CommeUnJeu · L1 PCSI
Determinants
The determinant is the canonical scalar attached to a family of \(n\) vectors in a chosen basis, obtained as the unique \(n\)-linear alternating form taking value \(1\) on that basis. Concretely, \(\det_\mathcal{B}(x_1, \dots, x_n) \ne 0\) if and only if \((x_1, \dots, x_n)\) is a basis. In \(\mathbb{R}^2\) this scalar is the signed area of the parallelogram spanned by \((x_1, x_2)\) ; in \(\mathbb{R}^3\) it is the signed volume of the parallelepiped spanned by \((x_1, x_2, x_3)\). Three avatars of the same object will appear in this chapter: \(\det_\mathcal{B}\) for a family of vectors in a basis \(\mathcal{B}\), \(\det(u)\) for an endomorphism \(u \in \mathcal{L}(E)\), and \(\det(A)\) for a square matrix \(A \in \mathcal{M}_n(\mathbb{K})\). All three are tied together by the bridge \(\det(u) = \det(\mathrm{Mat}_\mathcal{B}(u))\), and all three rely on a single foundational fact: the space of \(n\)-linear alternating forms on a \(\mathbb{K}\)-vector space of dimension \(n\) is a line. We admit the existence and uniqueness of this form --- the existence proof is beyond our scope --- and build everything from its two defining properties (linear in each variable, alternating) together with the normalisation \(\det_\mathcal{B}(\mathcal{B}) = 1\). The chapter is built in five movements: an algebraic study of \(n\)-linear alternating forms, opened by a geometric motivation in terms of oriented area and oriented volume; the determinant of a family of vectors in a basis; the determinant of an endomorphism; the determinant of a square matrix; and the computation techniques (pivot, expansion along a row or a column, Vandermonde). Convention. All algebraic statements hold over any field \(\mathbb{K}\) ; the geometric interpretation (signed area, signed volume, orientation) is specific to \(\mathbb{R}^2\) and \(\mathbb{R}^3\). We signal these specialisations explicitly each time they appear. Throughout, \(E\) is a finite-dimensional \(\mathbb{K}\)-vector space with \(\dim E = n \ge 1\), \(\mathcal{B} = (e_1, \dots, e_n)\) is a basis of \(E\), and \(\mathcal{X} = (x_1, \dots, x_n)\) denotes a family of \(n\) vectors of \(E\). We write \(\mathrm{Mat}_\mathcal{B}(\mathcal{X})\) for the matrix whose \(j\)-th column is the coordinates of \(x_j\) in \(\mathcal{B}\). The symbols \(\det_\mathcal{B}\), \(\det(u)\), and \(\det(A)\) are three overloaded uses of the same notation ; the argument disambiguates.
I
Multilinear alternating forms
Geometric motivation: oriented areas and volumes
Rigorously defining the area of an arbitrary surface or the volume of an arbitrary solid is hard, but for parallelograms in the plane and parallelepipeds in space the problem becomes algebraic once we fix a reference unit. Let us take, as the starting brick of our investigation, the data of an « oriented unit of area » in the plane and an « oriented unit of volume » in space:
- In the plane, an oriented unit of area is just a parallelogram on a direct basis \(\mathcal{B} = (e_1, e_2)\).
- In space, an oriented unit of volume is a parallelepiped on a direct basis \(\mathcal{B} = (e_1, e_2, e_3)\).
- Homogeneity: multiplying one edge by \(2\) multiplies the oriented area by \(2\).
$$ \textcolor{colorprop}{\det_\mathcal{B}(2u\,;\,v)} = 2 \textcolor{colordef}{\det_\mathcal{B}(u\,;\,v)}. $$ - Additivity: decomposing one edge as \(u + u'\) decomposes the oriented area in the same way.
$$ \det_\mathcal{B}(u + u'\,;\,v) = \textcolor{colordef}{\det_\mathcal{B}(u\,;\,v)} + \textcolor{colorprop}{\det_\mathcal{B}(u'\,;\,v)}. $$
- Characterisation of bases. \((u, v)\) is a basis of the plane \(\iff \det_\mathcal{B}(u, v) \ne 0\). Same in space: \((u, v, w)\) is a basis of space \(\iff \det_\mathcal{B}(u, v, w) \ne 0\). In other words, colinearity of two vectors in the plane is captured by the flatness of the parallelogram they span. Same principle in space.
- Antisymmetry. \(\det_\mathcal{B}\) changes sign whenever two vectors are swapped. In particular, \(\det_\mathcal{B}(u, u) = 0\) in the plane and \(\det_\mathcal{B}(u, v, w) = 0\) in space as soon as two of \(u, v, w\) are equal. We say that \(\det_\mathcal{B}\) is alternating.
- Multilinearity. \(\det_\mathcal{B}\) is linear in each of its variables. This is the formal name for the two rules suggested by figures B and C: scaling one entry by \(\lambda\) scales the result by \(\lambda\), and splitting one entry as \(u + u'\) splits the result accordingly.
- Characterisation of orientation. \((u, v)\) is a direct basis of the plane \(\iff \det_\mathcal{B}(u, v) > 0\). Same in space: \((u, v, w)\) is a direct basis of space \(\iff \det_\mathcal{B}(u, v, w) > 0\).
I.1
Multilinear and alternating forms
We now abstract the geometric picture: « multilinear » and « alternating » will become the defining algebraic properties, and the geometric link will be restored later in this chapter when we prove that there is a unique such form taking the value \(1\) on a chosen basis.
Definition — \(n\)-linear alternating form
Let \(E\) be a \(\mathbb{K}\)-vector space of dimension \(n\). A map \(f \colon E^n \to \mathbb{K}\) is an \(n\)-linear form on \(E\) if it is linear with respect to each variable: for every \(k \in \llbracket 1, n \rrbracket\) and every choice of the other arguments \((x_1, \dots, x_{k-1}, x_{k+1}, \dots, x_n)\) fixed, the partial map $$ x \longmapsto f(x_1, \dots, x_{k-1}, x, x_{k+1}, \dots, x_n) $$ is linear from \(E\) to \(\mathbb{K}\). Such an \(n\)-linear form is alternating if it vanishes whenever two of its arguments are equal: for every \((x_1, \dots, x_n) \in E^n\) and every \(i \ne j\) with \(x_i = x_j\), $$ f(x_1, \dots, x_n) = 0. $$ The cases \(n = 2\) and \(n = 3\) are called bilinear and trilinear. Example
The map \(f \colon (\mathbb{R}^2)^2 \to \mathbb{R}\) defined by \(f((u_1, u_2), (v_1, v_2)) = u_1 v_2 - u_2 v_1\) is a \(2\)-linear alternating form on \(\mathbb{R}^2\). Indeed, \(f\) is bilinear by direct inspection (linear in \((u_1, u_2)\) for \((v_1, v_2)\) fixed, and symmetrically), and \(f(u, u) = u_1 u_2 - u_2 u_1 = 0\), so it vanishes on a repeated argument. Numerically, \(f((2, 1), (1, 3)) = 2 \cdot 3 - 1 \cdot 1 = 5\) and \(f((1, 3), (2, 1)) = 1 \cdot 1 - 3 \cdot 2 = -5\) --- swapping the two arguments changes the sign, the antisymmetry established in the next Theorem. This \(f\) is exactly \(\det_\mathcal{B}\) in the canonical basis of \(\mathbb{R}^2\), and \(|f(u, v)|\) is the area of the parallelogram spanned by \(u\) and \(v\). Skills to practice
- Multilinear and alternating forms
I.2
Properties of \(n\)-linear alternating forms
Three properties follow at once from « \(n\)-linear alternating » and bind the whole chapter: such a form is zero on every linked family, it is invariant under the operation « replace one variable by itself plus a linear combination of the others » (the algebraic engine of the Gauss pivot), and it is antisymmetric --- swapping two arguments multiplies the value by \(-1\). These three properties, together with the normalisation \(\det_\mathcal{B}(\mathcal{B}) = 1\), are everything we need to compute determinants: they drive the pivot reduction, the triangular-determinant rule, and the explicit formulas in dimensions \(2\) and \(3\) derived in the next section.
Theorem — Properties of \(n\)-linear alternating forms
Let \(E\) be a \(\mathbb{K}\)-vector space and \(f\) an \(n\)-linear alternating form on \(E\). Then: - (i) Zero on linked families. For every linked family \((x_1, \dots, x_n)\) of \(E\), \(f(x_1, \dots, x_n) = 0\).
