8 Determinants

A linear map T : \mathbb{R}^n \to \mathbb{R}^n transforms geometric objects — lines become lines, planes become planes, parallelograms become parallelograms. Distances and angles generally change, but certain properties persist. Among these is a remarkable invariant: the factor by which T scales volumes.

This chapter develops the determinant, a function \det : M_{n}(\mathbb{F}) \to \mathbb{F} that encodes this volume-scaling behavior. We characterize it axiomatically, derive computational methods, establish its fundamental properties, and explore its geometric content.

Remark. The determinant is an instance of what is called an alternating multilinear n-form: a map \det: \mathcal{V}^n \to \mathbb{F} on an n-dimensional vector space \mathcal{V} that is linear in each slot, vanishes when two arguments agree, and changes sign when two arguments are swapped. More generally, alternating k-forms are k-linear alternating maps and play a central role in exterior algebra and differential geometry; viewing the determinant as the canonical n-form on \mathcal{V} explains its connection to oriented volume and is useful background for readers advancing to higher mathematics.

8.1 Motivation

Consider a linear map T : \mathbb{R}^2 \to \mathbb{R}^2 represented by matrix A \in M_{2}(\mathbb{R}). Apply T to the unit square [0,1]^2. The image is a parallelogram with vertices at 0, a_1, a_2, and a_1 + a_2, where a_1, a_2 are the columns of A.

The area of this parallelogram depends only on A — it measures how much T stretches or compresses area. If T doubles area, every region’s area is doubled under T. If T collapses the plane onto a line (\ker(T) \neq \{0\}), all areas become zero.

More generally, consider the unit ball B^n = \{x \in \mathbb{R}^n : \|x\| \le 1\}. A linear map T : \mathbb{R}^n \to \mathbb{R}^n transforms B^n into an ellipsoid. The n-dimensional volume of this ellipsoid, relative to the volume of B^n, is the volume scaling factor of T.

This scaling factor is intrinsic to T — it does not depend on which region we measure. If T scales all volumes by factor k, then for any measurable set E \subset \mathbb{R}^n, the image T(E) has volume |k| \cdot \text{vol}(E). The absolute value accounts for orientation: if k < 0, the map reverses orientation.

Invariance under volume-preserving transformations. If S : \mathbb{R}^n \to \mathbb{R}^n is an isometry, then S preserves volumes. Rotations and reflections are isometries. If T scales volume by factor k, then S \circ T \circ S^{-1} also scales volume by factor k — conjugation by isometries does not change the scaling factor. This suggests the volume scaling factor is intrinsic to T, not to any particular matrix representation: two matrices representing the same linear map in different orthonormal bases have the same determinant.

Why volume? Among all geometric quantities — length, angle, curvature — only volume has the multiplicativity property: if S scales volume by \alpha and T scales volume by \beta, then S \circ T scales volume by \alpha\beta. This multiplicativity, which we prove in Theorem 8.6, makes the determinant a powerful algebraic tool.

Projection and dimension collapse. If T is not injective, then \ker(T) \neq \{0\}. The image \operatorname{im}(T) is a proper subspace of \mathbb{R}^n with \dim \operatorname{im}(T) < n by Theorem 4.5. The ball B^n is flattened into an ellipsoid lying in a k-dimensional subspace where k = \operatorname{rank}(T) < n. An n-dimensional volume cannot be accommodated in a proper subspace — the n-dimensional volume of T(B^n) is zero. This geometric fact corresponds to the algebraic fact that \det(T) = 0 when T is not invertible.

Observation. The volume-scaling factor is multiplicative, suggesting \det(AB) = \det(A)\det(B).

Observation. If the columns of A are linearly dependent, the image of B^n lies in a lower-dimensional subspace and has zero n-dimensional volume. This suggests \det(A) = 0 \iff A is not invertible.

Observation. The scaling factor should depend continuously on the entries of A: small perturbations produce small changes in volume.

We seek a function with these properties. The challenge is to define it precisely and prove it exists uniquely.

