Determinant

The determinant is an operation on square matrices. For a square matrix $A$, $\det A$ denotes the determinant of the square matrix $A$.

This article introduces three definitions of the determinant. It can be proved that the definitions given here are equivalent.

Definition by permutations

Prerequisites: Permutations, inversion count.

This method can be used to calculate determinants of low order by hand; its time complexity is factorial in the order.

Let $\pi(j_1j_2\cdots j_n)$ denote the inversion count of the permutation $j_1j_2\cdots j_n$, and let $S_n$ be the set of all permutations of length $n$. Notation:

$$\begin{aligned} \det A &= \begin{vmatrix} a_{11} & a_{12} & \cdots & a_{1n}\\ a_{21} & a_{22} & \cdots & a_{2n}\\ \vdots & \vdots & & \vdots\\ a_{n1} & a_{n2} & \cdots & a_{nn}\\ \end{vmatrix} \\ &= \sum_{(j_1j_2\cdots j_n) \in S_n} (-1)^{\pi(j_1j_2\cdots j_n)} a_{1 j_1} a_{2 j_2}\dots a_{n j_n} \end{aligned}$$

The $n$-th order determinant represented above is the algebraic sum of $n!$ terms. These terms are, for every possible selection of distinct rows and distinct columns from the square matrix $A$, the products of $n$ elements, namely $a_{1j_1}a_{2j_2}\cdots a_{nj_n}$.

The sign preceding the term $a_{1j_1}a_{2j_2}\cdots a_{nj_n}$ is ${(-1)}^{\pi(j_1j_2\cdots j_n)}$. In other words, the sign is positive when $j_1j_2\cdots j_n$ is an even permutation, and negative when $j_1j_2\cdots j_n$ is an odd permutation.

The diagonal rule for second- and third-order determinants is, in fact, an application of the definition by permutations. Determinants of fourth order and above are no longer suitable for the diagonal rule for the same reason. In particular, a first-order determinant is simply the element itself.

Theorem: From an $n$-th order determinant, select elements from rows $i_1,i_2,\cdots,i_n$ and columns $j_1,j_2,\cdots,j_n$ and take their product

$$a_{i_1j_1}a_{i_2j_2}\cdots a_{i_nj_n}$$

Here, $i_1,i_2,\cdots,i_n$ and $j_1,j_2,\cdots,j_n$ are both permutations of the numbers $1,2,\cdots,n$; there are $n$ such numbers. Then the sign of this term in the determinant is ${(-1)}^{s+t}$, where

$$s=\pi(i_1i_2\cdots i_n)$$ $$t=\pi(j_1j_2\cdots j_n)$$

Theorem: A determinant is equal to its transpose.

Theorem: Suppose that every element of the determinant $\det A$ in row $i$ can be expressed as the sum of two terms:

$$\begin{vmatrix} a_{11} & a_{12} & \cdots & a_{1n}\\ \vdots & \vdots & & \vdots\\ b_{i1}+c_{i1} & b_{i2}+c_{i2} & \cdots & b_{in}+c_{in}\\ \vdots & \vdots & & \vdots\\ a_{n1} & a_{n2} & \cdots & a_{nn}\\ \end{vmatrix}$$

Then this determinant is equal to the sum of two determinants, $\det A_1$ and $\det A_2$. In $A_1$, row $i$ is $b_{i1},b_{i2},\cdots,b_{in}$, while in $A_2$, row $i$ is $c_{i1},c_{i2},\cdots,c_{in}$; all other rows of $A_1$ and $A_2$ are the same as those of $A$. The same property holds for columns.

Definition by induction

This method only describes an algebraic property of determinants. Its time complexity is also factorial in the order, so it is not suitable for calculation.

Algebraic cofactors

In an $n$-th order determinant $\det A$, choose any rows and columns of matrix $A$—specifically, $k$ rows and $k$ columns. The $k$-th order matrix formed by the elements at the intersections of these rows and columns is called a $k$-th order submatrix, and its determinant is called a $k$-th order minor.