- (ii) Invariance under linear-combination patch. For every \((x_1, \dots, x_n) \in E^n\), every \(k \in \llbracket 1, n \rrbracket\) and every \((\lambda_i)_{i \ne k}\) in \(\mathbb{K}\), $$ f\big(\dots, x_k + \sum_{i \ne k} \lambda_i x_i, \dots\big) = f(x_1, \dots, x_n). $$
- (iii) Antisymmetry. For every \((x_1, \dots, x_n) \in E^n\) and every \(i \ne j\), $$ f(\dots, x_j, \dots, x_i, \dots) = - f(\dots, x_i, \dots, x_j, \dots), $$ where on the left \(x_i\) and \(x_j\) have been swapped in slots \(i\) and \(j\).
The three properties share a common engine: linearize then exploit the alternating property to kill the terms that contain a repeated argument.
- (i) Zero on linked families. The family \((x_1, \dots, x_n)\) being linked, some \(x_k\) writes \(x_k = \sum_{i \ne k} \lambda_i x_i\). By \(n\)-linearity in the \(k\)-th slot, $$ \begin{aligned} f(x_1, \dots, x_n) &= f\big(\dots, \textstyle\sum_{i \ne k} \lambda_i x_i, \dots\big) && \text{(substitute)} \\ &= \sum_{i \ne k} \lambda_i \, f(\dots, x_i, \dots) && \text{(linearity in slot \(k\))} \\ &= 0 && \text{(each summand has \(x_i\) in slots \(i\) and \(k\), alternating).} \end{aligned} $$
- (ii) Invariance under linear-combination patch. Expand by \(n\)-linearity in the \(k\)-th slot: $$ \begin{aligned} f\big(\dots, x_k + \textstyle\sum_{i \ne k} \lambda_i x_i, \dots\big) &= f(\dots, x_k, \dots) + \sum_{i \ne k} \lambda_i \, f(\dots, x_i, \dots) && \text{(linearity in slot \(k\))} \\ &= f(\dots, x_k, \dots) && \text{(each summand has \(x_i\) twice).} \end{aligned} $$
- (iii) Antisymmetry. Fix \(i < j\) and the other slots. Apply the alternating property to the family obtained by replacing both slot-\(i\) and slot-\(j\) entries by \(x_i + x_j\): $$ \begin{aligned} 0 &= f(\dots, x_i + x_j, \dots, x_i + x_j, \dots) && \text{(alternating: equal arguments)} \\ &= f(\dots, x_i, \dots, x_i, \dots) + f(\dots, x_i, \dots, x_j, \dots) \\ &\quad + f(\dots, x_j, \dots, x_i, \dots) + f(\dots, x_j, \dots, x_j, \dots) && \text{(bilinearity in slots \(i, j\))} \\ &= 0 + f(\dots, x_i, \dots, x_j, \dots) + f(\dots, x_j, \dots, x_i, \dots) + 0 && \text{(alternating on the two equal-pair terms).} \end{aligned} $$ Isolating: \(f(\dots, x_j, \dots, x_i, \dots) = - f(\dots, x_i, \dots, x_j, \dots)\).
Skills to practice
- Properties of \(n\)-linear alternating forms
II
Determinant of a family of vectors in a basis
We now admit the central fact of the chapter: in dimension \(n\), after fixing a basis \(\mathcal{B}\), there is a unique \(n\)-linear alternating form \(f\) on \(E^n\) satisfying \(f(\mathcal{B}) = 1\). This unique form is the determinant in the basis \(\mathcal{B}\), denoted \(\det_\mathcal{B}\). We admit its existence and uniqueness --- this is the foundational result whose proof lies beyond our scope. The single property we extract and use everywhere afterwards is one-dimensionality: the space of \(n\)-linear alternating forms on \(E^n\) is the line generated by \(\det_\mathcal{B}\), so every \(n\)-linear alternating form \(f\) equals \(f(\mathcal{B}) \cdot \det_\mathcal{B}\). From this the rest of the chapter flows: the explicit formulas in dimensions \(2\) and \(3\), the characterisation of bases by \(\det \ne 0\), the intrinsic determinant of an endomorphism, and the determinant of a square matrix.
II.1
Definition in a basis
The existence, the uniqueness, and the one-dimensionality stated below are all admitted at this level (the existence proof is beyond our scope). One \(\det_\mathcal{B}\) throughout: there is a single object \(\det_\mathcal{B}\) --- the unique \(n\)-linear alternating form normalised by \(\det_\mathcal{B}(\mathcal{B}) = 1\). Once one-dimensionality is granted, the uniqueness is immediate, and every later computation reduces to evaluating \(\det_\mathcal{B}\) via its two defining properties (linear in each variable, alternating) and the normalisation.
Theorem — Determinant in a basis
Let \(E\) be a \(\mathbb{K}\)-vector space of dimension \(n \ge 1\) and \(\mathcal{B} = (e_1, \dots, e_n)\) a basis of \(E\). - (1) Existence (admitted). There exists an \(n\)-linear alternating form \(\det_\mathcal{B} \colon E^n \to \mathbb{K}\) such that \(\det_\mathcal{B}(\mathcal{B}) = 1\). It is called the determinant in the basis \(\mathcal{B}\).
- (2) Uniqueness. This form is unique: it is the only \(n\)-linear alternating form on \(E^n\) taking the value \(1\) on \(\mathcal{B}\).
- (3) One-dimensionality (admitted). Every \(n\)-linear alternating form \(f\) on \(E^n\) is a scalar multiple of \(\det_\mathcal{B}\), namely \(f = f(\mathcal{B}) \cdot \det_\mathcal{B}\).
Why uniqueness follows from one-dimensionality. Suppose \(g\) is another \(n\)-linear alternating form on \(E^n\) with \(g(\mathcal{B}) = 1\). By (3) applied to \(g\), we have \(g = g(\mathcal{B}) \cdot \det_\mathcal{B} = 1 \cdot \det_\mathcal{B} = \det_\mathcal{B}\). So the normalisation \(f(\mathcal{B}) = 1\) pins down a single form: the uniqueness (2) is exactly the one-dimensionality (3) read at the value \(1\). This is the only part of (1)--(3) we justify here; existence and one-dimensionality themselves are admitted.
Proposition — Determinant in dimensions \(2\) and \(3\)
Let \(\mathcal{B} = (e_1, \dots, e_n)\) be a basis of \(E\). - If \(\dim E = 2\). For \(x, y \in E\) with coordinates \((x_1, x_2)\) and \((y_1, y_2)\) in \(\mathcal{B}\), $$ \textcolor{colorprop}{ \det_\mathcal{B}(x, y) = x_1 y_2 - x_2 y_1. } $$
- If \(\dim E = 3\). For \(x, y, z \in E\) with coordinates \((x_1, x_2, x_3)\), \((y_1, y_2, y_3)\), \((z_1, z_2, z_3)\) in \(\mathcal{B}\), $$ \textcolor{colorprop}{ \det_\mathcal{B}(x, y, z) = \underbrace{x_1 y_2 z_3 + x_2 y_3 z_1 + x_3 y_1 z_2}_{\text{descending diagonals}} - \underbrace{\big(x_3 y_2 z_1 + x_2 y_1 z_3 + x_1 y_3 z_2\big)}_{\text{ascending diagonals}}. } $$
We derive each formula directly from the two defining properties of \(\det_\mathcal{B}\) --- linearity in each variable and the alternating character --- with no enumeration of permutations.
- Dimension \(2\). Write \(x = x_1 e_1 + x_2 e_2\) and \(y = y_1 e_1 + y_2 e_2\). Expand by bilinearity, slot by slot: $$ \begin{aligned} \det_\mathcal{B}(x, y) &= \det_\mathcal{B}(x_1 e_1 + x_2 e_2,\ y_1 e_1 + y_2 e_2) \\ &= x_1 y_1 \det_\mathcal{B}(e_1, e_1) + x_1 y_2 \det_\mathcal{B}(e_1, e_2) \\ &\quad + x_2 y_1 \det_\mathcal{B}(e_2, e_1) + x_2 y_2 \det_\mathcal{B}(e_2, e_2) && \text{(bilinearity).} \end{aligned} $$ The terms \(\det_\mathcal{B}(e_1, e_1)\) and \(\det_\mathcal{B}(e_2, e_2)\) vanish (alternating: a repeated argument). By antisymmetry, \(\det_\mathcal{B}(e_2, e_1) = - \det_\mathcal{B}(e_1, e_2)\), and \(\det_\mathcal{B}(e_1, e_2) = \det_\mathcal{B}(\mathcal{B}) = 1\). Hence \(\det_\mathcal{B}(x, y) = x_1 y_2 - x_2 y_1\).
