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Determinant

Also Determinante · det

The determinant det(A) of a square matrix A is the number that gives the signed volume of the parallelepiped spanned by its columns, i.e. the factor by which A scales volumes; det(A) ≠ 0 iff A is invertible.

Know these firstLinear MapMatrix Rank

Formula#

Closed forms for small nn (MML (4.5)–(4.8)):

  • n=2n = 2: det⁡(a11a12a21a22)=a11a22−a12a21\det\begin{pmatrix} a_{11} & a_{12} \\ a_{21} & a_{22} \end{pmatrix} = a_{11}a_{22} - a_{12}a_{21}.
  • n=3n = 3 (Sarrus’ rule): det⁡(A)=a11a22a33+a21a32a13+a31a12a23−a31a22a13−a11a32a23−a21a12a33\det(A) = a_{11}a_{22}a_{33} + a_{21}a_{32}a_{13} + a_{31}a_{12}a_{23} - a_{31}a_{22}a_{13} - a_{11}a_{32}a_{23} - a_{21}a_{12}a_{33}.
  • Triangular TT (all entries below, or all above, the diagonal are 00): det⁡(T)=∏i=1nTii\det(T) = \prod_{i=1}^n T_{ii}.

Properties#

See Invertible Matrix > Properties for the other equivalent conditions. Geometrically, det⁡(A)=0\det(A) = 0 means the columns are linearly dependent, so they span a flat parallelepiped with volume 00.

Signed volume (MML Example 4.2): ∣det⁡(A)∣|\det(A)| is the volume of the parallelepiped spanned by the columns a1,…,ana_1, \dots, a_n of AA, and for n=2n = 2 the area of the parallelogram with vertices 0,a1,a2,a1+a20, a_1, a_2, a_1 + a_2. The sign gives the orientation of (a1,…,an)(a_1, \dots, a_n) relative to the standard basis (e1,…,en)(e_1, \dots, e_n). Swapping two columns flips it.

Rules for A,B∈Rn×nA, B \in \mathbb{R}^{n \times n} (MML p. 103):

  • det⁡(AB)=det⁡(A)det⁡(B)\det(AB) = \det(A)\det(B).
  • det⁡(A⊤)=det⁡(A)\det(A^\top) = \det(A). So every statement about columns holds for rows as well.
  • det⁡(A−1)=1det⁡(A)\det(A^{-1}) = \dfrac{1}{\det(A)} if AA is invertible (from det⁡(A)det⁡(A−1)=det⁡(I)=1\det(A)\det(A^{-1}) = \det(I) = 1).
  • Similar matrices B=S−1ASB = S^{-1}AS have the same determinant. So det⁡\det is a property of the linear map, independent of the basis.
  • Adding a multiple of one row/column to another doesn’t change det⁡(A)\det(A).
  • Multiplying one row/column by λ∈R\lambda \in \mathbb{R} multiplies det⁡(A)\det(A) by λ\lambda. In particular det⁡(λA)=λndet⁡(A)\det(\lambda A) = \lambda^n \det(A).
  • Swapping two rows/columns changes the sign of det⁡(A)\det(A).
  • det⁡(A)=∏i=1nλi\det(A) = \prod_{i=1}^n \lambda_i, the product of the eigenvalues counted with algebraic multiplicity (MML Thm. 4.16).

Computing it: the last three row rules say how each step of Gaussian elimination changes det⁡\det. Reduce AA to triangular form, multiply the diagonal entries, and correct the sign once per row swap. This costs O(n3)O(n^3) operations.

Characteristic polynomial: pA(λ)=det⁡(A−λI)p_A(\lambda) = \det(A - \lambda I) is a polynomial in λ\lambda whose roots are the eigenvalues of AA (see Eigenvalue > Characteristic polynomial).

Example#

Intuition#

la determinant area
A matrix AA sends e1e_1 to its first column a1a_1 and e2e_2 to its second column a2a_2, so the unit square becomes the parallelogram spanned by a1,a2a_1, a_2. Its area ∣det⁡(A)∣|\det(A)| is the factor by which AA scales every area: the image of the grid consists of equal parallelograms, each ∣det⁡(A)∣|\det(A)| times as large as a grid cell, and any shape can be filled with ever finer grid cells. So the disk of area 0.790.79 becomes an ellipse of area 2.5⋅0.79=1.962.5 \cdot 0.79 = 1.96. So det⁡\det is not a property of the square but of the map.

la determinant sign
The sign tells whether AA keeps or flips orientation. Turning from e1e_1 to e2e_2 goes counterclockwise. If the turn from a1a_1 to a2a_2 does too, det⁡(A)>0\det(A) > 0 and the letter F stays readable. If the turn is clockwise (here: the columns swapped), det⁡(A)<0\det(A) < 0 and the F comes out mirrored. In between lies det⁡(A)=0\det(A) = 0: the columns are parallel (a2=12a1a_2 = \frac12 a_1), the whole plane is squashed onto a line, and that information can’t be undone. This is why det⁡(A)=0\det(A) = 0 means not invertible.

la determinant product
Applying BB scales areas by det⁡(B)\det(B), and applying AA afterwards scales the result by det⁡(A)\det(A). Together the area changes by det⁡(A)det⁡(B)\det(A)\det(B), which gives det⁡(AB)=det⁡(A)det⁡(B)\det(AB) = \det(A)\det(B) without any computation. The same picture gives det⁡(A−1)=1/det⁡(A)\det(A^{-1}) = 1/\det(A), because A−1A^{-1} must undo the scaling.

Sources#

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