Linear Algebra/Determinant

The determinant is a function which associates to a square matrix an element of the field on which it is defined (commonly the real or complex numbers). The determinant is required to hold these properties:

• It is linear on the rows of the matrix.

$det\left[\begin{array}{ccc}\ddots & ⋮& \dots \\ \lambda a_1+\mu b_1& \cdots & \lambda a_n+\mu b_n\\ \cdots & ⋮& \ddots \end{array}\right]=\lambda det\left[\begin{array}{ccc}\ddots & ⋮& \cdots \\ a_1& \cdots & a_n\\ \cdots & ⋮& \ddots \end{array}\right]+\mu det\left[\begin{array}{ccc}\ddots & ⋮& \cdots \\ b_1& \cdots & b_n\\ \cdots & ⋮& \ddots \end{array}\right]$

• If the matrix has two equal rows its determinant is zero.
• The determinant of the identity matrix is 1.

It is possible to prove that $detA=det{A}^T$, making the definition of the determinant on the rows equal to the one on the columns.

Properties

• The determinant is zero if and only if the rows are linearly dependent.
• Changing two rows changes the sign of the determinant:

$det\left[\begin{array}{c}\cdots \\ \text{row A}\\ \cdots \\ \text{row B}\\ \cdots \end{array}\right]=-det\left[\begin{array}{c}\cdots \\ \text{row B}\\ \cdots \\ \text{row A}\\ \cdots \end{array}\right]$

• The determinant is a multiplicative map in the sense that

$det(AB)=det(A)det(B)$ for all n-by-n matrices $A$ and $B$.

This is generalized by the Cauchy-Binet formula to products of non-square matrices.

• It is easy to see that $det(r{I}_n)=r^n$ and thus

$det(rA)=det(r{I}_n\cdot A)=r^{n}det(A)$ for all $n$-by-$n$ matrices $A$ and all scalars $r$.

• A matrix over a commutative ring R is invertible if and only if its determinant is a unit in R. In particular, if A is a matrix over a field such as the real or complex numbers, then A is invertible if and only if det(A) is not zero. In this case we have

$det(A^{-1})=det(A{)}^{-1}.$

Expressed differently: the vectors v₁,...,v_n in Rⁿ form a basis if and only if det(v₁,...,v_n) is non-zero.

A matrix and its transpose have the same determinant:

$det(A^{\mathrm{\top }})=det(A).$

The determinants of a complex matrix and of its conjugate transpose are conjugate:

$det(A^{\ast })=det(A{)}^{\ast }.$

Theorems

Existence

Using Laplace's formula for the determinant

Binet's theorem

$det(AB)=detA\cdot detB$