- Dimension \(3\). Write each vector on the basis and expand by trilinearity. Among the \(3^3 = 27\) resulting terms \(\det_\mathcal{B}(e_a, e_b, e_c)\), those with a repeated index vanish (alternating). The surviving six are the rearrangements of \((e_1, e_2, e_3)\). Each is reduced to \(\pm \det_\mathcal{B}(e_1, e_2, e_3) = \pm 1\) by antisymmetry, the sign being \(+1\) when the rearrangement is obtained by an even number of swaps of adjacent slots and \(-1\) otherwise: $$ \begin{aligned} \det_\mathcal{B}(e_1, e_2, e_3) &= +1, & \det_\mathcal{B}(e_2, e_3, e_1) &= +1, & \det_\mathcal{B}(e_3, e_1, e_2) &= +1, \\ \det_\mathcal{B}(e_1, e_3, e_2) &= -1, & \det_\mathcal{B}(e_2, e_1, e_3) &= -1, & \det_\mathcal{B}(e_3, e_2, e_1) &= -1. \end{aligned} $$ Collecting the coefficients \(x_a y_b z_c\) in front of each surviving rearrangement gives $$ \det_\mathcal{B}(x, y, z) = x_1 y_2 z_3 + x_2 y_3 z_1 + x_3 y_1 z_2 - x_1 y_3 z_2 - x_2 y_1 z_3 - x_3 y_2 z_1, $$ matching the descending-diagonal / ascending-diagonal split announced above.
Example
Compute \(\det_\mathcal{B}\big((2, 1), (1, 3)\big)\) and \(\det_\mathcal{B}\big((1, 0, 2), (0, 3, -1), (4, 2, 1)\big)\) in the canonical bases of \(\mathbb{R}^2\) and \(\mathbb{R}^3\).
$$ \begin{aligned} \det_\mathcal{B}\big((2, 1), (1, 3)\big) &= 2 \cdot 3 - 1 \cdot 1 && \text{(Sarrus in dim \(2\))} \\
&= 5. \end{aligned} $$ For the \(3 \times 3\) Sarrus: $$ \begin{aligned} \det_\mathcal{B}\big((1, 0, 2), (0, 3, -1), (4, 2, 1)\big) &= 1 \cdot 3 \cdot 1 + 0 \cdot (-1) \cdot 4 + 2 \cdot 0 \cdot 2 && \text{(descending diagonals)} \\
&\quad - 2 \cdot 3 \cdot 4 - 0 \cdot 0 \cdot 1 - 1 \cdot (-1) \cdot 2 && \text{(ascending diagonals)} \\
&= 3 + 0 + 0 - 24 - 0 + 2 \\
&= -19. \end{aligned} $$
Skills to practice
- Computing via Sarrus and the structural properties
II.2
Characterisation of bases and change-of-basis formula
Two consequences of the unidimensionality Theorem promote \(\det_\mathcal{B}\) from a curious algebraic gadget to a basis-detector: a family of \(n\) vectors is a basis iff its determinant is nonzero, and changing the basis multiplies all determinants by a single scalar --- the determinant of the new basis in the old one.
Proposition — Characterisation of bases
For \(E\) of dimension \(n\), \(\mathcal{B}\) a basis of \(E\), and \(\mathcal{X} = (x_1, \dots, x_n)\) a family of \(n\) vectors of \(E\), $$ \textcolor{colorprop}{ \mathcal{X} \text{ is a basis of } E \iff \det_\mathcal{B}(\mathcal{X}) \ne 0. } $$ - \((\Leftarrow)\) Contrapositive. If \(\mathcal{X}\) is not a basis, then --- in a vector space of dimension \(n\) --- a family of \(n\) vectors fails to be a basis iff it is linked. So \(\mathcal{X}\) is linked, and by the Properties Theorem (i) of the previous section, \(\det_\mathcal{B}(\mathcal{X}) = 0\).
- \((\Rightarrow)\) Direct. Suppose \(\mathcal{X}\) is a basis. Then by Existence (admis, point (1) of the Definition-Theorem above), the form \(\det_\mathcal{X}\) exists ; it is \(n\)-linear alternating, so by one-dimensionality (point (3) of the same Theorem) it is a multiple of \(\det_\mathcal{B}\): there exists \(c \in \mathbb{K}\) with \(\det_\mathcal{X} = c \cdot \det_\mathcal{B}\). Evaluating at \(\mathcal{X}\): $$ 1 = \det_\mathcal{X}(\mathcal{X}) = c \cdot \det_\mathcal{B}(\mathcal{X}). $$ The equation \(c \cdot \det_\mathcal{B}(\mathcal{X}) = 1\) forces \(\det_\mathcal{B}(\mathcal{X}) \ne 0\).
Proposition — Change-of-basis formula
For \(\mathcal{B}, \mathcal{B}'\) two bases of \(E\), $$ \textcolor{colorprop}{ \det_\mathcal{B} = \det_\mathcal{B}(\mathcal{B}') \cdot \det_{\mathcal{B}'} \qquad \text{and in particular} \qquad \det_\mathcal{B}(\mathcal{B}') \cdot \det_{\mathcal{B}'}(\mathcal{B}) = 1. } $$
The form \(\det_\mathcal{B}\) is \(n\)-linear alternating, so by the one-dimensionality Theorem above it is a scalar multiple of \(\det_{\mathcal{B}'}\): \(\det_\mathcal{B} = c \cdot \det_{\mathcal{B}'}\) for some \(c \in \mathbb{K}\). Evaluate at \(\mathcal{B}'\): $$ \det_\mathcal{B}(\mathcal{B}') = c \cdot \det_{\mathcal{B}'}(\mathcal{B}') = c \cdot 1 = c. $$ Hence \(\det_\mathcal{B} = \det_\mathcal{B}(\mathcal{B}') \cdot \det_{\mathcal{B}'}\). Swapping the roles of \(\mathcal{B}\) and \(\mathcal{B}'\) gives the symmetric identity \(\det_{\mathcal{B}'} = \det_{\mathcal{B}'}(\mathcal{B}) \cdot \det_\mathcal{B}\) as forms. Substituting the second identity into the first, $$ \det_\mathcal{B} = \det_\mathcal{B}(\mathcal{B}') \cdot \det_{\mathcal{B}'}(\mathcal{B}) \cdot \det_\mathcal{B}. $$ Since \(\det_\mathcal{B}\) is not the zero form (it takes value \(1\) at \(\mathcal{B}\)), the scalar factor must equal \(1\): \(\det_\mathcal{B}(\mathcal{B}') \cdot \det_{\mathcal{B}'}(\mathcal{B}) = 1\).
Method — Compute the determinant of a family in a basis
Two routes are available: - Direct, for small dimension. For \(n \le 3\), the explicit dimension-\(2\) / dimension-\(3\) formulas (Sarrus) apply: \(n = 2\) gives the two-term expression \(x_1 y_2 - x_2 y_1\), \(n = 3\) gives the six-term expression. For \(n \ge 4\) there is no comparably short formula --- switch to the matrix route.
- Via the matrix. Form \(A = \mathrm{Mat}_\mathcal{B}(\mathcal{X})\) and reduce to \(\det(A)\), computed by the matrix techniques developed below (pivot method, cofactor expansion, triangular determinant). This is the practical workhorse.
Skills to practice
- Characterising bases and changing basis
III
Determinant of an endomorphism
For \(u \in \mathcal{L}(E)\) and a basis \(\mathcal{B}\), the scalar \(\det_\mathcal{B}(u(e_1), \dots, u(e_n))\) measures how \(u\) scales signed volumes built on \(\mathcal{B}\). Remarkably, this scalar does not depend on the chosen basis \(\mathcal{B}\): it is an intrinsic invariant of \(u\), denoted \(\det(u)\). We prove this intrinsically, using only the one-dimensionality and change-of-basis results established earlier in this chapter --- without invoking matrices, which appear only in the next section. From this intrinsic Definition follow multiplicativity (\(\det(u \circ v) = \det(u) \det(v)\)), characterisation of automorphisms (\(u\) bijective iff \(\det(u) \ne 0\)), and the scalar identity \(\det(\lambda u) = \lambda^n \det(u)\).
