v) = 0$, then $\varphi(v) = 0$ for all $\varphi \in V^*$; taking $\varphi = e^j$ gives $e^j(v) = v_j = 0$ for every $j$, so $v = 0$ by the uniqueness of coordinates. Since $\dim V^{**} = \dim V^* = \dim V$ (Theorem 2.1 applied twice), injective plus equal dimensions gives surjective. $\blacksquare$

Definition. For a linear map $T: V \to W$, the dual map is $T^*: W^* \to V^*$, $T^*(\varphi) = \varphi \circ T$. It is linear, because $T^*(\alpha\varphi + \beta\psi)(v) = (\alpha\varphi + \beta\psi)(T(v)) = \alpha T^*(\varphi)(v) + \beta T^*(\psi)(v)$. Applying the construction twice gives $T^{**} = (T^*)^*: V^{**} \to W^{**}$.

Proposition 2.3 (naturality of $\operatorname{ev}$). For every linear map $T: V \to W$, the square

commutes: $T^{**} \circ \operatorname{ev}_V = \operatorname{ev}_W \circ T$.

Proof. Take $v \in V$ and $\varphi \in W^*$. Then $(T^{**}(\operatorname{ev}_V(v)))(\varphi) = (\operatorname{ev}_V(v))(T^*\varphi) = (T^*\varphi)(v) = \varphi(T(v)) = (\operatorname{ev}_W(T(v)))(\varphi)$: the first step is the definition of the dual of a dual map, the middle step is the definition of $T^*$, and the last is the definition of $\operatorname{ev}_W$. Since the two elements of $W^{**}$ agree on every $\varphi$, they are equal; and this holds for every $v$. $\blacksquare$

Why this is “natural.” The isomorphism $V \to V^{**}$ is canonical: it does not require choosing a basis. In contrast, the isomorphism $V \to V^*$ (represented, once a basis is chosen, by $v \mapsto \sum_i v_i e^i$) does require a basis. This is why the double dual is “better” than the dual: the isomorphism is not an artefact of coordinates. In the language of category theory, Proposition 2.3 says precisely that $\operatorname{ev}: \operatorname{Id} \Rightarrow (-)^{**}$ is a natural transformation from the identity functor to the double-dual functor: the square above is exactly the commuting square from the definition of a natural transformation.

Quick check.

1. What is the dual basis of $e_1, \dots, e_n$?
2. Why is $V \cong V^{**}$ canonical while $V \cong V^*$ is not?
3. Define the dual map $T^*$.
4. What does Proposition 2.3 say?
5. Compute $e^3(v)$ for $v = 2e_1 - e_3$.

Answers: 1. Functionals $e^j$ with $e^j(e_i) = \delta_{ij}$. 2. $\operatorname{ev}$ uses no choice of basis; the isomorphism $V \cong V^*$ needs one. 3. $T^*(\varphi) = \varphi \circ T$. 4. $T^{**} \circ \operatorname{ev}_V = \operatorname{ev}_W \circ T$: the square commutes. 5. $-1$.

Definition. Let $U \subseteq V$ be a subspace. The quotient space $V/U$ is the set of equivalence classes $v + U$ under the relation $v \sim v'$ iff $v - v' \in U$.

Theorem 3.1. $V/U$ is a vector space with $\dim(V/U) = \dim V - \dim U$.

Proof. Addition and scalar multiplication are well defined: if $v_1 + U = v_1' + U$ and $v_2 + U = v_2' + U$, then $(v_1 - v_1') + (v_2 - v_2') \in U$, so $(v_1 + v_2) + U = (v_1' + v_2') + U$; likewise for scalars. The vector space axioms hold because they hold for representatives. The zero vector is $0 + U = U$. The dimension formula follows from rank–nullity (Theorem 1.7) applied to the quotient map $\pi: V \to V/U$, $\pi(v) = v + U$: $\ker \pi = U$ by definition of the relation, and $\pi$ is surjective by construction, so $\dim V = \dim U + \dim(V/U)$. $\blacksquare$

The One-Line Reveal. The quotient space “sets $U$ to zero”: every vector in $U$ becomes the zero vector of $V/U$, and rank–nullity counts what you collapse ($\dim U$) plus what remains ($\dim V/U$).

