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:"

Let $A$ be a $n \times n$ triagular matrix with $n \in \mathbb{N}$ given by

$$
    A = \begin{pmatrix} a_{11} & \cdots &a_{1n}\\ & \ddots & \vdots \\ & & a_{nn} \end{pmatrix}.
$$

We claim that $\det(A) = a_{11} \cdot a_{22} \cdots a_{nn}$. We first check the claim for $n=1$ which is given by $\det(A) = a_{11}$. 

Now suppose for some $k \in \mathbb{N}$, the determinant of a $k \times k$ triangular $A_{k}$ is given by

$$
    \det(A_k) = a_1{11} \cdot a_{22} \cdots a_{kk}
$$

then by assumption 

$$
    \det(A_{k+1}) = \begin{pmatrix} A_k & a_{(k+1)1}\\& \vdots\\ 0 \cdots 0 & a_{(k+1)(k+1)}\end{pmatrix} = a_{(k+1)(k+1)} \det(A_k) + 0 = a_{11}a_1{11} \cdot a_{22} \cdots a_{kk} \cdot a_{(k+1)(k+1)}. 
$$

Hence if the claim holds for some $k \in \mathbb{N}$ then it also holds for $k+1$. The principle of natural induction implies now that for all $n \in \mathbb{N}$ we have

$$
    \det(A) = a_{11} \cdot a_{22} \cdots a_{nn}.
$$

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.

Lemma: let A be an n \times n matrix with n \in \mathbb{N}. If A_{jk} denotes the cofactor of a_{jk} for k \in \mathbb{N} then

a_{i1} A_{j1} + a_{i2} A_{j2} + \dots + a_{in} A_{jn} = \begin{cases} \det(A) &\text{ if } i = j,\ 0 &\text{ if } i \neq j.\end{cases}

??? note "Proof:"

If $i = j$ then we obtain the cofactor expansion of $\det(A)$ along the $i$th row of $A$. 

If $i \neq j$, let $A^*$ be the matrix obtained by replacing the $j$th row of $A$ by the $i$th row of $A$

$$
    A^* = \begin{pmatrix} a_{11} & a_{12} & \cdots & a_{1n}\\ \vdots \\ a_{11} & a_{12} & \cdots & a_{1n} \\ \vdots \\ a_{11} & a_{12} & \cdots & a_{1n} \\ \vdots \\ a_{n1} & a_{n2} & \cdots & a_{nn} \end{pmatrix} \begin{array}{ll} j\text{th row}\\ \\ \\ \\\end{array}
$$

since two rows of $A^*$ are the same its determinant must be zero. It follows from the cofactor expansion of $\det(A^*)$ along the $j$th row that

$$
\begin{align*}
    0 &= \det(A^*) = a_{i1} A_{j1}^* + a_{i2} A_{j2}^* + \dots + a_{in} A_{jn}^*, \\
    &= a_{i1} A_{j1} + a_{i2} A_{j2} + \dots + a_{in} A_{jn}.
\end{align*}
$$

Theorem: let E be an n \times n elementary matrix and A an n \times n matrix with n \in \mathbb{N} then we have

\det(E A) = \det(E) \det(A),

where

\det(E) = \begin{cases} -1 &\text{ if E is of type I},\ \alpha \in \mathbb{R}\backslash {0} &\text{ if E is of type II},\ 1 &\text{ if E is of type III}. \end{cases}

??? note "Proof:"

Will be added later.

Similar results hold for column operations, since for the elementary matrix E, E^T is also an elementary matrix and \det(A E) = \det((AE)^T) = \det(E^T A^T) = \det(E^T) \det(A^T) = \det(E) \det(A).

Theorem: an n \times n matrix A with n \in \mathbb{N} is singular if and only if

\det(A) = 0

??? note "Proof:"

Let $A$ be an $n \times n$ matrixwith $n \in \mathbb{N}$. Matrix $A$ can be reduced to row echelon form with a finite number of row operations obtaining

$$
    U = E_k E_{k-1} \cdots E_1 A,
$$

where $U$ is in $n \times n$ row echelon form and $E_i$ are $n \times n$ elementary matrices for $i \in \{1, \dots, k\}$. It follows then that 

$$
\begin{align*}
    \det(U) &= \det(E_k E_{k-1} \cdots E_1 A), \\
            &= \det(E_k) \det(E_{k-1}) \cdots \det(E_1) \det(A). 
\end{align*}
$$

Since the determinants of the elementary matrices are all nonzero, it follows that $\det(A) = 0$ if and only if $\det(U) = 0$. If $A$ is singular then $U$ has a row consisting entirely of zeros and hence $\det(U) = 0$. If $A$ is nonsingular then $U$ is triangular with 1's along the diagonal and hence $\det(U) = 1$.

From this theorem we may pose a method for computing \det(A) by taking

\det(A) = \Big(\det(E_k) \det(E_{k-1} \cdots \det(E_1)\Big)^{-1}.

Theorem: let A and B be n \times n matrices with n \in \mathbb{N} then

\det(AB) = \det(A) \det(B)

??? note "Proof:"

Will be added later.