mathematics-physics-wiki/docs/en/mathematics/linear-algebra/determinants.md

Determinants

Definition

With each n \times n matrix A with n \in \mathbb{N} it is possible to associate a scalar, the determinant of A denoted by \det (A) or |A|.

Definition: let A = (a_{ij}) be an n \times n matrix and let M_{ij} denote the (n-1) \times (n-1) matrix obtained from A by deleting the row and column containing a_{ij} with n \in \mathbb{N} and (i,j) \in \{1, \dots, n\} \times \{1, \dots, n\}. The determinant of M_{ij} is called the minor of a_{ij}. We define the cofactor of A_{ij} of a_{ij} by

A_{ij} = (-1)^{i+j} \det(M_{ij}).

This definition is necessary to formulate a definition for the determinant, as may be observed below.

Definition: the determinant of an n \times n matrix A with n \in \mathbb{N}, denoted by \det (A) or |A| is a scalar associated with the matrix A that is defined inductively as

\det (A) = \begin{cases}a_{11} &\text{ if } n = 1 \ a_{11} A_{11} + a_{12} A_{12} + \dots + a_{1n} A_{1n} &\text{ if } n > 1\end{cases}

where

A_{1j} = (-1)^{1+j} \det (M_{1j})

with j \in \{1, \dots, n\} are the cofactors associated with the entries in the first row of A.

Theorem: if A is an n \times n matrix with n \in \mathbb{N} \backslash \{1\} then \det(A) cam be expressed as a cofactor expansion using any row or column of A.

??? note "Proof:"

Will be added later.

We then have for a n \times n matrix A with n \in \mathbb{N} \backslash \{1\}

\begin{align*} \det(A) &= a_{i1} A_{i1} + a_{i2} A_{i2} + \dots + a_{in} A_{in}, \ &= a_{1j} A_{1j} + a_{2j} A_{2j} + \dots + a_{nj} A_{nj}, \end{align*}

with i,j \in \mathbb{N}.

For example, the determinant of a 4 \times 4 matrix A given by

A = \begin{pmatrix} 0 & 2 & 3 & 0\ 0 & 4 & 5 & 0\ 0 & 1 & 0 & 3\ 2 & 0 & 1 & 3\end{pmatrix}

may be determined using the definition and the theorem above

\det(A) = 2 \cdot (-1)^5 \det\begin{pmatrix} 2 & 3 & 0\ 4 & 5 & 0\ 1 & 0 & 3\end{pmatrix} = -2 \cdot 3 \cdot (-1)^6 \det\begin{pmatrix} 2 & 3 \ 4 & 5\end{pmatrix} = 12.

Properties of determinants

Theorem: if A is an n \times n matrix then \det (A^T) = \det (A).

??? note "Proof:"

It may be observed that the result holds for $n=1$. Assume that the results holds for all $k \times k$ matrices and that $A$ is a $(k+1) \times (k+1)$ matrix for some $k \in \mathbb{N}$. Expanding $\det (A)$ along the first row of $A$ obtains

$$
    \det(A) = a_{11} \det(M_{11}) - a_{12} \det(M_{12}) + \dots + (-1)^{k+2} a_{1(k+1)} \det(M_{1(k+1)}),
$$

since the minors are all $k \times k$ matrices it follows from the principle of natural induction that

$$
    \det(A) = a_{11} \det(M_{11}^T) - a_{12} \det(M_{12}^T) + \dots + (-1)^{k+2} a_{1(k+1)} \det(M_{1(k+1)}^T).
$$

The right hand side of the above equation is the expansion by minors of $\det(A^T)$ using the first column of $A^T$, therefore $\det(A^T) = \det(A)$.

Theorem: if A is an n \times n triangular matrix with n \in \mathbb{N}, then the determinant of A equals the product of the diagonal elements of A.

??? note "Proof:"

Will be added later.

Theorem: let A be an n \times n matrix

1. if A has a row or column consisting entirely of zeros, then \det(A) = 0.
2. if A has two identical rows or two identical columns, then \det(A) = 0.

??? note "Proof:"

Will be added later.