Determinants and Permutations

Multilinear and Alternating Maps

Definition 66: Multilinear Map

Let $𝑉$ be a vector space over a field $𝐹$ . A function $𝑓 : 𝑉^{𝑛} \to 𝐹$ is $𝒏$ -linear on $𝑉$ if it is linear in each argument when the other arguments are held fixed.

That is, if $𝑓$ is $𝑛$ -linear, then for each $𝑖 \in {1, \dots, 𝑛}$ , if we fix the vectors $𝐯_{1}, \dots, 𝐯_{𝑖 - 1}, 𝐯_{𝑖 + 1}, \dots, 𝐯_{𝑛} \in 𝑉$ , the function defined by $𝐯_{𝑖} \mapsto 𝑓 (𝐯_{1}, \dots, 𝐯_{𝑖}, \dots, 𝐯_{𝑛})$ is a linear map, such that for all $𝐯_{𝑗}, 𝐯_{𝑘} \in 𝑉$ and all scalars $𝛼, 𝛽 \in 𝐹$ , we have:

𝑓 (𝐯_{1}, \dots, 𝛼 𝐯_{𝑗} + 𝛽 𝐯_{𝑘}, \dots, 𝐯_{𝑛}) = 𝛼 𝑓 (𝐯_{1}, \dots, 𝐯_{𝑗}, \dots, 𝐯_{𝑛}) + 𝛽 𝑓 (𝐯_{1}, \dots, 𝐯_{𝑘}, \dots, 𝐯_{𝑛}) .

Terminology

A function $𝑓 : 𝑉^{𝑛} \to 𝐹$ is a multilinear map if it is $𝑛$ -linear for some $𝑛 \in ℤ^{+}$ .

Notation

The set of all $𝑛$ -linear maps on $𝑉$ is denoted $𝑉^{(𝑛)}$ , and $𝑉^{(𝑛)} \subseteq ℒ (𝑉^{𝑛}, 𝐹)$ .

Proposition 16

$𝑉^{(𝑛)}$ is a subspace of $ℒ (𝑉^{𝑛}, 𝐹)$ .

Definition 67: Alternating Map

Let $𝑉$ be a vector space over a field $𝐹$ . A function $𝑓 : 𝑉^{𝑛} \to 𝐹$ is alternating if $𝑓 (𝐯_{1}, \dots, 𝐯_{𝑛}) = 0$ if $𝐯_{𝑖} = 𝐯_{𝑗}$ for some $𝑖 \neq 𝑗$ .

That is, in an alternating map, if any two arguments are equal, the output is zero.

Notation

The set of all alternating $𝑛$ -linear maps on $𝑉$ is denoted $𝑉_{alt}^{(𝑛)}$ , and $𝑉_{alt}^{(𝑛)} \subseteq 𝑉^{(𝑛)}$ .

Theorem 56: Alternating Multilinear Maps and Linear Dependence

Let $𝑉$ be a vector space, and let $𝑓 \in 𝑉_{alt}^{(𝑛)}$ . Then for all linearly dependent sets ${𝐯_{1}, \dots, 𝐯_{𝑛}}$ , it holds that $𝑓 (𝐯_{1}, \dots, 𝐯_{𝑛}) = 0$ .

Proof: Theorem 56.

Let

𝑉

be a vector space, and let

𝑓 \in 𝑉_{alt}^{(𝑛)}

Theorem 57: Alternating $𝑛$ -Linear Maps where $𝑛 > \dim (𝑉)$

Let $𝑉$ be a vector space over a field $𝐹$ , and let $𝑓 \in 𝑉_{alt}^{(𝑛)}$ . If $𝑛 > \dim (𝑉)$ , then $𝑓 (𝐯_{1}, \dots, 𝐯_{𝑛}) = 0$ for all vectors $𝐯_{1}, \dots, 𝐯_{𝑛} \in 𝑉$ .

