Determinants and volume

A linear map distorts space: it might expand, contract, or collapse it. The determinant is the number that measures that distortion. It is a single scalar that encodes whether a matrix is invertible and by how much it scales volumes.

Definition via volume

For a square matrix $A \in \mathbb{R}^{n \times n}$, the determinant $\det A$ (also written $|A|$) is the signed volume of the parallelepiped spanned by the column vectors of $A$, relative to the unit cube's volume of $1$[1].

Two properties characterize the determinant completely (up to normalization):

1. Multilinearity in each column. Scaling one column by $c$ scales the determinant by $c$. Splitting a column into a sum of two columns splits the determinant into a sum of two determinants.
2. Alternating. Swapping any two columns changes the sign of the determinant.

The normalization $\det I = 1$ (the identity does not distort volume) pins it down. From these three requirements, one can derive the cofactor-expansion formula and every other property.

Key properties

Product rule:[2]

Derivation. Both sides depend multilinearly on the columns of $A$ and $B$. The cleanest proof uses the fact that any matrix is a product of elementary matrices (row operations), whose determinants multiply; here we state it as the fundamental multiplicativity law.

Transpose:

So everything that holds for columns holds for rows: swapping rows changes sign, scaling a row scales the determinant, and a zero row gives $\det = 0$.

Triangular matrices: $\det A = a_{11} a_{22} \cdots a_{nn}$ (the product of diagonal entries). Follows from multilinearity and alternating: off-diagonal entries in a triangular matrix contribute zero.

Block diagonal: $\det\begin{pmatrix} A & 0 \\ 0 & B \end{pmatrix} = \det(A)\det(B)$.

Determinant = 0 means singular

$\det A = 0$ if and only if the columns of $A$ are linearly dependent, i.e. the map $x \mapsto Ax$ collapses $\mathbb{R}^n$ into a lower-dimensional subspace. The collapsed image has zero $n$-dimensional volume, so the determinant (a volume scale factor) is zero. A matrix with nonzero determinant is invertible (or nonsingular), and

The sign and magnitude read off directly from what the map does to area in $2\times 2$:

• $\begin{pmatrix} 2 & 0 \\ 0 & 2\end{pmatrix}$ doubles each axis, so it scales area by $\det = 4$: the unit square maps to a $2\times 2$ square.
• A shear $\begin{pmatrix} 1 & 3 \\ 0 & 1\end{pmatrix}$ slants the square into a parallelogram of the same base and height, so $\det = 1$: area is untouched even though the shape is not.
• $\begin{pmatrix} 0.5 & 0 \\ 0 & 1\end{pmatrix}$ squeezes one axis to half, so $\det = 0.5$: area halved.
• A reflection $\begin{pmatrix} 1 & 0 \\ 0 & -1\end{pmatrix}$ preserves area but flips orientation, so $\det = -1$: the negative sign is the orientation reversal, not a smaller area. A rotation, which neither stretches nor flips, has $\det = 1$ at every angle.
• The singular $\begin{pmatrix} 1 & 2 \\ 2 & 4\end{pmatrix}$ sends both columns onto the same line, collapsing the square to a segment of zero area: $\det = 0$.

Computing the determinant

For $2 \times 2$:

For $n \times n$, the cofactor expansion along column $j$ gives a recursion:

where $M_{ij}$ is the determinant of the $(n-1)\times(n-1)$ matrix obtained by deleting row $i$ and column $j$. This is $O(n!)$ naively, but the LU factorization computes the determinant in $O(n^3)$ as the product of the LU diagonal entries (with sign from the row-swap parity).

Check it on arrays

import numpy as np

A = np.array([[2.0, 1.0, 0.0],
              [1.0, 3.0, 1.0],
              [0.0, 1.0, 2.0]])

d = np.linalg.det(A)
print(f"det(A) = {d:.6f}")   # should be 2*5-1*2 = 8 (cofactor along row 1)

# Product rule: det(AB) = det(A)*det(B)
B = np.array([[1.0, 2.0, 0.0],
              [0.0, 1.0, 3.0],
              [2.0, 0.0, 1.0]])
print(f"det(A)*det(B) = {d * np.linalg.det(B):.6f}")
print(f"det(A@B)      = {np.linalg.det(A @ B):.6f}")   # same

# Singular matrix: det = 0
C = np.array([[1.0, 2.0, 3.0],
              [2.0, 4.0, 6.0],
              [1.0, 0.0, 1.0]])
print(f"det(C) = {np.linalg.det(C):.2e}")   # ~0 (row 2 = 2*row 1)

# Triangular: det = product of diagonal
T = np.array([[3.0, 1.0, 5.0],
              [0.0, 2.0, 7.0],
              [0.0, 0.0, 4.0]])
print(f"det(T) = {np.linalg.det(T):.4f}")  # 3*2*4 = 24
print(f"diagonal product = {3*2*4}")

The product rule holds to floating-point precision; the singular matrix gives a determinant near zero; the triangular matrix matches its diagonal product.

Exercises

Exercise 1. Compute by hand the determinant of $A = \begin{pmatrix} 2 & 1 & 0 \\ 1 & 3 & 1 \\ 0 & 1 & 2 \end{pmatrix}$ using cofactor expansion along the first row.

Exercise 2. Using only multilinearity, the alternating property, and $\det I = 1$, prove that a matrix with two equal columns has determinant $0$.

Exercise 3 (code). A determinant is multiplied by $c^n$ when the whole $n\times n$ matrix is scaled by $c$ (each of the $n$ columns contributes one factor of $c$ by multilinearity). Verify this for a random $4\times 4$ matrix and $c = 3$, and confirm the orientation-reversal fact that swapping two columns flips the sign of the determinant.

import numpy as np

rng = np.random.default_rng(0)
A = rng.normal(size=(4, 4))
n, c = 4, 3.0

# Scaling the whole matrix by c multiplies det by c**n.
assert np.isclose(np.linalg.det(c * A), c**n * np.linalg.det(A))

# Swapping two columns flips the sign of the determinant.
A_swapped = A[:, [1, 0, 2, 3]]
assert np.isclose(np.linalg.det(A_swapped), -np.linalg.det(A))

print("c**n factor:", c**n)            # 81
print("det(A):", round(np.linalg.det(A), 6))

With the determinant linking to singularity, the next lesson asks: which vectors does $A$ only scale, without rotating?