III.1
Intrinsic definition and basis-independence
We start from the scalar that an endomorphism \(u\) produces in a fixed basis \(\mathcal{B}\), namely \(\det_\mathcal{B}(u(e_1), \dots, u(e_n))\), and show that the same scalar appears regardless of the basis chosen. This justifies the notation \(\det(u)\), an intrinsic invariant of \(u\).
Proposition — Effect of an endomorphism on a family
For \(u \in \mathcal{L}(E)\) and a basis \(\mathcal{B}\) of \(E\), there exists a unique scalar \(\lambda_\mathcal{B}(u) \in \mathbb{K}\) such that, for every family \(\mathcal{X} = (x_1, \dots, x_n)\) of \(n\) vectors of \(E\), $$ \textcolor{colorprop}{ \det_\mathcal{B}\big(u(x_1), \dots, u(x_n)\big) = \lambda_\mathcal{B}(u) \cdot \det_\mathcal{B}(x_1, \dots, x_n). } $$ Moreover, \(\lambda_\mathcal{B}(u) = \det_\mathcal{B}(u(e_1), \dots, u(e_n))\).
Define \(\varphi \colon E^n \to \mathbb{K}\) by \(\varphi(x_1, \dots, x_n) = \det_\mathcal{B}(u(x_1), \dots, u(x_n))\).
- \(\varphi\) is \(n\)-linear. For each \(k\), \(u\) being linear, the partial map \(x \mapsto u(x)\) is linear in slot \(k\) ; composing with \(\det_\mathcal{B}\) (which is \(n\)-linear) keeps the slot-\(k\) partial map linear.
- \(\varphi\) is alternating. If \(x_i = x_j\) for \(i \ne j\), then \(u(x_i) = u(x_j)\), and \(\det_\mathcal{B}\) vanishes on a family with two equal arguments.
Theorem — Determinant of an endomorphism
For \(u \in \mathcal{L}(E)\), the scalar \(\lambda_\mathcal{B}(u)\) of the preceding Proposition does not depend on the basis \(\mathcal{B}\). This scalar is called the determinant of \(u\) and denoted \(\det(u)\). For every basis \(\mathcal{B}\) and every family \(\mathcal{X} \in E^n\), $$ \textcolor{colorprop}{ \det_\mathcal{B}\big(u(\mathcal{X})\big) = \det(u) \cdot \det_\mathcal{B}(\mathcal{X}). } $$
Let \(\mathcal{B}, \mathcal{B}'\) be two bases of \(E\). Apply the preceding Proposition's identity in each basis, with the family \(\mathcal{X} = \mathcal{B}'\): $$ \begin{aligned} \det_\mathcal{B}\big(u(\mathcal{B}')\big) &= \lambda_\mathcal{B}(u) \cdot \det_\mathcal{B}(\mathcal{B}') && \text{(in basis \(\mathcal{B}\), \(\mathcal{X} = \mathcal{B}'\))} \\
\det_{\mathcal{B}'}\big(u(\mathcal{B}')\big) &= \lambda_{\mathcal{B}'}(u) \cdot \det_{\mathcal{B}'}(\mathcal{B}') = \lambda_{\mathcal{B}'}(u) && \text{(in basis \(\mathcal{B}'\), \(\mathcal{X} = \mathcal{B}'\)).} \end{aligned} $$ By the change-of-basis formula established earlier, \(\det_\mathcal{B}\big(u(\mathcal{B}')\big) = \det_\mathcal{B}(\mathcal{B}') \cdot \det_{\mathcal{B}'}\big(u(\mathcal{B}')\big) = \det_\mathcal{B}(\mathcal{B}') \cdot \lambda_{\mathcal{B}'}(u)\). Equating with the first line: $$ \lambda_\mathcal{B}(u) \cdot \det_\mathcal{B}(\mathcal{B}') = \det_\mathcal{B}(\mathcal{B}') \cdot \lambda_{\mathcal{B}'}(u). $$ Now \(\det_\mathcal{B}(\mathcal{B}') \ne 0\) since \(\mathcal{B}'\) is a basis (Characterisation of bases, established earlier), so we can cancel: \(\lambda_\mathcal{B}(u) = \lambda_{\mathcal{B}'}(u)\).
Skills to practice
- Computing the determinant of an endomorphism
III.2
Multiplicativity and characterisation of automorphisms
Once \(\det(u)\) is known to be a well-defined scalar attached to \(u\), its three structural properties --- identity, multiplicativity, scalar exponent \(\lambda^n\) --- follow from the defining identity. The most useful consequence is the characterisation \(u \in \mathrm{GL}(E) \iff \det(u) \ne 0\), which transports to the matrix side in the next section.
Proposition — Properties of \(\det\) on \(\mathcal{L}(E)\)
For \(u, v \in \mathcal{L}(E)\) and \(\lambda \in \mathbb{K}\): - (0) Identity. \(\det(\mathrm{Id}_E) = 1\).
- (i) Multiplicativity. \(\det(u \circ v) = \det(u) \det(v)\).
- (ii) Characterisation of automorphisms. \(u\) is an automorphism of \(E\) if and only if \(\det(u) \ne 0\). In this case, \(\det(u^{-1}) = \det(u)^{-1}\).
- (iii) Scalar. \(\det(\lambda u) = \lambda^n \det(u)\).
We work intrinsically (apply the Definition-Theorem of \(\det\) twice or once as needed).
- (0) Identity. Direct from the Definition-Theorem: $$ \det(\mathrm{Id}_E) = \det_\mathcal{B}(\mathrm{Id}_E(e_1), \dots, \mathrm{Id}_E(e_n)) = \det_\mathcal{B}(e_1, \dots, e_n) = 1. $$
- (i) Multiplicativity. Apply the defining identity to \(v(\mathcal{X})\), then to \(u(v(\mathcal{X}))\): $$ \begin{aligned} \det_\mathcal{B}\big((u \circ v)(\mathcal{X})\big) &= \det_\mathcal{B}\big(u(v(\mathcal{X}))\big) && \text{(definition of \(u \circ v\))} \\ &= \det(u) \cdot \det_\mathcal{B}(v(\mathcal{X})) && \text{(Def-Theorem on \(u\), \(v(\mathcal{X})\))} \\ &= \det(u) \cdot \det(v) \cdot \det_\mathcal{B}(\mathcal{X}) && \text{(Def-Theorem on \(v\), \(\mathcal{X}\)).} \end{aligned} $$ The Def-Theorem also says, applied to \(u \circ v\) directly, that \(\det_\mathcal{B}\big((u \circ v)(\mathcal{X})\big) = \det(u \circ v) \cdot \det_\mathcal{B}(\mathcal{X})\). The two expressions for \(\det_\mathcal{B}\big((u \circ v)(\mathcal{X})\big)\) are therefore equal for every \(\mathcal{X} \in E^n\) ; evaluating at \(\mathcal{X} = \mathcal{B}\) (so that \(\det_\mathcal{B}(\mathcal{X}) = 1\)) extracts the scalar identity \(\det(u \circ v) = \det(u) \det(v)\).
- (ii) Characterisation of automorphisms. \((\Rightarrow)\) If \(u\) is bijective, \(u \circ u^{-1} = \mathrm{Id}_E\). By (0) and (i), \(\det(u) \det(u^{-1}) = \det(\mathrm{Id}_E) = 1\) ; in particular both \(\det(u)\) and \(\det(u^{-1})\) are nonzero, and \(\det(u^{-1}) = \det(u)^{-1}\). \((\Leftarrow)\) Suppose \(\det(u) \ne 0\). Then \(\det_\mathcal{B}(u(\mathcal{B})) = \det(u) \cdot \det_\mathcal{B}(\mathcal{B}) = \det(u) \ne 0\). By the Characterisation of bases established earlier, \(u(\mathcal{B}) = (u(e_1), \dots, u(e_n))\) is a basis of \(E\), hence \(u\) is surjective. On a finite-dimensional space, a surjective endomorphism is bijective.