Theorem 58: Antisymmetry in Alternating Multilinear Maps

Suppose $𝑓 \in 𝑉_{alt}^{(𝑛)}$ is an alternating multilinear map. Then:

𝑓 (\dots, 𝐯_{𝑖}, \dots, 𝐯_{𝑗}, \dots,) = - 𝑓 (\dots, 𝐯_{𝑗}, \dots, 𝐯_{𝑖}, \dots)

for all $𝑖 \neq 𝑗$ and all vectors $𝐯_{1}, \dots, 𝐯_{𝑛} \in 𝑉$ . That is, swapping any two arguments of $𝑓$ negates the output of $𝑓$ .

Proof: Theorem 58.

Suppose $𝑓 \in 𝑉_{alt}^{(𝑛)}$ is an alternating multilinear map, and choose some arbitrary $𝐯_{𝑖}, 𝐯_{𝑗} \in 𝑉$ . Then:

\begin{matrix} 𝑓 (\dots, 𝐯_{𝑖} + 𝐯_{𝑗}, \dots, 𝐯_{𝑖} + 𝐯_{𝑗}, \dots) & = 0 & by alternation \\ 𝑓 (\dots, 𝐯_{𝑖}, \dots, 𝐯_{𝑖} + 𝐯_{𝑗}, \dots) + 𝑓 (\dots, 𝐯_{𝑗}, \dots, 𝐯_{𝑖} + 𝐯_{𝑗}, \dots) & = 0 & by multilinearity \\ 𝑓 (\dots, 𝐯_{𝑖}, \dots, 𝐯_{𝑖}, \dots) + 𝑓 (\dots, 𝐯_{𝑖}, \dots, 𝐯_{𝑗}, \dots) + 𝑓 (\dots, 𝐯_{𝑗}, \dots, 𝐯_{𝑖}, \dots) + 𝑓 (\dots, 𝐯_{𝑗}, \dots, 𝐯_{𝑗}, \dots) & = 0 & by multilinearity \\ 0 + 𝑓 (\dots, 𝐯_{𝑖}, \dots, 𝐯_{𝑗}, \dots) + 𝑓 (\dots, 𝐯_{𝑗}, \dots, 𝐯_{𝑖}, \dots) + 0 & = 0 & by alternation. \end{matrix}

Finally, we can rearrange the equation to get:

𝑓 (\dots, 𝐯_{𝑖}, \dots, 𝐯_{𝑗}, \dots) = - 𝑓 (\dots, 𝐯_{𝑗}, \dots, 𝐯_{𝑖}, \dots),

as desired. ∎

Axioms of Determinants

Definition 68: Determinant Function

Let $𝑉$ be a vector space over a field $𝐹$ , and fix an ordered basis $ℬ = {𝐛_{1}, \dots, 𝐛_{𝑛}}$ for $𝑉$ .

A function $𝐷 : 𝑉^{𝑛} \to 𝐹$ is a determinant function on $𝑉$ with respect to the basis $ℬ$ if the following properties hold:

Multilinearity. $𝐷$ is an $𝑛$ -linear map. That is,
$𝐷 (𝐯_{1}, \dots, 𝛼 𝐯_{𝑖} + 𝛽 𝐯_{𝑗}, \dots, 𝐯_{𝑛}) = 𝛼 𝐷 (𝐯_{1}, \dots, 𝐯_{𝑖}, \dots, 𝐯_{𝑛}) + 𝛽 𝐷 (𝐯_{1}, \dots, 𝐯_{𝑗}, \dots, 𝐯_{𝑛}) .$
Alternation. $𝐷$ is alternating. That is, if $𝐯_{𝑖} = 𝐯_{𝑗}$ for some $𝑖 \neq 𝑗$ , then $𝐷 (𝐯_{1}, \dots, 𝐯_{𝑛}) = 0$ .
Normalization. $𝐷 (𝐛_{1}, \dots, 𝐛_{𝑛}) = 1$ .

That is, a alternating multilinear map $𝐷$ is a determinant function if it evaluates to the multiplicative identity $1$ on the basis vectors of $𝑉$ .