For an $n$-th order determinant $\det A$, the element $a_{ij}$ has a minor matrix $M_{ij}$: in the original matrix $A$, after deleting the row and column containing $a_{ij}$, it is the remaining $n-1$-th order submatrix; its determinant $\det M_{ij}$ is called the minor.

For an $n$-th order determinant $\det A$, the element $a_{ij}$ has minor $\det M_{ij}$; after the minor is multiplied by the sign ${(-1)}^{i+j}$, it is called the algebraic cofactor of $a_{ij}$, denoted by $A_{ij}$.

The following conclusion can be derived from the definition by permutations in the previous section:

Theorem: If, in an $n$-th order determinant $\det A$, all elements in row $i$ or column $j$ except $a_{ij}$ are $0$, then the determinant is equal to the product of $a_{ij}$ and its algebraic cofactor $A_{ij}$.

Expansion of determinants

Since transposing a square matrix does not change its determinant, it is enough to introduce either row expansion or column expansion.

The determinant $\det A$ is defined as the sum of the products of all elements in any one of its rows (or columns) and their corresponding algebraic cofactors.

In other words, a determinant can be defined recursively by its expansion along a row (or column):

$$\begin{aligned} \det A &= a_{i1}A_{i1}+a_{i2}A_{i2}+\cdots+a_{in}A_{in} \\ &= \sum_{j = 1}^{n} a_{ij}A_{ij} \\ &= \sum_{j = 1}^{n} (-1)^{i + j} a_{ij} \det M_{ij} \end{aligned}$$ $$\begin{aligned} \det A &= a_{1j}A_{1j}+a_{2j}A_{2j}+\cdots+a_{nj}A_{nj} \\ &= \sum_{i = 1}^{n} a_{ij}A_{ij} \\ &= \sum_{i = 1}^{n} (-1)^{i + j} a_{ij} \det M_{ij} \end{aligned}$$

The recursion terminates at the determinant of a first-order matrix, which is simply its only element.

Therefore, we have the following conclusion:

Theorem: The sum of the products of the elements in one row (or column) of a determinant $\det A$ and the corresponding algebraic cofactors of another row (or column) is equal to $0$.

In other words, when $i\neq j$:

$$a_{i1}A_{j1}+a_{i2}A_{j2}+\cdots+a_{in}A_{jn}=0$$ $$a_{1i}A_{1j}+a_{2i}A_{2j}+\cdots+a_{ni}A_{nj}=0$$

Axiomatic definition

The axiomatic definition says that an operation satisfying certain properties can only be the determinant.

Prerequisite: Elementary operations.

Let $D_i(k)$ denote a scaling matrix, $P_{ij}$ a transposition matrix, and $T_{ij}(k)$ a shear matrix.

For an $n$-th order matrix $A$, the operation $\det$, if it satisfies the following four properties, is called a determinant:

• Multiplying all elements in one row or one column of a determinant by a number $k$ simultaneously is equivalent to multiplying the determinant by $k$.

  $$\det(D_i(k)A) = \det(AD_i(k)) = k \det A$$

• Swapping two rows or two columns of a determinant changes the sign of the determinant.

  $$\det(P_{ij}A) = \det(AP_{ij}) = -\det A$$

• Multiplying the elements in one row or one column of a determinant by the same number and adding them to the corresponding elements in another row or column leaves the determinant unchanged.

  $$\det(T_{ij}(k)A) = \det(AT_{ij}(k))= \det A$$

• The determinant of the identity matrix is $1$.

  $$\det I = 1$$

Using the properties of determinants under elementary operations makes it convenient to calculate higher-order determinants by hand. The determinant calculation using Gaussian elimination method also uses this property, with time complexity $O(n^3)$.

The above properties also have several corollaries:

• A common factor of one row or one column of a determinant can be taken outside the determinant symbol.
• If all elements in one row or one column of a determinant are $0$, then the determinant is $0$.
• If corresponding elements in two rows or two columns of a determinant are proportional, then the determinant is $0$.
• If two rows or two columns of a determinant are identical, then the determinant is $0$.

These corollaries are very commonly used when calculating determinants by hand.