- (iii) Scalar. Apply the defining identity: $$ \begin{aligned} \det_\mathcal{B}\big((\lambda u)(\mathcal{X})\big) &= \det_\mathcal{B}\big(\lambda \cdot u(x_1), \dots, \lambda \cdot u(x_n)\big) && \text{(definition of \(\lambda u\))} \\ &= \lambda^n \det_\mathcal{B}\big(u(x_1), \dots, u(x_n)\big) && \text{(\(n\)-linearity, factor \(\lambda\) out of each slot)} \\ &= \lambda^n \cdot \det(u) \cdot \det_\mathcal{B}(\mathcal{X}) && \text{(Definition-Theorem).} \end{aligned} $$ Comparing with \(\det_\mathcal{B}\big((\lambda u)(\mathcal{X})\big) = \det(\lambda u) \cdot \det_\mathcal{B}(\mathcal{X})\) at \(\mathcal{X} = \mathcal{B}\): \(\det(\lambda u) = \lambda^n \det(u)\).
Example
Two geometric endomorphisms of \(E\) make the volume-scaling reading of \(\det(u)\) concrete. - Homothety. For \(u = \lambda \, \mathrm{Id}_E\) with \(\lambda \in \mathbb{K}\), by (iii) of the Properties Proposition, \(\det(u) = \lambda^n\). Geometric reading in \(\mathbb{R}^n\): a homothety of ratio \(\lambda\) scales every signed \(n\)-volume by \(\lambda^n\).
- Projector onto a proper subspace. If \(p \in \mathcal{L}(E)\) is a projector onto a subspace of dimension \(< n\), then \(p\) is not surjective, hence not bijective ; by (ii), \(\det(p) = 0\). Geometric reading: a projector flattens \(E\) onto a strictly lower-dimensional subspace, collapsing every \(n\)-volume to \(0\).
Skills to practice
- Using multiplicativity and the characterisation of automorphisms
IV
Determinant of a square matrix
The matrix side of the determinant. Identifying a matrix \(A \in \mathcal{M}_n(\mathbb{K})\) with the family of its columns in the canonical basis \(\mathcal{B}_n\) of \(\mathbb{K}^n\), we define \(\det(A) = \det_{\mathcal{B}_n}(C_1, \dots, C_n)\). It inherits at once the two defining properties of \(\det_{\mathcal{B}_n}\): it is \(n\)-linear and alternating with respect to the columns. The link with the endomorphism determinant is the bridge \(\det(u) = \det(\mathrm{Mat}_\mathcal{B}(u))\). The main Theorem of the section bundles five properties: \(n\)-linear alternating by columns, multiplicativity \(\det(AB) = \det(A) \det(B)\), characterisation of invertibility \(A \in \mathrm{GL}_n(\mathbb{K}) \iff \det(A) \ne 0\), invariance by similarity \(\det(P^{-1} A P) = \det(A)\), and \(\det(A^T) = \det(A)\) (which makes \(\det\) \(n\)-linear alternating by rows as well). The geometric interpretation in \(\mathbb{R}^2\) and \(\mathbb{R}^3\) --- \(|\det(A)|\) is the absolute area / volume of the image parallelogram / parallelepiped --- is then a direct consequence in those special cases.
IV.1
Definition and link with the basis-determinant
We import the definition of \(\det_\mathcal{B}\) to matrices by reading \(A \in \mathcal{M}_n(\mathbb{K})\) as the family of its columns in the canonical basis \(\mathcal{B}_n\) of \(\mathbb{K}^n\). Two bridging Propositions express the resulting scalar in two equivalent ways: in terms of the column family in a basis (Family in a basis vs.\ matrix), and in terms of an endomorphism through its representing matrix (Bridge \(\det(u) = \det(\mathrm{Mat}_\mathcal{B}(u))\)).
Definition — Determinant of a square matrix
For \(A = (a_{ij}) \in \mathcal{M}_n(\mathbb{K})\) with columns \((C_1, \dots, C_n)\) in the canonical basis \(\mathcal{B}_n = (E_1, \dots, E_n)\) of \(\mathbb{K}^n\), the determinant of \(A\) is the scalar $$ \det(A) = \det_{\mathcal{B}_n}(C_1, \dots, C_n). $$ Equivalently, \(\det\) is the unique map \(\mathcal{M}_n(\mathbb{K}) \to \mathbb{K}\) that is linear with respect to each column, alternating in the columns, and equal to \(1\) on the identity matrix: \(\det(I_n) = 1\). Proposition — Family in a basis vs.\ matrix
For \(E\) a \(\mathbb{K}\)-vector space of dimension \(n\), \(\mathcal{B}\) a basis of \(E\), and \(\mathcal{X}\) a family of \(n\) vectors of \(E\), $$ \textcolor{colorprop}{ \det_\mathcal{B}(\mathcal{X}) = \det\big(\mathrm{Mat}_\mathcal{B}(\mathcal{X})\big). } $$
Let \(A = \mathrm{Mat}_\mathcal{B}(\mathcal{X}) = (C_1, \dots, C_n)\), whose \(j\)-th column \(C_j \in \mathbb{K}^n\) is the coordinate vector of \(x_j\) in \(\mathcal{B}\). By definition, \(\det(A) = \det_{\mathcal{B}_n}(C_1, \dots, C_n)\). Now the coordinate isomorphism \(E \to \mathbb{K}^n\), \(v \mapsto \mathrm{coord}_\mathcal{B}(v)\), sends \(\mathcal{B}\) to the canonical basis \(\mathcal{B}_n\) and \(x_j\) to \(C_j\). The map \(\mathcal{X} = (x_1, \dots, x_n) \mapsto \det_{\mathcal{B}_n}(\mathrm{coord}_\mathcal{B}(x_1), \dots, \mathrm{coord}_\mathcal{B}(x_n))\) is \(n\)-linear (composition of the linear coordinate map with the \(n\)-linear \(\det_{\mathcal{B}_n}\)), alternating (it inherits the alternating character through the injective coordinate map), and takes the value \(1\) at \(\mathcal{B}\) (since \(\mathrm{coord}_\mathcal{B}(\mathcal{B}) = \mathcal{B}_n\) and \(\det_{\mathcal{B}_n}(\mathcal{B}_n) = 1\)). By the uniqueness of \(\det_\mathcal{B}\), this map is \(\det_\mathcal{B}\). Evaluating at \(\mathcal{X}\) gives \(\det_\mathcal{B}(\mathcal{X}) = \det_{\mathcal{B}_n}(C_1, \dots, C_n) = \det(A)\).
Proposition — Bridge between endomorphism and matrix determinant
For every \(u \in \mathcal{L}(E)\) and every basis \(\mathcal{B}\) of \(E\), $$ \textcolor{colorprop}{ \det(u) = \det\big(\mathrm{Mat}_\mathcal{B}(u)\big). } $$
By definition of \(\det(u)\) (Definition-Theorem of the previous section), \(\det(u) = \det_\mathcal{B}(u(e_1), \dots, u(e_n))\). The matrix of the family \((u(e_1), \dots, u(e_n))\) in the basis \(\mathcal{B}\) is exactly \(\mathrm{Mat}_\mathcal{B}(u)\) --- this is the definition of the matrix of an endomorphism. By the previous Proposition (Family-in-basis vs.\ matrix), $$ \det_\mathcal{B}(u(e_1), \dots, u(e_n)) = \det\big(\mathrm{Mat}_\mathcal{B}(u(e_1), \dots, u(e_n))\big) = \det\big(\mathrm{Mat}_\mathcal{B}(u)\big). $$ Hence \(\det(u) = \det\big(\mathrm{Mat}_\mathcal{B}(u)\big)\). As a consequence, \(\det\big(\mathrm{Mat}_\mathcal{B}(u)\big)\) is independent of \(\mathcal{B}\) --- equivalently, similar matrices have the same determinant (also obtained as (iv) of the Main Theorem below).
Skills to practice
- Definition and link with the basis-determinant
IV.2
Multiplicativity\(\virgule\) transposition and the group morphism
This subsection collects the structural properties of the matrix determinant:
- \(n\)-linear and alternating in the columns (hence also in the rows, via transposition);
- multiplicative on products;
- invariant under similarity;
- non-vanishing characterises invertibility.