Theorem 59: Existence and Uniqueness of the Determinant Function

Let $𝑉$ be a vector space over a field $𝐹$ , and fix an ordered basis $ℬ = {𝐛_{1}, \dots, 𝐛_{𝑛}}$ for $𝑉$ . Then there is exactly one determinant function on $𝑉$ with respect to the basis $ℬ$ .

Partial Proof: Theorem 59 (Uniqueness).

Let $𝑉$ be a vector space over a field $𝐹$ , and fix an ordered basis $ℬ = {𝐛_{1}, \dots, 𝐛_{𝑛}}$ for $𝑉$ . Assume that there exist two determinant functions $𝐷_{1}, 𝐷_{2} : 𝑉^{𝑛} \to 𝐹$ on $𝑉$ with respect to the basis $ℬ$ .

Lemma: On any ordering of the basis vectors

𝐛_{𝑗_{1}}, \dots, 𝐛_{𝑗_{𝑛}}

𝐷_{1} (𝐛_{𝑗_{1}}, \dots, 𝐛_{𝑗_{𝑛}}) = 𝐷_{2} (𝐛_{𝑗_{1}}, \dots, 𝐛_{𝑗_{𝑛}})

Let $(𝑗_{1}, \dots, 𝑗_{𝑛})$ be an ordering (permutation) of the indices $(1, \dots, 𝑛)$ . Then $(𝑗_{1}, \dots, 𝑗_{𝑛})$ can be achieved by finite $𝑘$ number of swaps of adjacent indices (see Theorem 62).

By the antisymmetry of alternating maps (Theorem 58), each swap of adjacent indices negates the output of $𝐷_{1}$ and $𝐷_{2}$ . Therefore, after $𝑘$ swaps, we have:

𝐷_{1} (𝐛_{𝑗_{1}}, \dots, 𝐛_{𝑗_{𝑛}}) = {(- 1)}^{𝑘} 𝐷_{1} (𝐛_{1}, \dots, 𝐛_{𝑛}) = {(- 1)}^{𝑘} 𝐷_{2} (𝐛_{1}, \dots, 𝐛_{𝑛}) = 𝐷_{2} (𝐛_{𝑗_{1}}, \dots, 𝐛_{𝑗_{𝑛}}),

as desired. ∎

For arbitrary vectors $𝐯_{1}, \dots, 𝐯_{𝑛} \in 𝑉$ , we can express each vector as a linear combination of the basis vectors:

𝐯_{𝑖} = \sum_{𝑗 = 1}^{𝑛} 𝛼_{𝑖 𝑗} 𝐛_{𝑗},

where $𝛼_{𝑖 𝑗} \in 𝐹$ are the coefficients of the linear combination.

We have:

\begin{matrix} 𝐷_{1} (𝐯_{1}, \dots, 𝐯_{𝑛}) & = 𝐷_{1} (\sum_{𝑗 = 1}^{𝑛} 𝛼_{1 𝑗} 𝐛_{𝑗}, \dots, \sum_{𝑗 = 1}^{𝑛} 𝛼_{𝑛 𝑗} 𝐛_{𝑗}) & since 𝐯_{𝑖} = \sum_{𝑗 = 1}^{𝑛} 𝛼_{𝑖 𝑗} 𝐛_{𝑗} \\ = \sum_{𝑗_{1} = 1}^{𝑛} \dots \sum_{𝑗_{𝑛} = 1}^{𝑛} 𝛼_{1 𝑗_{1}} \dots 𝛼_{𝑛 𝑗_{𝑛}} 𝐷_{1} (𝐛_{𝑗_{1}}, \dots, 𝐛_{𝑗_{𝑛}}) & by multilinearity \\ = \sum_{𝑗_{1} = 1}^{𝑛} \dots \sum_{𝑗_{𝑛} = 1}^{𝑛} 𝛼_{1 𝑗_{1}} \dots 𝛼_{𝑛 𝑗_{𝑛}} 𝐷_{2} (𝐛_{𝑗_{1}}, \dots, 𝐛_{𝑗_{𝑛}}) & because 𝐷_{1} (𝐛_{𝑗_{1}}, \dots, 𝐛_{𝑗_{𝑛}}) = 𝐷_{2} (𝐛_{𝑗_{1}}, \dots, 𝐛_{𝑗_{𝑛}}) \\ = 𝐷_{2} (\sum_{𝑗 = 1}^{𝑛} 𝛼_{1 𝑗} 𝐛_{𝑗}, \dots, \sum_{𝑗 = 1}^{𝑛} 𝛼_{𝑛 𝑗} 𝐛_{𝑗}) & by multilinearity \\ = 𝐷_{2} (𝐯_{1}, \dots, 𝐯_{𝑛}) & by substitution of linear combinations, \end{matrix}