8.2 Signed Volume and Orientation

In \mathbb{R}^n, the unit cube [0,1]^n has volume 1. A linear map T : \mathbb{R}^n \to \mathbb{R}^n sends this cube to a parallelepiped — the set \{ t_1 a_1 + \cdots + t_n a_n : 0 \le t_i \le 1 \} where a_1, \ldots, a_n are the columns of A = [T]_{\mathcal{E}} in the standard basis. If \{a_1, \ldots, a_n\} are linearly dependent, the parallelepiped is degenerate and its n-dimensional volume is zero.

We assign signed volume: a sign depending on the orientation of the parallelepiped. The standard basis \{e_1, \ldots, e_n\} defines the positive orientation. If T preserves orientation, \det(T) > 0; if T reverses orientation, \det(T) < 0; if T collapses dimension, \det(T) = 0.

Orientation in \mathbb{R}^n. Two ordered bases of \mathbb{R}^n have the same orientation if one can be continuously deformed into the other through invertible linear maps. Equivalently, two bases have the same orientation if the change-of-basis matrix between them has positive determinant.

In \mathbb{R}^2, the standard basis \{e_1, e_2\} defines counterclockwise orientation; swapping to \{e_2, e_1\} reverses it to clockwise. In \mathbb{R}^3, \{e_1, e_2, e_3\} defines right-handed orientation.

Examples in \mathbb{R}^2.

Identity: I = \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix}. The unit square is unchanged, so \det(I) = 1.
Scaling: A = \begin{pmatrix} k & 0 \\ 0 & k \end{pmatrix} scales both directions by k, multiplying area by k^2. Thus \det(A) = k^2.
Shear: A = \begin{pmatrix} 1 & 1 \\ 0 & 1 \end{pmatrix} sends the unit square to a parallelogram with base 1 and height 1, area 1. Thus \det(A) = 1.
Reflection: A = \begin{pmatrix} -1 & 0 \\ 0 & 1 \end{pmatrix} reflects across the y-axis, reversing orientation. Area is unchanged, so \det(A) = -1.
Rotation: A = \begin{pmatrix} \cos\theta & -\sin\theta \\ \sin\theta & \cos\theta \end{pmatrix} rotates by \theta, preserving area and orientation. Indeed \det(A) = \cos^2\theta + \sin^2\theta = 1.
Projection: A = \begin{pmatrix} 1 & 0 \\ 0 & 0 \end{pmatrix} collapses the plane onto the x-axis, giving zero 2-dimensional area. Thus \det(A) = 0.

These examples suggest that \det encodes both magnitude (how much volume changes) and sign (whether orientation is preserved).

Invariant subspaces. If T leaves a subspace W invariant, we can choose a basis adapted to this decomposition, yielding a block-triangular matrix. The determinant of such a matrix factors into determinants of the diagonal blocks, reflecting the independent scalings on each piece. We return to this in Theorem 8.9 and more fully in Chapter 8.

8.3 Axiomatic Characterization

We define the determinant by specifying the properties it must satisfy, then prove such a function exists and is unique.

Definition 8.1 (Determinant (axiomatic)) A function \det : M_{n}(\mathbb{F}) \to \mathbb{F} is a determinant function if it satisfies:

(Multilinearity in columns) For each j = 1, \ldots, n, the function is linear in the j-th column when the other columns are held fixed: \det(a_1, \ldots, ca_j + da_j', \ldots, a_n) = c\det(a_1, \ldots, a_j, \ldots, a_n) + d\det(a_1, \ldots, a_j', \ldots, a_n).
(Alternating) Swapping any two columns negates the determinant: \det(\ldots, a_i, \ldots, a_j, \ldots) = -\det(\ldots, a_j, \ldots, a_i, \ldots).
(Normalization) \det(I_n) = 1.

We write the matrix as a tuple of its columns: A = (a_1, \ldots, a_n) with a_j \in \mathbb{F}^n. The multilinearity axiom reflects that volume scales linearly with each edge of the parallelepiped. The alternating property encodes orientation. Normalization fixes the scale: the identity preserves volume exactly.

