Determinants

Definition of $\det A$

Permutations

A permutation of $(1,2,\dots,n)$ is a sequence of $n$ integers $i_1, \dots, i_n\in\{1,\dots,n\}$ such that $i_k\neq i_l$ for all $k\neq l$.

For example, all possible permutations of $(1,2,3)$ are

$(1, 2, 3), \quad (1, 3, 2), \quad (2, 1, 3), \quad (2, 3, 1), \quad (3, 1, 2), \quad (3, 2, 2).$

There are $n!$ permutations of $(1,2,\dots ,n)$.

Sign of a permutation

By definition, the number of inversions of a permutation $(i_1,\dots, i_n)$ is

$\delta(i_1, \dots, i_n) \stackrel{\rm def}{=} \# \{k<l \mid i_k>i_l\}$

A permutation $(i_1, i_2, \dots, i_n)$ is called even if $\delta(i_1,\dots, i_n)$ is even.

A permutation $(i_1, i_2, \dots, i_n)$ is called odd if $\delta(i_1,\dots, i_n)$ is odd.

The sign of the permutation $(i_1, \dots, i_n)$ is defined to be

$\epsilon_{i_1i_2\cdots i_n} = (-1)^{\delta(i_1, i_2, \dots, i_n)} = \begin{cases} +1&\text{if } \delta(i_1, \dots, i_n) \text{ is even}\\ -1&\text{if } \delta(i_1, \dots, i_n) \text{ is odd}\\ \end{cases}$

The quantity $\epsilon_{i_1i_2\cdots i_n}$ is called the Levi-Civita symbol.

The determinant

$\det A \stackrel{\rm def}{=} \sum_{i_1, \dots, i_n} \epsilon_{i_1i_2\cdots i_n} \,a_{1i_1}a_{2i_2}\cdots a_{ni_n}$

Properties of the sign $\epsilon_{i_1i_2\cdots i_n}$

• $\epsilon_{i_1\cdots i_k\cdots i_\ell\cdots i_n} = -\epsilon_{i_1\cdots i_\ell\cdots i_k\cdots i_n}$
• if $i_1, \dots, i_k < i_{k+1}, \dots, i_n$ then $\epsilon_{i_1i_2\cdots i_n} =\epsilon_{i_1\cdots i_k} \epsilon_{i_{k+1}\cdots i_n}$

What is the sign of $a_{14}a_{22}a_{35}a_{43}a_{51}$ when you expand a $5\times 5$ determinant

The question can be rephrased as: compute $\epsilon_{42531}$

$\left|\begin{matrix} a_{11}&a_{12}&a_{13}&a_{14}&a_{15} \\ a_{21}&a_{22}&a_{23}&a_{24}&a_{25} \\ a_{31}&a_{32}&a_{33}&a_{34}&a_{35} \\ a_{41}&a_{42}&a_{43}&a_{44}&a_{45} \\ a_{51}&a_{52}&a_{53}&a_{54}&a_{55} \end{matrix} \right|$

Special cases

$2\times2$ determinants

$\left| \begin{matrix} a & b \\ c & d \end{matrix} \right| = ad-bc$

$3\times3$ determinants

$\left| \begin{matrix} a_1 & a_2 & a_3 \\ b_1 & b_2 & b_3 \\ c_1 & c_2 & c_3 \end{matrix} \right| = a_1b_2c_3 -a_1b_3c_2 - a_2b_1c_3 + a_2b_3c_1 +a_3b_1c_2-a_3b_2c_1$

Determinant of upper triangular matrices

If all entries of a matrix below its diagonal are zero, then the determinant of the matrix is the product of its diagonal entries:

$\left|\begin{matrix} a_{11}&a_{12}&\cdots &\cdots&a_{1n} \\ 0&a_{22}&\cdots &\cdots&a_{2n} \\ 0 & 0&a_{33} &\cdots&a_{3n} \\ \cdots& \cdots&\cdots &\cdots&\cdots \\ 0& 0&\cdots & 0 &a_{nn} \end{matrix}\right|= a_{11}a_{22}\cdots a_{nn}$

Determinant of the identity matrix

$\det I = 1.$

Determinant of block triangular matrices

If $A$ is a $k\times k$ matrix, $B$ an $k\times l$ matrix, and $C$ an $l\times l$ matrix then

$\left|\begin{matrix} A & B \\ 0 & C \end{matrix}\right| = (\det A)(\det C)$

Properties

Determinant of the transpose

$\det A^\top = \det A$

Swapping rows or columns changes the sign

If $B$ is the matrix you get by swapping two rows, or by swapping two columns in the matrix $A$, then $\det A = -\det B$