Theorem 3.2 (universal property of the quotient). Let $\pi: V \to V/U$ be the quotient map. A linear map $T: V \to W$ with $U \subseteq \ker T$ induces a unique linear map $\bar{T}: V/U \to W$ with $\bar{T} \circ \pi = T$.

Proof. Define $\bar{T}(v + U) = T(v)$. This is well defined: if $v + U = v' + U$, then $v - v' \in U \subseteq \ker T$, so $T(v) = T(v')$. It is linear because $T$ is: $\bar{T}((v + U) + (v' + U)) = \bar{T}((v + v') + U) = T(v + v') = T(v) + T(v')$, and likewise for scalars. By construction $\bar{T} \circ \pi = T$. For uniqueness: every element of $V/U$ has the form $\pi(v)$ (surjectivity, Theorem 3.1), so any $\bar{S}$ with $\bar{S} \circ \pi = T$ must send $\pi(v)$ to $T(v)$; there is no freedom left. $\blacksquare$

Quick check.

1. When do two cosets coincide?
2. Why is the kernel of the quotient map equal to U?
3. State the universal property of the quotient.
4. What is the dimension of the quotient?
5. What makes a map out of the quotient well defined?

Answers: 1. v + U = v’ + U exactly when v - v’ lies in U. 2. The quotient map sends v to U exactly when v lies in U. 3. A map killing U factors uniquely through the quotient map (Theorem 3.2). 4. dim V minus dim U. 5. It must take the same value on every representative: its pre-composition must kill U.

This section constructs the determinant honestly: not as an opaque formula, but as the unique normalised alternating multilinear form, with the formula derived rather than decreed. The only combinatorial input is the sign of a permutation, which is proved from scratch.

Definition. A permutation of $\{1, \dots, n\}$ is a bijection $\sigma: \{1, \dots, n\} \to \{1, \dots, n\}$. A transposition is a permutation that swaps two elements and fixes all others.

Lemma 4.1 (cycle decomposition). Every permutation is a product of transpositions.

Proof. Powers of $\sigma$ act injectively: if $\sigma^i(x) = \sigma^j(x)$ with $i < j$, applying $\sigma^{-i}$ (which exists, $\sigma$ being a bijection) gives $\sigma^{j-i}(x) = x$. So among $x, \sigma(x), \dots, \sigma^n(x)$ (that is $n + 1$ values in an $n$-element set), two coincide, and by injectivity of the powers the first coincidence is a return: $\sigma^r(x) = x$ for some $1 \leq r \leq n$. The permutation $c$ that cycles the orbit $x \mapsto \sigma(x) \mapsto \cdots \mapsto \sigma^{r-1}(x) \mapsto x$ and fixes everything else is a product of transpositions, $c = (x \ \sigma^{r-1}(x)) \cdots (x \ \sigma(x))$: following any element through the factors confirms this (check the orbit elements; all others are fixed by every factor). Now let $\rho$ agree with $\sigma$ off the orbit of $x$ and fix the orbit pointwise; then $\sigma = c \circ \rho$ (apply $\rho$ first: on the orbit it is the identity and $c$ reproduces $\sigma$; off the orbit $c$ is the identity and $\rho$ reproduces $\sigma$), and $\rho$ moves strictly fewer elements than $\sigma$. Induct on the number of moved elements. $\blacksquare$

Lemma 4.2 (the sign polynomial). Let $P = \prod_{1 \leq i < j \leq n} (x_i - x_j)$, a polynomial in $x_1, \dots, x_n$. Swapping the variables $x_p$ and $x_q$ (with $p < q$) carries $P$ to $-P$.