Theorem 8.1 If \det satisfies axioms (1)–(3), then:

If two columns of A are equal, then \det(A) = 0
If the columns of A are linearly dependent, then \det(A) = 0
Adding a multiple of one column to another does not change \det(A)
Scaling a column by c multiplies \det(A) by c

Proof.

If a_i = a_j for i \neq j, swapping columns i and j leaves A unchanged but the alternating property gives \det(A) = -\det(A), so 2\det(A) = 0. Over \mathbb{R} or \mathbb{C} (characteristic \neq 2), this forces \det(A) = 0.
Suppose a_j = \sum_{i \neq j} c_i a_i. By multilinearity in column j, \det(A) = \sum_{i \neq j} c_i \det(a_1, \ldots, \underbrace{a_i}_{\text{col }j}, \ldots, a_n). Each term has two identical columns (the i-th and the j-th both equal a_i), so by (a) each vanishes.
By multilinearity, \det(\ldots, a_i + ca_j, \ldots, a_j, \ldots) = \det(\ldots, a_i, \ldots, a_j, \ldots) + c\det(\ldots, a_j, \ldots, a_j, \ldots). The second term vanishes by (a).
Immediate from multilinearity. \square

The determinant vanishes precisely when the columns are linearly dependent — consistent with the geometric interpretation that degenerate parallelepipeds have zero n-dimensional volume.

Remark. Part (c) is the column version of the row replacement operation from Gaussian elimination. Once we establish \det(A^T) = \det(A) (see Theorem 8.7), it follows immediately that row replacement operations likewise leave the determinant unchanged.

8.4 Permutations and the Leibniz Formula

To derive an explicit formula, we expand the determinant using multilinearity and the alternating property.

Definition 8.2 (Permutation) A permutation of \{1, \ldots, n\} is a bijection \sigma : \{1, \ldots, n\} \to \{1, \ldots, n\}. The set of all permutations forms the symmetric group S_n under composition, where (\sigma \circ \tau)(i) = \sigma(\tau(i)).

A transposition is a permutation swapping exactly two elements. Every permutation decomposes into a product of transpositions.

Definition 8.3 (Sign of a permutation) The sign of \sigma \in S_n, denoted \operatorname{sgn}(\sigma), is +1 if \sigma can be expressed as a product of an even number of transpositions and -1 if odd. This is well-defined: every decomposition of \sigma into transpositions has the same parity.

The sign is multiplicative: \operatorname{sgn}(\sigma \circ \tau) = \operatorname{sgn}(\sigma)\operatorname{sgn}(\tau). In particular, \operatorname{sgn}(\sigma^{-1}) = \operatorname{sgn}(\sigma), since 1 = \operatorname{sgn}(\text{id}) = \operatorname{sgn}(\sigma \circ \sigma^{-1}) = \operatorname{sgn}(\sigma)\operatorname{sgn}(\sigma^{-1}). Every transposition has sign -1.

Examples. The identity has sign +1. In S_3, the 3-cycle (1\,2\,3) sending 1 \to 2 \to 3 \to 1 has exactly two inversions — the pairs (1,3) and (2,3) where a larger index precedes a smaller value — so it is even with \operatorname{sgn}((1\,2\,3)) = +1. The transposition (1\,2) fixing 3 has one inversion, so \operatorname{sgn}((1\,2)) = -1.

Now we derive the formula. Write each column in the standard basis: a_j = \sum_{i=1}^n a_{ij} e_i. Applying multilinearity n times, \det(A) = \sum_{i_1, \ldots, i_n = 1}^n a_{i_1 1} \cdots a_{i_n n} \,\det(e_{i_1}, \ldots, e_{i_n}). If any two indices i_j coincide, the determinant of the resulting tuple of standard basis vectors vanishes by Theorem 8.1(a). Only terms where (i_1, \ldots, i_n) is a permutation contribute. Writing i_j = \sigma(j) for \sigma \in S_n: \det(A) = \sum_{\sigma \in S_n} a_{\sigma(1),1} \cdots a_{\sigma(n),n} \,\det(e_{\sigma(1)}, \ldots, e_{\sigma(n)}).