which implies that $𝐷_{1} = 𝐷_{2}$ . Therefore, the determinant function is unique. ∎

Remark

Although there can be many alternating multilinear maps on $𝑉$ , by fixing an ordered basis $ℬ$ and enforcing normalization over $ℬ$ , we can guarantee that there is exactly one determinant function on $𝑉$ with respect to $ℬ$ .

Definition 69: Determinant of a Matrix

Let $𝐴 \in 𝐹^{𝑛 \times 𝑛}$ be a square matrix over a field $𝐹$ . The determinant of $𝐴$ , denoted $\det (𝐴)$ or $| 𝐴 |$ , is the determinant function $𝐷$ on $𝐹^{𝑛}$ evaluated on the columns of $𝐴$ with respect to the standard basis $ℰ_{𝑛} = {𝐞_{1}, \dots, 𝐞_{𝑛}}$ for $𝐹^{𝑛}$ . That is, $\det : 𝐹^{𝑛 \times 𝑛} \to 𝐹$ such that:

\det (𝐴) = 𝐷 (𝐚_{1}, 𝐚_{2}, \dots, 𝐚_{𝑛}),

where $𝐚_{𝑖}$ is the $𝑖$ ^th column of $𝐴$ .

Corollary 33

$\det (𝐼_{𝑛}) = 𝐷 (𝐞_{1}, 𝐞_{2}, \dots, 𝐞_{𝑛}) = 1$ by normalization.

Corollary 34

If $𝐴 \in 𝐹^{𝑛 \times 𝑛}$ has two identical columns, then $\det (𝐴) = 0$ by alternation.

Notation

We will use $\det$ instead of $𝐷$ for any determinant function from now on. For example, $\det (𝐴) = \det (𝐚_{1}, 𝐚_{2}, \dots, 𝐚_{𝑛})$ .

Theorem 60: Determinant of a $2 \times 2$ Matrix

Let $𝐴 = (\begin{matrix} 𝑎 & 𝑏 \\ 𝑐 & 𝑑 \end{matrix}) \in 𝐹^{2 \times 2}$ . Then the determinant of $𝐴$ is given by:

\det (𝐴) = 𝑎 𝑑 - 𝑏 𝑐 .

Proof: Determinant of a

2 \times 2

Matrix.

Let $𝐴 = (\begin{matrix} 𝑎 & 𝑏 \\ 𝑐 & 𝑑 \end{matrix}) \in 𝐹^{2 \times 2}$ . Then the determinant of $𝐴$ is given by:

\begin{matrix} \det (𝐴) & = \det (𝐚_{1}, 𝐚_{2}) & by definition of determinant of a matrix \\ = \det ((\begin{matrix} 𝑎 \\ 𝑐 \end{matrix}), (\begin{matrix} 𝑏 \\ 𝑑 \end{matrix})) & by substitution of columns \\ = \det ((\begin{matrix} 𝑎 \\ 0 \end{matrix}), (\begin{matrix} 0 \\ 𝑑 \end{matrix})) + \det ((\begin{matrix} 0 \\ 𝑐 \end{matrix}), (\begin{matrix} 𝑏 \\ 0 \end{matrix})) & by multilinearity \\ = 𝑎 𝑑 \det ((\begin{matrix} 1 \\ 0 \end{matrix}), (\begin{matrix} 0 \\ 1 \end{matrix})) + 𝑏 𝑐 \det ((\begin{matrix} 0 \\ 1 \end{matrix}), (\begin{matrix} 1 \\ 0 \end{matrix})) & by multilinearity \\ = 𝑎 𝑑 \det (𝐞_{1}, 𝐞_{2}) + 𝑏 𝑐 \det (𝐞_{2}, 𝐞_{1}) & by substitution of standard basis vectors \\ = 𝑎 𝑑 \det (𝐞_{1}, 𝐞_{2}) - 𝑏 𝑐 \det (𝐞_{1}, 𝐞_{2}) & by antisymmetry of alternating maps \\ = 𝑎 𝑑 - 𝑏 𝑐 & by normalization, \end{matrix}

as desired. ∎

Property

If the columns of $𝐴$ are linearly dependent, then $\det (𝐴) = 0$ .

Proposition 17: Determinant of Elementary Row Operations

Suppose $𝐴 \in 𝐹^{𝑛 \times 𝑛}$ . Then:

Row swap. If $𝐴^{'}$ is obtained by applying $𝑅_{𝑖} \leftrightarrow 𝑅_{𝑗}$ to $𝐴$ , then $\det (𝐴^{'}) = - \det (𝐴)$ .
Row scaling. If $𝐴^{'}$ is obtained by applying $𝑅_{𝑖} \leftarrow 𝑐 𝑅_{𝑖}$ to $𝐴$ , then $\det (𝐴^{'}) = 𝑐 \det (𝐴)$ .
Row replacement. If $𝐴^{'}$ is obtained by applying $𝑅_{𝑖} \leftarrow 𝑅_{𝑖} + 𝑐 𝑅_{𝑗}$ to $𝐴$ , then $\det (𝐴^{'}) = \det (𝐴)$ .

Lemma

If $𝐴 \in 𝐹^{𝑛 \times 𝑛}$ and $𝐸$ is an elementary matrix, then $\det (𝐸 𝐴) = \det (𝐸) \det (𝐴)$ .

Corollary 35: Determinant of Elementary Matrices

$\det (𝐸 [𝑅_{𝑖} \leftrightarrow 𝑅_{𝑗}]) = - 1$ , $\det (𝐸 [𝑅_{𝑖} \leftarrow 𝑐 𝑅_{𝑖}]) = 𝑐$ , and $\det (𝐸 [𝑅_{𝑖} \leftarrow 𝑅_{𝑖} + 𝑐 𝑅_{𝑗}]) = 1$ .

Definition 70: Determinant of a Linear Map

Let $𝑇 : 𝑉 \to 𝑉$ be a linear map on a finite-dimensional vector space $𝑉$ over a field $𝐹$ . The determinant of $𝑇$ , denoted $\det (𝑇)$ , is the determinant of the standard matrix of $𝑇$ with respect to any basis $ℬ$ for $𝑉$ :

\det (𝑇) = \det ({[𝑇]}_{ℬ ℬ}) .

Theorem 61: Basis Independence of Determinants of Linear Maps

Let $𝑇 : 𝑉 \to 𝑉$ be a linear map on a finite-dimensional vector space $𝑉$ over a field $𝐹$ . Then the determinant of $𝑇$ is independent of the choice of basis for $𝑉$ .

Permutations

Definition 71: Permutation

Suppose $𝑚$ is a positive integer. A permutation of the set ${1, 2, \dots, 𝑚}$ (shortened to $[𝑚]$ ) is a bijective mapping $𝜎 : [𝑚] \to [𝑚]$ . The set of all permutations of $[𝑚]$ is denoted $perm 𝑚$ .

Example.

Define a permutation

𝜎 : [3] \to [3]

𝜎 (1) = 2

𝜎 (2) = 3

, and

𝜎 (3) = 1

. Then

𝜎 \in perm 3

Nonexample.