The determinant as a function of its rows

If we have $n$ row vectors $a_1, \dots ,a_n\in\F^n$, given by

$\begin{aligned} a_1 &= \mat a_{11} & a_{12} & \cdots & a_{1n} \rix, \\ a_2 &= \mat a_{21} & a_{22} & \cdots & a_{2n} \rix, \\ &\qquad\vdots \\ a_n &= \mat a_{n1} & a_{n2} & \cdots & a_{nn} \rix, \end{aligned}$

then we define

$\det(a_1, a_2, \dots, a_n) = \left|\begin{matrix} a_{11}&a_{12}&\cdots&a_{1n} \\ a_{21}&a_{22}&\cdots&a_{2n} \\ \vdots&\vdots& &\vdots \\ a_{n1}&a_{n2}&\cdots&a_{nn} \end{matrix} \right|$

One has for all $a, b, a_1, a_2, \dots, a_n\in \F^n$ and $t\in \F$:

$\begin{aligned} \det(a+b, a_2, a_3, \dots, a_n) &= \det(a, a_2, a_3, \dots, a_n)+\det(b, a_2, a_3, \dots, a_n)\\ \det(ta, a_2, a_3, \dots, a_n) &= t\det(a, a_2, a_3, \dots, a_n) \\ \det(a_1, \dots,a_i, \dots, a_j,\dots a_n) &= - \det(a_1, \dots,a_j, \dots, a_i,\dots a_n)\\ \det(a_1, \dots,a_i, \dots, a_i,\dots a_n) &= 0\\ \det(a_1, \dots,a_i, \dots, a_j,\dots a_n) &= \det(a_1, \dots,a_i+ta_j, \dots, a_j,\dots a_n) \end{aligned}$

Consequence: if $\{a_1, \dots, a_n\}$ is linearly dependent, then $\det(a_1, \dots, a_n)=0$

Cofactor expansion

The $ij$-minor of an $n\times n$ matrix $A$ is the $(n-1)\times(n-1)$ matrix obtained by deleting the $i^{\rm th}$ row and $j^{\rm th}$ column from $A$. Our textbook uses the notation $\tilde A_{ij}$ for the $ij$-minor of $A$.

The $ij$-cofactor of the matrix $A$ is the number

$c_{ij} = (-1)^{i+j}\det \tilde A_{ij}$

Cofactor Expansion Theorem. If $A=(a_{ij})$ is an $n\times n$ matrix, then one has

for any $i\in\{1,2, \dots, n\}$.

A consequence of the cofactor expansion theorem is that if $i\neq j$ then

$a_{j1}c_{i1} + a_{j2}c_{i2} + \cdots + a_{jn}c_{in} = 0.$

Example

Expanding a $3\times3$ determinant along its middle row:

$\begin{aligned} \deter1 & 2 & 3 \\ 3&2&1 \\ -2&4&3\minant &= a_{21}c_{21} + a_{22}c_{22} + a_{23}c_{23}\\ &=-3\cdot\deter2 &1\\ 4&3\minant +2\cdot\deter1&3\\-2&3\minant -1\cdot\deter1&2\\-2&4\minant=\cdots \end{aligned}$

A formula to invert a matrix and Cramer's rule

Theorem. For any $n\times n$ matrix $A$ one has

Cramer's rule. For any $y\in\F^n$ the solution of $Ax=y$ is given by

$\begin{gathered} x_1=\frac{\deter y_1 & a_{12} &\dots& a_{1n}\\ y_2 & a_{22} &\dots& a_{2n}\\ & &\dots& \\ y_n & a_{n2} &\dots& a_{nn} \minant}{\deter a_{11} & a_{12} &\dots& a_{1n}\\ a_{12} & a_{22} &\dots& a_{2n}\\ & &\dots& \\ a_{1n} & a_{n2} &\dots& a_{nn} \minant},\quad x_2=\frac{\deter a_{11} &y_1 &\dots& a_{1n}\\ a_{12} &y_2 &\dots& a_{2n}\\ & &\dots& \\ a_{1n} &y_n &\dots& a_{nn} \minant}{\deter a_{11} & a_{12} &\dots& a_{1n}\\ a_{12} & a_{22} &\dots& a_{2n}\\ & &\dots& \\ a_{1n} & a_{n2} &\dots& a_{nn} \minant},\quad\dots, \\ x_n=\frac{\deter a_{11} & a_{12} &\dots&y_1 \\ a_{12} & a_{22} &\dots&y_2 \\ & &\dots& \\ a_{1n} & a_{n2} &\dots&y_n \minant}{\deter a_{11} & a_{12} &\dots& a_{1n}\\ a_{12} & a_{22} &\dots& a_{2n}\\ & &\dots& \\ a_{1n} & a_{n2} &\dots& a_{nn} \minant},\quad \end{gathered}$

Example the inverse of a $2\times 2$ matrix

$\mat a&b\\c&d\rix^{-1} = \frac{1}{ad-bc}\mat d &-b \\ -c & a\rix$

Determinant of a matrix product

Theorem. For any two $n\times n$ matrices $A$ and $B$ one has

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

Theorem. $A$ is invertible $\iff \det A\neq 0$

Theorem. If $a_1, \dots, a_n\in\F^n$ then