Any permuted tuple (e_{\sigma(1)}, \ldots, e_{\sigma(n)}) can be rearranged to (e_1, \ldots, e_n) by a sequence of transpositions; the alternating property contributes a factor of \operatorname{sgn}(\sigma) for each swap, giving \det(e_{\sigma(1)}, \ldots, e_{\sigma(n)}) = \operatorname{sgn}(\sigma)\det(I_n) = \operatorname{sgn}(\sigma).

This yields \det(A) = \sum_{\sigma \in S_n} \operatorname{sgn}(\sigma)\, a_{\sigma(1),1} \cdots a_{\sigma(n),n}.

This is the natural form produced by column expansion. The more standard convention indexes by row: reindex via \tau = \sigma^{-1}, noting a_{\sigma(j),j} = a_{\tau^{-1}(j),j}. Setting i = \sigma(j), equivalently j = \tau(i), the product \prod_j a_{\sigma(j),j} becomes \prod_i a_{i,\tau(i)}. Since \operatorname{sgn}(\tau) = \operatorname{sgn}(\sigma^{-1}) = \operatorname{sgn}(\sigma) and \sigma \mapsto \tau = \sigma^{-1} is a bijection on S_n, we arrive at:

Theorem 8.2 (Leibniz formula) \det(A) = \sum_{\sigma \in S_n} \operatorname{sgn}(\sigma)\, a_{1\sigma(1)}\, a_{2\sigma(2)} \cdots a_{n\sigma(n)}.

Each term selects one entry from each row and one from each column: a_{i,\sigma(i)} lies in row i, and the bijectivity of \sigma ensures distinct columns. The sum has n! terms.

Since this formula was forced uniquely by axioms (1)–(3), and since one can verify directly that this sum satisfies all three axioms, the Leibniz formula simultaneously establishes existence and uniqueness of the determinant.

Example: n = 2. Two permutations: \text{id} with sign +1 and (1\,2) with sign -1. Thus \det\begin{pmatrix} a & b \\ c & d \end{pmatrix} = ad - bc.

Example: n = 3. Six permutations in S_3, three even and three odd: \det\begin{pmatrix} a & b & c \\ d & e & f \\ g & h & i \end{pmatrix} = aei + bfg + cdh - ceg - afh - bdi.

8.5 Computational Methods

The Leibniz formula is impractical for large n — computing n! terms is infeasible. We develop efficient methods based on row operations.

8.5.1 Row Operations and Determinants

Theorem 8.3 Let A \in M_{n}(\mathbb{F}).

If B is obtained from A by swapping two rows, then \det(B) = -\det(A)
If B is obtained from A by multiplying a row by scalar c, then \det(B) = c\det(A)
If B is obtained from A by adding a multiple of one row to another, then \det(B) = \det(A)

Proof. We work directly from the Leibniz formula.

(1) Swapping rows i and j produces B with b_{i\ell} = a_{j\ell}, b_{j\ell} = a_{i\ell}, and b_{k\ell} = a_{k\ell} for k \notin \{i,j\}. The term for permutation \sigma in \det(B) is b_{1\sigma(1)} \cdots b_{i\sigma(i)} \cdots b_{j\sigma(j)} \cdots b_{n\sigma(n)} = a_{1\sigma(1)} \cdots a_{j\sigma(i)} \cdots a_{i\sigma(j)} \cdots a_{n\sigma(n)}, which equals a_{1\tau(1)} \cdots a_{n\tau(n)} for the permutation \tau = \sigma \circ (i\,j) — that is, \sigma with its values at positions i and j swapped: \tau(i) = \sigma(j), \tau(j) = \sigma(i), \tau(k) = \sigma(k) otherwise. Since \operatorname{sgn}(\tau) = \operatorname{sgn}(\sigma)\operatorname{sgn}(i\,j) = -\operatorname{sgn}(\sigma), and \sigma \mapsto \sigma \circ (i\,j) is a bijection on S_n, summing gives \det(B) = -\det(A).