Define a function

𝜏 : [3] \to [3]

𝜏 (1) = 2

𝜏 (2) = 2

, and

𝜏 (3) = 1

. Then

𝜏 \notin perm 3

since it is not bijective.

Property

For any $𝑚 \in ℤ^{+}$ , $| perm 𝑚 | = 𝑚!$ .

Definition 72: Standard Sequence of a Permutation

The sequence $𝛔 = (𝜎 (1), 𝜎 (2), \dots, 𝜎 (𝑚)) \in {[𝑚]}^{𝑚}$ is called the standard sequence of $𝜎$ .

Example.

The permutation

𝜎 \in perm 3

defined by

𝜎 (1) = 2

𝜎 (2) = 3

, and

𝜎 (3) = 1

has standard sequence

𝛔 = (1, 2, 3)

Corollary 36: Standard Sequence of the Identity Permutation

The identity permutation ${id}_{[𝑚]} \in perm 𝑚$ defined by $𝑘 \mapsto 𝑘$ has standard sequence $(1, 2, \dots, 𝑚)$ .

Definition 73: Transposition

A transposition is a permutation $𝜏 \in perm 𝑚$ such that $𝜏 (𝑖) = 𝑗$ and $𝜏 (𝑗) = 𝑖$ for some $𝑖 \neq 𝑗$ , and $𝜏 (𝑘) = 𝑘$ for all $𝑘 \in [𝑚] \ {𝑖, 𝑗}$ . That is, it is the permutation whose standard sequence is the result of swapping exactly one pair of elements in the identity sequence $(1, 2, \dots, 𝑚)$ .

Example.

The permutation

𝜏 \in perm 3

defined by

𝜏 (1) = 2

𝜏 (2) = 1

, and

𝜏 (3) = 3

is a transposition.

Example.

The permutation whose standard sequence is

𝛕 = (1, 4, 3, 2, 5)

is a transposition.

Nonexample.

The permutation whose standard sequence is

𝛔 = (1, 4, 3, 2)

is not a transposition.

Theorem 62: Permutation Decomposition

Every permutation $𝜎 \in perm 𝑚$ can be expressed as a finite composition of transpositions. That is, there exists a finite sequence of transpositions $𝜏_{1}, \dots, 𝜏_{𝑘} \in perm 𝑚$ such that $𝜎 = 𝜏_{1} \circ \dots \circ 𝜏_{𝑘}$ .

Lemma: Parity of a Permutation

Let $𝜎 \in perm 𝑚$ be a permutation. The number of transpositions in any decomposition of $𝜎$ is either always even or always odd.

Definition 74: Sign of a Permutation

Let $𝜎 \in perm 𝑚$ be a permutation. The sign of $𝜎$ , denoted $sgn (𝜎)$ , is defined as:

sgn (𝜎) = {(- 1)}^{𝑘},

where $𝑘$ is the number of transpositions in any decomposition of $𝜎$ .

That is, the sign of a permutation is $1$ if it can be expressed as an even number of transpositions, and $- 1$ if it can be expressed as an odd number of transpositions.

Terminology

A permutation $𝜎$ is even if $sgn (𝜎) = 1$ , and odd if $sgn (𝜎) = - 1$ .

Corollary 37

For any $𝜎 \in perm 𝑚$ and transposition $𝜏$ , $sgn (𝜏 \circ 𝜎) = - sgn (𝜎)$ .

Theorem 63: Leibniz Formula for Determinants

For any matrix $𝐴 \in 𝐹^{𝑛 \times 𝑛}$ , the unique determinant of $𝐴$ is given by:

\det (𝐴) = \sum_{𝜎 \in perm 𝑛} sgn (𝜎) \prod_{𝑖 = 1}^{𝑛} 𝑎_{𝑖, 𝜎 (𝑖)} .

Proof: Leibniz Formula for Determinants.