Theorem — Main properties of the matrix determinant
For \(A, B \in \mathcal{M}_n(\mathbb{K})\), \(P \in \mathrm{GL}_n(\mathbb{K})\), \(\lambda \in \mathbb{K}\): - (i) Multilinear alternating by columns. \(A \mapsto \det(A)\) is \(n\)-linear alternating with respect to the columns of \(A\). In particular, \(\det(\lambda A) = \lambda^n \det(A)\).
- (ii) Multiplicativity. \(\det(AB) = \det(A) \det(B)\).
- (iii) Characterisation of invertibility. \(A \in \mathrm{GL}_n(\mathbb{K})\) if and only if \(\det(A) \ne 0\). In this case, \(\det(A^{-1}) = \det(A)^{-1}\).
- (iv) Invariance by similarity. \(\det(P^{-1} A P) = \det(A)\).
- (v) Transposition. \(\det(A^T) = \det(A)\). As a consequence, \(\det\) is also \(n\)-linear alternating with respect to the rows of \(A\).
- (i) Multilinear alternating by columns. Immediate from the definition: \(\det(A) = \det_{\mathcal{B}_n}(C_1, \dots, C_n)\), and \(\det_{\mathcal{B}_n}\) is \(n\)-linear alternating. For \(\lambda A\), each of the \(n\) columns is scaled by \(\lambda\), so by \(n\)-linearity \(\det(\lambda A) = \lambda^n \det(A)\).
- (ii) Multiplicativity. Define \(\varphi \colon (\mathbb{K}^n)^n \to \mathbb{K}\) by \(\varphi(C_1, \dots, C_n) = \det_{\mathcal{B}_n}(A C_1, \dots, A C_n)\). The map \(\varphi\) is \(n\)-linear (each slot \(C_k \mapsto A C_k\) is linear, composed with the \(n\)-linear \(\det_{\mathcal{B}_n}\)) and alternating (if \(C_i = C_j\), then \(A C_i = A C_j\) and \(\det_{\mathcal{B}_n}\) vanishes on a repeated argument). By one-dimensionality (Theorem « Determinant in a basis », point (3)), \(\varphi = \varphi(\mathcal{B}_n) \cdot \det_{\mathcal{B}_n}\). The constant is \(\varphi(\mathcal{B}_n) = \det_{\mathcal{B}_n}(A E_1, \dots, A E_n) = \det_{\mathcal{B}_n}(\text{columns of } A) = \det(A)\), since \(A E_i\) is the \(i\)-th column of \(A\). Evaluating \(\varphi\) at the columns \((B_1, \dots, B_n)\) of \(B\): $$ \det(AB) = \det_{\mathcal{B}_n}(A B_1, \dots, A B_n) = \varphi(B_1, \dots, B_n) = \det(A) \cdot \det_{\mathcal{B}_n}(B_1, \dots, B_n) = \det(A) \det(B). $$
- (iii) Characterisation of invertibility. First \(\det(I_n) = \det_{\mathcal{B}_n}(E_1, \dots, E_n) = \det_{\mathcal{B}_n}(\mathcal{B}_n) = 1\), straight from the defining normalisation --- no closed-form formula needed. \((\Rightarrow)\) If \(A\) is invertible, \(A A^{-1} = I_n\) gives, by (ii), \(\det(A) \det(A^{-1}) = \det(I_n) = 1\) ; both are nonzero and \(\det(A^{-1}) = \det(A)^{-1}\). \((\Leftarrow)\) If \(\det(A) \ne 0\), the columns \(C_1, \dots, C_n\) of \(A\) satisfy \(\det_{\mathcal{B}_n}(C_1, \dots, C_n) \ne 0\), so by « Characterisation of bases » they form a basis of \(\mathbb{K}^n\). The map \(X \mapsto A X\) sends \(E_i\) to \(C_i\), so its image contains a basis, hence equals \(\mathbb{K}^n\) ; a surjective endomorphism of a finite-dimensional space is bijective, so \(A\) is invertible.
- (iv) Invariance by similarity. By (ii) and (iii), $$ \det(P^{-1} A P) = \det(P^{-1}) \det(A) \det(P) = \det(P)^{-1} \det(P) \det(A) = \det(A). $$
- (v) Transposition. We admit the identity \(\det(A^T) = \det(A)\) as a foundational result (its démonstration is non exigible at this level). It is precisely this transposition invariance that licenses transferring every column statement to the rows: \(\det\) is \(n\)-linear alternating in the columns of \(A^T\) by (i), i.e.\ in the rows of \(A\) ; combined with \(\det(A) = \det(A^T)\), the determinant is therefore also \(n\)-linear alternating with respect to the rows of \(A\). In particular every column statement in this chapter has a row counterpart.
Proposition — \(\det \colon \mathrm{GL}_n(\mathbb{K}) \to \mathbb{K}^*\) is a surjective group morphism
The restriction \(\det \colon \mathrm{GL}_n(\mathbb{K}) \to \mathbb{K}^*\) is a surjective group morphism: for \(A, B \in \mathrm{GL}_n(\mathbb{K})\), \(\det(AB) = \det(A) \det(B)\), \(\det(I_n) = 1\), \(\det(A^{-1}) = \det(A)^{-1}\), and every \(\lambda \in \mathbb{K}^*\) is the determinant of some invertible matrix.
The morphism part is the restriction of (ii) and (iii) of the Main Theorem to \(\mathrm{GL}_n(\mathbb{K})\): \(\det(AB) = \det(A) \det(B)\), \(\det(I_n) = 1\), \(\det(A^{-1}) = \det(A)^{-1}\). For surjectivity, fix \(\lambda \in \mathbb{K}^*\) and set \(D = \mathrm{diag}(\lambda, 1, \dots, 1)\). It is invertible (inverse \(\mathrm{diag}(\lambda^{-1}, 1, \dots, 1)\)). Its columns are \((\lambda E_1, E_2, \dots, E_n)\), so by \(n\)-linearity of \(\det\) in the first column (point (i) of the Main Theorem) we may factor \(\lambda\) out of column \(1\): $$ \det(D) = \det(\lambda E_1, E_2, \dots, E_n) = \lambda \det(E_1, E_2, \dots, E_n) = \lambda \det(I_n) = \lambda, $$ using \(\det(I_n) = 1\). No general triangular-determinant result is needed. Hence every \(\lambda \in \mathbb{K}^*\) is reached.
Example
\(\det(I_n) = 1\). The diagonal matrix \(\mathrm{diag}(\lambda_1, \dots, \lambda_n)\) has determinant \(\lambda_1 \cdots \lambda_n\) (it is triangular, so its determinant is the product of its diagonal entries). Non-example. \(\det\) is not linear: in general \(\det(A + B) \ne \det(A) + \det(B)\). Counter-example in dimension \(2\): \(A = \begin{pmatrix} 1 & 0 \\ 0 & 0 \end{pmatrix}\), \(B = \begin{pmatrix} 0 & 0 \\ 0 & 1 \end{pmatrix}\) have \(\det(A) = \det(B) = 0\), yet \(A + B = I_2\) and \(\det(A + B) = 1\). Skills to practice
- Using multiplicativity and transposition
V
Computation of determinants
We now turn to practical computation. Three classes of techniques cover essentially every determinant met at this level: elementary row / column operations and the Gauss pivot, exploitation of triangular structure, and cofactor expansion along a well-chosen row or column ; the Vandermonde determinant is a classical fourth case. All of them rest on the same foundation --- the definition of \(\det\) and its structural properties (multilinearity by rows / columns, the alternating character, and the triangular-determinant rule) --- with no recourse to an explicit summation formula.
V.1
Effect of elementary operations
Three elementary operations on the columns (or symmetrically on the rows) of \(A\) change \(\det(A)\) in a predictable way. Combining them strategically clears the matrix to triangular form, where the determinant reads off as the product of diagonal entries.
Proposition — Determinant and elementary operations
For \(A \in \mathcal{M}_n(\mathbb{K})\), \(i \ne j\), \(\lambda \in \mathbb{K}\): - Transvection. The operation \(L_i \leftarrow L_i + \lambda L_j\) (or \(C_i \leftarrow C_i + \lambda C_j\)) does not change the determinant.
- Dilation. The operation \(L_i \leftarrow \lambda L_i\) (or \(C_i \leftarrow \lambda C_i\)) multiplies the determinant by \(\lambda\).