(2) Multiplying row i by c gives b_{i\ell} = ca_{i\ell} and b_{k\ell} = a_{k\ell} for k \neq i. Each term in the Leibniz formula for \det(B) contains exactly one factor from row i, namely b_{i\sigma(i)} = ca_{i\sigma(i)}. Thus every term is scaled by c, giving \det(B) = c\det(A).

(3) Adding c times row j to row i (where i \neq j) gives b_{i\ell} = a_{i\ell} + ca_{j\ell} and b_{k\ell} = a_{k\ell} for k \neq i. In the Leibniz formula, the factor from row i in each term splits as b_{i\sigma(i)} = a_{i\sigma(i)} + ca_{j\sigma(i)}, so \det(B) = \det(A) + c\det(A'), where A' is the matrix equal to A except with row i replaced by row j. Now A' has rows i and j both equal to row j of A. Swapping those two equal rows leaves A' unchanged, but by part (1) it negates the determinant. Thus \det(A') = -\det(A'), giving \det(A') = 0, and \det(B) = \det(A). \square

8.5.2 Determinants of Triangular Matrices

Theorem 8.4 If A is upper or lower triangular, then \det(A) = a_{11} a_{22} \cdots a_{nn}.

Proof. For upper triangular A, we have a_{ij} = 0 whenever i > j. In the Leibniz formula, the term for permutation \sigma vanishes unless a_{i,\sigma(i)} \neq 0 for all i, which requires \sigma(i) \ge i for all i. But a bijection \sigma : \{1,\ldots,n\} \to \{1,\ldots,n\} satisfying \sigma(i) \ge i for all i must be the identity: since \sigma is a bijection of \{1,\ldots,n\}, we have \sum_{i=1}^n \sigma(i) = \sum_{i=1}^n i. Combined with \sigma(i) \ge i for each i, this forces \sigma(i) = i for all i. Thus only the identity term survives, giving \det(A) = a_{11} a_{22} \cdots a_{nn}. The lower triangular case is symmetric: a_{ij} = 0 for i < j requires \sigma(i) \le i for all i, and \sum_i \sigma(i) = \sum_i i with \sigma(i) \le i again forces \sigma = \text{id}. \square

8.5.3 Gaussian Elimination Algorithm

To compute \det(A), reduce A to row echelon form U by row operations, tracking their effect:

each row swap multiplies the running determinant by -1;
scaling a row by c multiplies by c;
row replacement does not change the determinant.

Since U is upper triangular, \det(U) = u_{11} u_{22} \cdots u_{nn}. Adjusting for accumulated signs and scalings gives \det(A).

Corollary. A is invertible if and only if \det(A) \neq 0. By Theorem 7.11, A is invertible iff it has n pivots, iff all diagonal entries of U are nonzero, iff \det(U) \neq 0, iff \det(A) \neq 0.

8.6 Cofactor Expansion

An alternative method expands the determinant recursively.

Definition 8.4 (Minor and cofactor) The (i,j)-minor M_{ij} of A \in M_n(\mathbb{F}) is the determinant of the (n-1) \times (n-1) matrix obtained by deleting row i and column j. The (i,j)-cofactor is C_{ij} = (-1)^{i+j} M_{ij}.

The sign (-1)^{i+j} accounts for the sign changes incurred by moving the (i,j) entry to the top-left position via row and column swaps.

Theorem 8.5 (Cofactor expansion) For any j \in \{1, \ldots, n\} (expansion along column j): \det(A) = \sum_{i=1}^{n} a_{ij} C_{ij}.

For any i \in \{1, \ldots, n\} (expansion along row i): \det(A) = \sum_{j=1}^{n} a_{ij} C_{ij}.

Proof. We first prove expansion along column 1; expansion along other columns then follows by analogous row permutations.