From the uniqueness part of Theorem 59, the determinant of a matrix is unique. We will now show that the Leibniz formula satisfies the properties of a determinant function.

For some $𝐴 = (\begin{matrix} 𝐚_{1} & 𝐚_{2} & \dots & 𝐚_{𝑛} \end{matrix}) \in 𝐹^{𝑛 \times 𝑛}$ , let $\det (𝐴) = \det (𝐚_{1}, 𝐚_{2}, \dots, 𝐚_{𝑛})$ as in Definition 69.

Multilinearity.
Alternation.
Normalization.

Example.

Let $𝑀 = (\begin{matrix} 𝑎 & 𝑏 & 𝑐 \\ 𝑑 & 𝑒 & 𝑓 \\ 𝑔 & ℎ & 𝑖 \end{matrix}) \in 𝐹^{3 \times 3}$ . Realize that in terms of their standard sequences:

perm 3 = {(1, 2, 3), (1, 3, 2), (2, 1, 3), (2, 3, 1), (3, 1, 2), (3, 2, 1)} .

The permutations $(1, 2, 3)$ , $(2, 3, 1)$ , and $(3, 1, 2)$ have even parity (require either $0$ or $2$ transpositions), so they have a positive sign. The permutations $(1, 3, 2)$ , $(2, 1, 3)$ , and $(3, 2, 1)$ have odd parity (require $1$ transposition), so they have a negative sign.

Thus, by the Leibniz formula, we can compute the determinant of $𝑀$ as follows:

\begin{matrix} \det (𝑀) & = \sum_{𝜎 \in perm 3} sgn (𝜎) \prod_{𝑖 = 1}^{3} 𝑚_{𝑖, 𝜎 (𝑖)} \\ = 𝑚_{11} 𝑚_{22} 𝑚_{33} + 𝑚_{12} 𝑚_{23} 𝑚_{31} + 𝑚_{13} 𝑚_{21} 𝑚_{32} - 𝑚_{13} 𝑚_{22} 𝑚_{31} - 𝑚_{12} 𝑚_{21} 𝑚_{33} - 𝑚_{11} 𝑚_{23} 𝑚_{32} \\ = 𝑎 𝑒 𝑖 + 𝑏 𝑓 𝑔 + 𝑐 𝑑 ℎ - 𝑐 𝑒 𝑔 - 𝑏 𝑑 𝑖 - 𝑎 𝑓 ℎ . \end{matrix}

Remark

The Leibniz formula requires $𝑛 \cdot 𝑛!$ matrix accesses to compute the determinant of an $𝑛 \times 𝑛$ matrix, which is not efficient for large $𝑛$ .

Proof: Theorem 59 (Existence).

We will now complete the proof of Theorem 59 by showing that the Leibniz formula can be extended to determinant functions on arbitrary vector spaces.

Define a determinant function $𝐷 : 𝑉^{𝑛} \to 𝐹$ on an arbitrary $𝑛$ -dimensional vector field $𝑉$ with respect to a fixed ordered basis $ℬ = {𝐛_{1}, \dots, 𝐛_{𝑛}}$ as follows:

𝐷 (𝐯_{1}, \dots, 𝐯_{𝑛}) = \det (𝐴),

where $𝐴 \in 𝐹^{𝑛 \times 𝑛}$ is the matrix whose $𝑖$ ^th column is the coordinate vector of $𝐯_{𝑖}$ with respect to the basis $ℬ$ . That is,

𝐴 = (\begin{matrix} {[𝐯_{1}]}_{ℬ} & {[𝐯_{2}]}_{ℬ} & \dots & {[𝐯_{𝑛}]}_{ℬ} \end{matrix}) .

By Theorem 63, $\det (𝐴)$ is well-defined, and thus $𝐷 (𝐯_{1}, \dots, 𝐯_{𝑛})$ must also be well-defined. ∎

Exercises

Exercise 72.

Let

𝑛 \in ℤ^{+}

. Show that

\dim (𝑉^{(𝑛)}) = {(\dim 𝑉)}^{𝑛}