Proof. Inspect the factors of $P$ in three groups. Factors $(x_i - x_j)$ with $\{i, j\} \cap \{p, q\} = \emptyset$ are unchanged. The factor $(x_p - x_q)$ becomes $(x_q - x_p) = -(x_p - x_q)$. Each remaining factor involves exactly one of $p, q$ and some third index $a$, and the two factors involving $a$ multiply to the same value before and after the substitution: if $a < p$ they are $(x_a - x_p)(x_a - x_q)$, which becomes $(x_a - x_q)(x_a - x_p)$, the same product; if $p < a < q$ they are $(x_p - x_a)(x_a - x_q)$, which becomes $(x_q - x_a)(x_a - x_p) = (-(x_a - x_q))(-(x_p - x_a))$, the same product; if $a > q$ they are $(x_p - x_a)(x_q - x_a)$, which becomes $(x_q - x_a)(x_p - x_a)$, again the same product. So every factor is preserved except $(x_p - x_q)$, which negates; the total product negates. $\blacksquare$

Definition (action on polynomials). For $\sigma \in S_n$ and a polynomial $Q$ in $x_1, \dots, x_n$, let $\sigma Q$ be the polynomial $(\sigma Q)(x_1, \dots, x_n) = Q(x_{\sigma^{-1}(1)}, \dots, x_{\sigma^{-1}(n)})$. With this convention $\sigma(\tau Q) = (\sigma\tau)Q$, since both sides substitute $x_{\tau^{-1}(\sigma^{-1}(i))} = x_{(\sigma\tau)^{-1}(i)}$ for $x_i$.

Proposition 4.3 (the sign). For every $\sigma \in S_n$ there is a sign $\operatorname{sgn}(\sigma) \in \{+1, -1\}$ with $\sigma P = \operatorname{sgn}(\sigma)\, P$. Moreover: (i) if $\sigma$ is a product of $m$ transpositions, then $\operatorname{sgn}(\sigma) = (-1)^m$, so the parity of $m$ is the same for every factorisation; (ii) $\operatorname{sgn}(\sigma\tau) = \operatorname{sgn}(\sigma)\operatorname{sgn}(\tau)$.

Proof. The polynomial $\sigma P = \prod_{i<j}(x_{\sigma^{-1}(i)} - x_{\sigma^{-1}(j)})$ runs over each unordered pair $\{a, b\}$ exactly once, because $\sigma^{-1}$ is a bijection; each factor is $(x_a - x_b)$ up to an overall sign. So $\sigma P = \pm P$. Also $P \neq 0$: substituting $x_i = i$ makes every factor $(i - j)$ with $i < j$ a nonzero number. So $\operatorname{sgn}(\sigma)$ is well defined. For (i), induct on $m$: the case $m = 0$ is $\sigma = \operatorname{id}$, and $\operatorname{id}\, P = P$ gives $\operatorname{sgn}(\operatorname{id}) = +1$; if $\tau$ is a transposition swapping $x_p, x_q$, then $\tau P$ is $P$ with those variables swapped, which is $-P$ by Lemma 4.2, so $\operatorname{sgn}(\tau) = -1$; and the induction step follows from $\sigma(\tau Q) = (\sigma\tau)Q$ together with linearity of the action, $\sigma(cQ) = c(\sigma Q)$. For (ii): $(\sigma\tau)P = \sigma(\tau P) = \sigma(\operatorname{sgn}(\tau) P) = \operatorname{sgn}(\tau)(\sigma P) = \operatorname{sgn}(\tau)\operatorname{sgn}(\sigma) P$, and cancelling the nonzero $P$ gives the claim. $\blacksquare$

Definition. Let $V$ be a vector space and $k \geq 1$. An alternating $k$-form on $V$ is a function $\omega: V^k \to k$, linear in each argument separately (making it multilinear), with $\omega(v_1, \dots, v, \dots, v, \dots, v_k) = 0$ whenever two arguments coincide. The alternating $k$-forms form a vector space under pointwise operations: every axiom is checked argument by argument, exactly as in Theorem 1.1.

Lemma 4.4 (swapping negates; the converse). An alternating form negates under the exchange of any two arguments; conversely, a $k$-linear form whose values negate under every exchange of two arguments is alternating.