Writing a_1 = \sum_{i=1}^n a_{i1} e_i and applying multilinearity in the first column: \det(A) = \sum_{i=1}^{n} a_{i1} \det(e_i, a_2, \ldots, a_n).

For i = 1: \det(e_1, a_2, \ldots, a_n) has a 1 in position (1,1) and zeros below in the first column. By multilinearity expanding the remaining columns over rows 2,\ldots,n, this equals \det(\tilde{A}_{11}) = M_{11}. Since (-1)^{1+1} = 1, this term is a_{11}C_{11}.

For i > 1: move e_i from row i to row 1 by performing i-1 consecutive upward row swaps (swapping row i with row i-1, then i-1 with i-2, and so on). Each swap negates the determinant, contributing a factor of (-1)^{i-1}. After these swaps, the matrix has a 1 in position (1,1) and zeros elsewhere in column 1, with rows 1, 2, \ldots, i-1, i+1, \ldots, n of A occupying rows 2, 3, \ldots, n. The determinant of this matrix is M_{i1} (the minor deleting row i and column 1 of A). Therefore \det(e_i, a_2, \ldots, a_n) = (-1)^{i-1} M_{i1} = (-1)^{i+1} M_{i1} = C_{i1}.

Thus: \det(A) = \sum_{i=1}^{n} a_{i1} C_{i1}.

Row expansion follows from column expansion via the transpose: since \det(A^T) = \det(A) (proved in Theorem 8.7), and the column i of A^T is row i of A, applying column expansion to A^T along its i-th column gives \det(A) = \det(A^T) = \sum_{j=1}^{n} (A^T)_{ji}\, C^{A^T}_{ji} = \sum_{j=1}^{n} a_{ij}\, C_{ij}. \quad \square

Cofactor expansion trades one n \times n determinant for n determinants of size (n-1)\times(n-1). For small n this is practical; Gaussian elimination is far more efficient for large n, running in O(n^3) rather than the O(n!) of naive cofactor recursion.

8.7 Fundamental Properties

8.7.1 Multiplicativity

Theorem 8.6 For A, B \in M_{n}(\mathbb{F}), \det(AB) = \det(A)\det(B).

Proof. We handle the two cases separately.

If \det(A) = 0: Then A is not invertible, so \operatorname{rank}(A) < n. Since \operatorname{rank}(AB) \le \operatorname{rank}(A) < n, the product AB is also not invertible, giving \det(AB) = 0 = \det(A)\det(B).

If \det(A) \neq 0: Define f : M_n(\mathbb{F}) \to \mathbb{F} by f(B) = \det(AB)/\det(A). We verify f satisfies the three determinant axioms.

Multilinearity: Since AB = (Ab_1, \ldots, Ab_n) and b_j \mapsto Ab_j is linear, multilinearity of \det in columns gives multilinearity of f in the columns of B.

Alternating: Swapping columns j and k of B swaps the corresponding columns of AB, negating \det(AB), hence negating f(B).

Normalization: f(I_n) = \det(A \cdot I_n)/\det(A) = 1.

By uniqueness of the determinant (Theorem 8.2), f = \det. Thus \det(AB)/\det(A) = \det(B), giving \det(AB) = \det(A)\det(B). \square

This theorem formalizes the geometric observation that volume scaling factors compose multiplicatively.

Corollary. If A is invertible, \det(A^{-1}) = 1/\det(A), since \det(A)\det(A^{-1}) = \det(I_n) = 1.

8.7.2 Transpose

Theorem 8.7 For any A \in M_{n}(\mathbb{F}), \det(A^T) = \det(A).