- Swap. The operation \(L_i \leftrightarrow L_j\) (or \(C_i \leftrightarrow C_j\)) multiplies the determinant by \(-1\).
By transposition-invariance (point (v) of the Main Theorem above), it suffices to treat the column operations ; the row identities follow by applying the column ones to \(A^T\).
- Transvection. By multilinearity in column \(i\), $$ \begin{aligned} \det(\dots, C_i + \lambda C_j, \dots, C_j, \dots) &= \det(\dots, C_i, \dots, C_j, \dots) \\ &\quad + \lambda \det(\dots, C_j, \dots, C_j, \dots) && \text{(linearity in column \(i\))} \\ &= \det(\dots, C_i, \dots, C_j, \dots) + 0 && \text{(alternating: \(C_j\) twice)} \\ &= \det(A). \end{aligned} $$
- Dilation. By linearity in column \(i\), \(\det(\dots, \lambda C_i, \dots) = \lambda \det(\dots, C_i, \dots) = \lambda \det(A)\).
- Swap. Direct from antisymmetry of \(\det\) in columns (Properties Theorem (iii) of the first section, inherited by \(\det\)).
Remark (dilation with \(\lambda = 0\)). The identity \(\det(\dots, \lambda C_i, \dots) = \lambda \det(A)\) holds for every \(\lambda \in \mathbb{K}\), including \(\lambda = 0\) (both sides become \(0\)). When the dilation is used as an elementary, reversible row operation in the Gauss pivot algorithm, however, we must impose \(\lambda \ne 0\) --- otherwise the operation collapses a row to zero and is not invertible. The identity is a statement of \(n\)-linearity ; the algorithm imposes the invertibility constraint.
Method — Compute a determinant by the pivot method
- Step 1. Pick a row or column with a maximum number of zeros if possible --- or create zeros by a sequence of transvections \(L_i \leftarrow L_i + \lambda L_j\) (which leave \(\det\) unchanged).
- Step 2. Iterate to bring the matrix to upper- or lower-triangular form.
- Step 3. Read \(\det(A)\) as the product of the diagonal entries (Triangular determinant Proposition of the next subsection).
Example
Compute \(\displaystyle\det\begin{pmatrix} 1 & 2 & 3 \\ 4 & 5 & 6 \\ 7 & 8 & 10 \end{pmatrix}\) by row reduction.
$$ \begin{aligned} \det\begin{pmatrix} 1 & 2 & 3 \\
4 & 5 & 6 \\
7 & 8 & 10 \end{pmatrix} &= \det\begin{pmatrix} 1 & 2 & 3 \\
0 & -3 & -6 \\
0 & -6 & -11 \end{pmatrix} && \text{(\(L_2 \leftarrow L_2 - 4 L_1\), \(L_3 \leftarrow L_3 - 7 L_1\))} \\
&= \det\begin{pmatrix} 1 & 2 & 3 \\
0 & -3 & -6 \\
0 & 0 & 1 \end{pmatrix} && \text{(\(L_3 \leftarrow L_3 - 2 L_2\))} \\
&= 1 \cdot (-3) \cdot 1 && \text{(triangular, product of diagonal entries)} \\
&= -3. \end{aligned} $$
Skills to practice
- Computing by the pivot method
V.2
Triangular determinants
For a triangular matrix the determinant is simply the product of the diagonal entries. This is exactly why the Gauss pivot computes determinants: clear the matrix to triangular form, then read off the product of the pivots. We prove it directly by induction, using only the \(n\)-linearity and the alternating character of \(\det\) (factor the diagonal entry out of the first column, then clear that column by transvections) --- no permutation enumeration, no cofactor expansion.
Proposition — Triangular determinant
For \(A = (a_{ij}) \in \mathcal{M}_n(\mathbb{K})\) upper-triangular (resp.\ lower-triangular), $$ \textcolor{colorprop}{ \det(A) = \prod_{i=1}^{n} a_{ii}. } $$ In particular the determinant of a diagonal matrix is the product of its diagonal entries.
Treat the upper-triangular case ; the lower-triangular case follows from \(\det(A^T) = \det(A)\) (point (v) of the Main Theorem), since the transpose of a lower-triangular matrix is upper-triangular with the same diagonal. We use only the two defining properties of \(\det\) --- \(n\)-linearity and the alternating character (with the normalisation \(\det(I_n) = 1\)) --- and argue by induction on \(n\), with no cofactor expansion and no minor. The base case \(n = 1\) is \(\det\big((a_{11})\big) = a_{11}\). Let \(A = (C_1, \dots, C_n)\) be upper-triangular of size \(n \ge 2\), with columns \(C_j = \sum_{i \le j} a_{ij} E_i\) in the canonical basis \(\mathcal{B}_n = (E_1, \dots, E_n)\) of \(\mathbb{K}^n\). The first column is \(C_1 = a_{11} E_1\). By \(n\)-linearity in the first column, factor \(a_{11}\) out: $$ \det(A) = \det(a_{11} E_1, C_2, \dots, C_n) = a_{11} \, \det(E_1, C_2, \dots, C_n). $$ Now subtract from each later column \(C_j\) (\(j \ge 2\)) its \(E_1\)-component \(a_{1j} E_1\). Each such operation \(C_j \leftarrow C_j - a_{1j} E_1\) is a transvection of the form « add a multiple of column \(1\) to column \(j\) », which leaves the determinant unchanged: by linearity in column \(j\) it adds \(-a_{1j}\det(E_1, \dots, E_1, \dots)\), a term with \(E_1\) appearing twice, hence \(0\) by the alternating property. After these operations every column \(C_j' = C_j - a_{1j} E_1 = \sum_{2 \le i \le j} a_{ij} E_i\) lies in \(\mathrm{span}(E_2, \dots, E_n)\), so $$ \det(A) = a_{11} \, \det(E_1, C_2', \dots, C_n'). $$ The family \((C_2', \dots, C_n')\), read in the coordinates \((E_2, \dots, E_n)\), is exactly an upper-triangular matrix of size \(n - 1\) with diagonal \((a_{22}, \dots, a_{nn})\) ; and \(\det(E_1, C_2', \dots, C_n')\) depends \((n-1)\)-linearly and alternatingly on \((C_2', \dots, C_n')\) while taking the value \(1\) at \((E_1, E_2, \dots, E_n)\). By the induction hypothesis applied to this size-\((n-1)\) block, \(\det(E_1, C_2', \dots, C_n') = a_{22} \cdots a_{nn}\). Hence \(\det(A) = a_{11} a_{22} \cdots a_{nn}\).
Skills to practice
- Computing triangular determinants
V.3
Expansion along a row or a column
A determinant of size \(n\) can be expressed as a signed linear combination of determinants of size \(n - 1\) (the minors), via expansion along a chosen row or column. It turns the computation of an \(n \times n\) determinant into \(n\) determinants of size \(n - 1\), and is most efficient along a row or column carrying many zeros. We admit the expansion formula at this level (its proof is not required) and use it as a computational workhorse.
Definition — Minor and cofactor
For \(A \in \mathcal{M}_n(\mathbb{K})\) with \(n \ge 2\) and \(i, j \in \llbracket 1, n \rrbracket\): - the minor of position \((i, j)\) is \(\Delta_{ij}(A) = \det A^{(i, j)}\), where \(A^{(i, j)} \in \mathcal{M}_{n-1}(\mathbb{K})\) is the matrix obtained by deleting row \(i\) and column \(j\) of \(A\) ;
- the cofactor of position \((i, j)\) is the signed minor \((-1)^{i+j} \Delta_{ij}(A)\).
Theorem — Expansion along a row or column
For \(A \in \mathcal{M}_n(\mathbb{K})\) with \(n \ge 2\): - Expansion along row \(i\). For every \(i \in \llbracket 1, n \rrbracket\), $$ \textcolor{colorprop}{ \det(A) = \sum_{j=1}^{n} (-1)^{i+j} a_{ij} \, \Delta_{ij}(A). } $$
- Expansion along column \(j\). For every \(j \in \llbracket 1, n \rrbracket\), $$ \textcolor{colorprop}{ \det(A) = \sum_{i=1}^{n} (-1)^{i+j} a_{ij} \, \Delta_{ij}(A). } $$
Method — Expand along a well-chosen row or column
- Step 1. Pick a row or column with the maximum number of zeros --- it reduces the number of nonzero cofactors to compute.