Proof. Put $v + w$ into two positions of the form and expand by multilinearity in those positions: $0 = \omega(v + w, v + w) = \omega(v, v) + \omega(v, w) + \omega(w, v) + \omega(w, w) = \omega(v, w) + \omega(w, v)$, the two pure terms vanishing by alternation; so the value swaps sign under the exchange. Conversely, if every exchange negates the value, then $\omega(\dots, v, \dots, v, \dots) = -\omega(\dots, v, \dots, v, \dots)$, so $2\,\omega(\dots, v, \dots, v, \dots) = 0$, and since the field’s characteristic is not $2$ (Scope), $\omega(\dots, v, \dots, v, \dots) = 0$. $\blacksquare$

Lemma 4.5 (sorting). For $\sigma \in S_n$ and any alternating $n$-form $\omega$ on $V$ with basis $e_1, \dots, e_n$: $\omega(e_{\sigma(1)}, \dots, e_{\sigma(n)}) = \operatorname{sgn}(\sigma)\, \omega(e_1, \dots, e_n)$.

Proof. By Lemma 4.1 write $\sigma = \tau_1 \cdots \tau_m$ as a product of transpositions, and induct on $m$. The case $m = 0$ is trivial. For the step, let $\tau$ be a transposition and suppose the claim for $\sigma'$; then the tuple $(e_{(\tau\sigma')(1)}, \dots, e_{(\tau\sigma')(n)})$ agrees with $(e_{\sigma'(1)}, \dots, e_{\sigma'(n)})$ except that the entries in the two positions moved by $\tau$ are exchanged, so Lemma 4.4 gives $\omega(e_{(\tau\sigma')(1)}, \dots) = -\omega(e_{\sigma'(1)}, \dots) = -\operatorname{sgn}(\sigma')\,\omega(e_1, \dots, e_n)$. By Proposition 4.3(ii) and $\operatorname{sgn}(\tau) = -1$, $\operatorname{sgn}(\tau\sigma') = -\operatorname{sgn}(\sigma')$, completing the induction. $\blacksquare$

Theorem 4.6 (determination). Let $\omega$ be an alternating $n$-form on $V$ with basis $e_1, \dots, e_n$, and write $v_i = \sum_j a_{ji} e_j$ for the coordinates. Then

In particular, an alternating $n$-form is determined by the single scalar $\omega(e_1, \dots, e_n)$.

Proof. Expand the arguments one at a time by multilinearity:

the sum running over all ordered $n$-tuples of indices. If two $p_i$ coincide, the term vanishes by alternation. The surviving terms have all $p_i$ distinct; an injective self-map of a finite set is a bijection, so $(p_1, \dots, p_n)$ is a permutation $\sigma$, and Lemma 4.5 evaluates $\omega(e_{\sigma(1)}, \dots, e_{\sigma(n)})$ as $\operatorname{sgn}(\sigma)\,\omega(e_1, \dots, e_n)$. Summing the surviving terms gives the display. $\blacksquare$

Theorem 4.7 (existence: the Leibniz form). The function

is an alternating $n$-linear form with $D(e_1, \dots, e_n) = 1$.

Proof. $n$-linearity. In the $\sigma$-term exactly one factor, $a_{\sigma(i) i}$, involves $v_i$, and it is the coordinate of $v_i$ along $e_{\sigma(i)}$, a linear function of $v_i$. A sum of functions each linear in $v_i$ is linear in $v_i$; this holds for every $i$.

Alternation. Let $\tau = (i\ j)$ be a transposition and compare $D(v_{\tau(1)}, \dots, v_{\tau(n)})$ with $D(v_1, \dots, v_n)$ term by term. The $\sigma$-term of the former is $\operatorname{sgn}(\sigma) \prod_l a_{\sigma(l), \tau(l)}$ (the $l$-th argument is $v_{\tau(l)}$, whose coordinates are indexed by $\tau(l)$). Reindex the product with $l = \tau(l')$: since $\tau^2 = \operatorname{id}$, the product becomes $\prod_{l'} a_{\sigma(\tau(l')), l'} = \prod_{l'} a_{\sigma'(l'), l'}$ where $\sigma' = \sigma\tau$. And $\operatorname{sgn}(\sigma) = \operatorname{sgn}(\sigma'\tau) = \operatorname{sgn}(\sigma')\operatorname{sgn}(\tau) = -\operatorname{sgn}(\sigma')$ by Proposition 4.3(ii). So the terms of $D(v_{\tau(1)}, \dots)$ are exactly the negatives of the terms of $D(v_1, \dots)$, and $D(v_{\tau(1)}, \dots, v_{\tau(n)}) = -D(v_1, \dots, v_n)$; taking $(i, j)$ to be any two positions proves alternation via Lemma 4.4’s equivalence.