Proof. The (i,j) entry of A^T is a_{ji}, so \det(A^T) = \sum_{\sigma \in S_n} \operatorname{sgn}(\sigma)\, a_{\sigma(1),1}\, a_{\sigma(2),2} \cdots a_{\sigma(n),n}. Reindex by \tau = \sigma^{-1}. Setting i = \sigma(j), equivalently j = \tau(i), the product \prod_j a_{\sigma(j),j} becomes \prod_i a_{i,\tau(i)}. Since \operatorname{sgn}(\tau) = \operatorname{sgn}(\sigma) and \sigma \mapsto \tau = \sigma^{-1} is a bijection on S_n, \det(A^T) = \sum_{\tau \in S_n} \operatorname{sgn}(\tau)\, a_{1\tau(1)} \cdots a_{n\tau(n)} = \det(A). \quad \square

This theorem justifies treating rows and columns symmetrically: every property stated for columns holds equally for rows. In particular, the determinant is multilinear and alternating in rows, which underpins Theorem 8.3.

8.8 Geometric Applications

8.8.1 Volume of Parallelepipeds

Theorem 8.8 Let v_1, \ldots, v_n \in \mathbb{R}^n and A = (v_1 \;\cdots\; v_n). The n-dimensional volume of the parallelepiped \mathcal{P} = \{ \sum_i t_i v_i : 0 \le t_i \le 1 \} is |\det(A)|.

The absolute value accounts for orientation — volume is nonnegative, but the determinant carries a sign.

Example. In \mathbb{R}^2, vectors v_1 = (a,c)^T and v_2 = (b,d)^T span a parallelogram with area |ad - bc| = |\det(A)| for A = \begin{pmatrix} a & b \\ c & d \end{pmatrix}.

8.8.2 Change of Variables in Integration

For a differentiable map T : \mathbb{R}^n \to \mathbb{R}^n with Jacobian J_T(x), \int_{T(E)} f(y)\,dy = \int_E f(T(x))\,|\det(J_T(x))|\,dx. For a linear map T(x) = Ax, the Jacobian is the constant matrix A, and this becomes \int_{T(E)} f\,dy = |\det(A)|\int_E f(Ax)\,dx. The factor |\det(A)| is precisely the volume scaling factor that motivated the entire development.

8.9 Block Matrices and Determinants

Theorem 8.9 If A = \begin{pmatrix} B & C \\ 0 & D \end{pmatrix} with B a k \times k block and D an (n-k) \times (n-k) block, then \det(A) = \det(B)\det(D).

Proof. In the Leibniz formula, the term for \sigma vanishes unless a_{i,\sigma(i)} \neq 0 for all i. The lower-left block of A is zero: a_{ij} = 0 whenever i > k and j \le k. For a nonzero term we therefore need \sigma(i) > k for every i > k. Since \sigma is a bijection of \{1, \ldots, n\} and the n-k indices \{k+1, \ldots, n\} must map into \{k+1, \ldots, n\} (a set of the same size), \sigma restricts to a bijection \sigma_B on \{1, \ldots, k\} and a bijection \sigma_D on \{k+1, \ldots, n\}. The product factorizes: \prod_{i=1}^n a_{i,\sigma(i)} = \prod_{i=1}^k b_{i,\sigma_B(i)} \cdot \prod_{i=k+1}^n d_{i-k,\,\sigma_D(i)-k}, and \operatorname{sgn}(\sigma) = \operatorname{sgn}(\sigma_B)\operatorname{sgn}(\sigma_D). Summing over all valid \sigma factorizes the sum as \det(B)\det(D). \square

8.10 Closing Remarks

The determinant packages several fundamental properties of a linear map into a single scalar: it detects invertibility, measures volume scaling, encodes orientation, and satisfies multiplicativity. Its axiomatic characterization — multilinearity, alternating, normalization — uniquely determines an explicit formula (Leibniz), from which computational methods follow.

Crucially, the determinant is well-defined on linear operators rather than merely on matrices: if B = P^{-1}AP, then \det(B) = \det(P^{-1})\det(A)\det(P) = \det(A), since \det(P^{-1})\det(P) = \det(I_n) = 1. This means \det(T) is an intrinsic property of the operator T, independent of the choice of basis used to represent it. In the next chapter, we develop eigenvalue theory, where the determinant reappears as the product of eigenvalues and as the constant term of the characteristic polynomial.