- Step 2. If no row / column has zeros to start with, use the pivot method described above to create them first.
- Step 3. For small \(n \le 4\), expansion is often quicker than full pivot reduction ; for \(n \ge 5\), prefer the pivot.
Example
Compute \(\det\begin{pmatrix} \textcolor{colordef}{1} & 4 & 7 \\ \textcolor{colordef}{2} & 5 & 8 \\ \textcolor{colordef}{3} & 6 & 9 \end{pmatrix}\) by expansion along the first column.
The first column entries \(\textcolor{colordef}{1, 2, 3}\) are pulled out, each multiplying its associated \(2 \times 2\) minor \(\textcolor{colorprop}{\det(\cdot)}\). The sign pattern for column \(1\) is \(+, -, +\): $$ \begin{aligned} \det\begin{pmatrix} \textcolor{colordef}{1} & 4 & 7 \\
\textcolor{colordef}{2} & 5 & 8 \\
\textcolor{colordef}{3} & 6 & 9 \end{pmatrix} &= \textcolor{colordef}{1} \cdot \textcolor{colorprop}{\det\begin{pmatrix} 5 & 8 \\
6 & 9 \end{pmatrix}} - \textcolor{colordef}{2} \cdot \textcolor{colorprop}{\det\begin{pmatrix} 4 & 7 \\
6 & 9 \end{pmatrix}} + \textcolor{colordef}{3} \cdot \textcolor{colorprop}{\det\begin{pmatrix} 4 & 7 \\
5 & 8 \end{pmatrix}} && \text{(expansion along column \(1\))} \\
&= \textcolor{colordef}{1} \cdot \textcolor{colorprop}{(45 - 48)} - \textcolor{colordef}{2} \cdot \textcolor{colorprop}{(36 - 42)} + \textcolor{colordef}{3} \cdot \textcolor{colorprop}{(32 - 35)} && \text{(\(2 \times 2\) Sarrus on each minor)} \\
&= -3 + 12 - 9 \\
&= 0. \end{aligned} $$ The vanishing reflects that the three columns of the original matrix are linked: \(C_3 = 2 C_2 - C_1\).
Example
Compute the same determinant \(\det\begin{pmatrix} 1 & \textcolor{colordef}{4} & 7 \\ 2 & \textcolor{colordef}{5} & 8 \\ 3 & \textcolor{colordef}{6} & 9 \end{pmatrix}\) by expansion along the second column.
The second column entries \(\textcolor{colordef}{4, 5, 6}\) are pulled out, each multiplying its associated \(2 \times 2\) minor \(\textcolor{colorprop}{\det(\cdot)}\). The sign pattern for column \(2\) is \(-, +, -\): $$ \begin{aligned} \det\begin{pmatrix} 1 & \textcolor{colordef}{4} & 7 \\
2 & \textcolor{colordef}{5} & 8 \\
3 & \textcolor{colordef}{6} & 9 \end{pmatrix} &= -\textcolor{colordef}{4} \cdot \textcolor{colorprop}{\det\begin{pmatrix} 2 & 8 \\
3 & 9 \end{pmatrix}} + \textcolor{colordef}{5} \cdot \textcolor{colorprop}{\det\begin{pmatrix} 1 & 7 \\
3 & 9 \end{pmatrix}} - \textcolor{colordef}{6} \cdot \textcolor{colorprop}{\det\begin{pmatrix} 1 & 7 \\
2 & 8 \end{pmatrix}} && \text{(expansion along column \(2\))} \\
&= -\textcolor{colordef}{4} \cdot \textcolor{colorprop}{(18 - 24)} + \textcolor{colordef}{5} \cdot \textcolor{colorprop}{(9 - 21)} - \textcolor{colordef}{6} \cdot \textcolor{colorprop}{(8 - 14)} && \text{(\(2 \times 2\) Sarrus on each minor)} \\
&= -4 \cdot (-6) + 5 \cdot (-12) - 6 \cdot (-6) \\
&= 24 - 60 + 36 \\
&= 0. \end{aligned} $$ As expected, the value \(0\) is independent of the column chosen --- both developments yield the same answer because the matrix is the same. The choice of column is a matter of computational convenience.
Skills to practice
- Expanding along a row or column
V.4
Vandermonde determinant
A classical \(n \times n\) determinant whose value is a tidy product of pairwise differences ; it underlies Lagrange polynomial interpolation and appears in many synthesis exercises.
Proposition — Vandermonde determinant
For \(x_1, \dots, x_n \in \mathbb{K}\), $$ \textcolor{colorprop}{ V_n(x_1, \dots, x_n) = \det\begin{pmatrix} 1 & 1 & \cdots & 1 \\
x_1 & x_2 & \cdots & x_n \\
\vdots & & & \vdots \\
x_1^{n-1} & x_2^{n-1} & \cdots & x_n^{n-1} \end{pmatrix} = \prod_{1 \le i < j \le n} (x_j - x_i). } $$ In particular, \(V_n \ne 0\) if and only if \(x_1, \dots, x_n\) are pairwise distinct. - Base cases. \(V_1(x_1) = \det(1) = 1\) (empty product by convention). \(V_2(x_1, x_2) = \det\begin{pmatrix} 1 & 1 \\ x_1 & x_2 \end{pmatrix} = x_2 - x_1\). The induction below stops at \(V_2\).
- Repeated-root case. If two of the \(x_i\) are equal, then the Vandermonde matrix has two equal columns, so \(V_n = 0\). The right-hand product also has a factor \((x_j - x_i) = 0\) for that pair, so the identity reads \(0 = 0\).
- Distinct-root case. Assume \(x_1, \dots, x_n\) pairwise distinct. View \(V_{n+1}(x_1, \dots, x_n, X)\) as a polynomial in \(X\). By expansion along the last column, \(V_{n+1}\) has degree \(\le n\) in \(X\), with leading coefficient (coefficient of \(X^n\)) equal to \(V_n(x_1, \dots, x_n)\) (the cofactor in row \(n+1\), column \(n+1\) is \(V_n(x_1, \dots, x_n)\) with sign \((-1)^{(n+1)+(n+1)} = 1\)). For \(X \in \{x_1, \dots, x_n\}\) the matrix has two equal columns, hence \(V_{n+1} = 0\) --- so the polynomial \(V_{n+1}(\cdot, X)\) in \(X\) has \(n\) distinct known roots. By the factor theorem (a polynomial of degree \(\le n\) with \(n\) distinct roots is a scalar multiple of \(\prod(X - x_k)\), the scalar being the leading coefficient), $$ V_{n+1}(x_1, \dots, x_n, X) = V_n(x_1, \dots, x_n) \prod_{k=1}^{n} (X - x_k). $$ Iterate down to the \(V_2\) base case: \(V_2(x_1, x_2) = x_2 - x_1\), then \(V_3(x_1, x_2, x_3) = V_2(x_1, x_2)(x_3 - x_1)(x_3 - x_2) = (x_2 - x_1)(x_3 - x_1)(x_3 - x_2)\), and so on. The general expression is \(V_n = \prod_{1 \le i < j \le n} (x_j - x_i)\).
Method — Recognize a Vandermonde structure
Triggers: rows or columns of the form \((1, x_j, x_j^2, \dots, x_j^{n-1})\) ; a determinant of an interpolation matrix ; an exercise that asks « show that distinct \(x_1, \dots, x_n\) make the matrix invertible » in degree-\(\le n-1\) polynomial interpolation. The Vandermonde formula reduces the determinant to a product of pairwise differences --- and the invertibility criterion to « pairwise distinct ». Example
\(V_3(1, 2, 3) = (2 - 1)(3 - 1)(3 - 2) = 1 \cdot 2 \cdot 1 = 2\). Lagrange interpolation. If \(x_1, \dots, x_n\) are pairwise distinct, the Vandermonde determinant is nonzero, so the Vandermonde matrix is invertible ; for every prescribed values \(y_1, \dots, y_n \in \mathbb{K}\) there is then a unique polynomial \(P \in \mathbb{K}[X]\) of degree \(\le n - 1\) satisfying \(P(x_i) = y_i\) for \(i = 1, \dots, n\). This is the algebraic core of Lagrange interpolation, treated in the chapter Polynomials. Skills to practice
- Computing Vandermonde determinants and applications
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