Normalisation. The coordinates of $e_i$ are $a_{ji} = \delta_{ji}$. The $\sigma$-term is $\operatorname{sgn}(\sigma) \prod_l \delta_{\sigma(l), l}$: nonzero only if $\sigma(l) = l$ for every $l$, i.e. $\sigma = \operatorname{id}$, contributing $+1 \cdot 1 = 1$. $\blacksquare$

Definition (determinant). The determinant of an $n \times n$ matrix $B = (b_{ij})$ is the value of the Leibniz form on its columns: $\det B = \sum_{\sigma} \operatorname{sgn}(\sigma) b_{\sigma(1) 1} \cdots b_{\sigma(n) n}$. The determinant of a linear map $T: V \to V$ is $\det T := \det B$, where $B$ is the matrix of $T$ in any basis; Corollary 4.9 shows the choice of basis does not matter.

Convention (matrices). The matrix of $T$ in a basis is the array whose $j$-th column holds the coordinates of $T e_j$; matrices act on coordinate columns by producing coordinates, and the matrix of a composition is the product of the matrices, because the $j$-th column of the product $BC$ is $B$ applied to the $j$-th column of $C$, which is exactly the image of $e_j$ under the composition $S \circ T$ computed in coordinates. This columnwise check is the only matrix-composition fact used below.

Theorem 4.8 (multiplicativity). For linear maps $S, T: V \to V$, $\det(S \circ T) = \det(S) \det(T)$.

Proof. Fix $S$ and consider $\omega(v_1, \dots, v_n) = D(Sv_1, \dots, Sv_n)$, an alternating $n$-linear form: composing each argument with the linear map $S$ preserves multilinearity, and repeated arguments still give zero. By Theorem 4.6, $\omega$ is determined by its value on the basis: $\omega = c\, D$ with $c = \omega(e_1, \dots, e_n) = D(Se_1, \dots, Se_n) = \det S$. So for all arguments,

Apply this twice:

The left side is itself an alternating $n$-form in the $v_i$, so by Theorem 4.6 it equals its own basis value times $D$, and that basis value is $D(STe_1, \dots, STe_n) = \det(ST)$. Cancelling the nonzero form $D$ (Theorem 4.7 gives $D(e_1, \dots, e_n) = 1$) yields $\det(ST) = \det(S)\det(T)$. $\blacksquare$

Corollary 4.9. A linear map $T: V \to V$ is invertible if and only if $\det T \neq 0$; and $\det T$ does not depend on the basis used to compute it.

Proof. If $T$ is invertible with inverse $T^{-1}$, then $1 = \det(\operatorname{id}) = \det(T^{-1} T) = \det(T^{-1})\det(T)$ (the Leibniz sum of the identity matrix is $1$, and multiplicativity applies), so $\det T \neq 0$. Conversely, if $\det T = 0$ and an inverse $S$ existed, the same computation would give the absurdity $1 = \det(S)\det(T) = 0$; so no inverse exists. For basis independence: if $B$ and $B'$ are the matrices of $T$ in bases $e$ and $e'$, let $S$ be the matrix whose $j$-th column holds the $e$-coordinates of $e'_j$. Then the $j$-th column of $BS$ holds the $e$-coordinates of $Te'_j$, and applying $S^{-1}$ converts $e$-coordinates to $e'$-coordinates; so $B' = S^{-1} B S$. Hence $\det B' = \det(S^{-1})\det B\det S = \det B$, since $\det(S^{-1})\det(S) = \det(S^{-1}S) = \det(\operatorname{id}) = 1$ by the invertibility computation above. $\blacksquare$