Higher-dimensional determinants

Matrix calculus: Determinants

Higher-dimensional determinants

The determinant of an $(n\times n)$ -matrix is a number. This number depends on the matrix, and in particular on the rows of the matrix. The dependence of the rows is the basis for the definition of a so-called determinantal function.

Determinantal function

A determinantal function on $\mathbb{R}^n$ is a function $D$ that assigns a number to each $n$ -tuple of vectors $\vec{a}_1,\ldots ,\vec{a}_n$ such that the following properties are satisfied:

Multilinearity: for $i=1,\ldots,n$ we have $\begin{array}{l} D(\vec{a}_1, \ldots , \vec{a}_{i-1}, \sum_{k=1}^m\beta_k\vec{b}_k,\vec{a}_{i+1}, \ldots ,\vec{a}_n) =\sum_{k=1}^m\beta_k D(\vec{a}_1, \ldots , \vec{a}_{i-1},\vec{b}_k,\vec{a}_{i+1},\ldots ,\vec{a}_n) \end{array}$
Antisymmetry: when two vectors of the arguments are interchanged, the value of the determinantal function turns into its negative.
Normalization: $D(\vec{e}_1,\vec{e}_2,\ldots ,\vec{e}_n)=1$ .

The multilinearity says that, for each $i$ , the function $D$ is linear in the $i$ -th argument (while all other arguments are kept constant). Therefore, the multinearity states that $D$ is linear in each argument.

The above definition depends on $n$ . Moreover, it is not apparent from the definition that, for each $n$ , there is indeed a determinantal function. This is the case, as will be shown below.

The antisymmetry may also be formulated as $D(\vec{a}_1, \ldots , \vec{a}_n) = 0$ when two arguments are equal (that is, when $\vec{a}_i=\vec{a}_j$ for mutually different $i$ and $j$ ). See Determinantal functions disappear on dependent vectors below.

Even without normalization, multilinear antisymmetric functions can be characterized, as will become clear from the last part of theorem Characterization of the determinant below.

Argument as a matrix

The function $D$ can also be seen as a function $D(A)$ having as an argument the $(n\times n)$ -matrix $A$ whose $i$ -th row is $\vec{a}_i$ .

Let $D(\vec{x},\vec{y}) = x_1y_2-x_2y_1$ for $\vec{x} = \rv{x_1,x_2}$ and $\vec{y} = \rv{y_1,y_2}$ in $\mathbb{R}^2$ . Then $D$ is a determinantal function on $\mathbb{R}^2$ :

Multilinearity: in each term of the sum in the definition, a coordinate of each argument occurs once.
Antisymmetry: $D(\vec{x},\vec{y}) = x_1y_2-x_2y_1= -(y_1x_2-y_2x_1)=-D(\vec{y},\vec{x})$
Normalization: $D(\vec{e}_1,\vec{e}_2)=1\cdot 1 - 0\cdot 0 =1$ .

If we write $\vec{x} = \rv{a,b}$ and $\vec{y} = \rv{c,d}$ , it becomes clear that this definition corresponds to the determinant of the matrix $\matrix{a&b\\ c &d}$ from an earlier definition.

Dimension 3

Let $D(\vec{x},\vec{y},\vec{z}) = \dotprod{\vec{x}}{(\vec{y}\times \vec{z})}$ for $\vec{x} = \rv{x_1,x_2,x_3}$ , $\vec{y} = \rv{y_1,y_2,y_3}$ , and $\vec{z} = \rv{z_1,z_2,z_3}$ in $\mathbb{R}^3$ , where $\dotprod{}{}$ represents the inner product and $\times$ the cross product. Then $D$ is a determinantal function on $\mathbb{R}^3$ :

Multilinearity: each of the vectors $\vec{x}$ , $\vec{y}$ , $\vec{z}$ occurs exactly once, and both the dot product and the cross product are linear in each of their arguments.
Antisymmetry: We show that if two of the three arguments are equal, then $D(\vec{x},\vec{y},\vec{z})=0$ follows. The cross product is antisymmetric, so $D(\vec{x},\vec{y},\vec{y})= \dotprod{\vec{x}}{(\vec{y}\times \vec{y})} =\dotprod{\vec{x}}{\vec{0}}= 0$ . Because $\vec{y}\times \vec{z}$ is perpendicular to both $\vec{y}$ and $\vec{z}$ , we have both $D(\vec{y},\vec{y},\vec{z}) = \dotprod{\vec{y}}{(\vec{y}\times \vec{z})} = 0$ and $D(\vec{z},\vec{y},\vec{z}) = \dotprod{\vec{z}}{(\vec{y}\times \vec{z})} = 0$ .
Normalization: $D(\vec{e}_1,\vec{e}_2,\vec{e}_3)=\dotprod{\vec{e}_1}{(\vec{e}_2\times \vec{e}_3)}=\dotprod{\vec{e}_1}{\vec{e}_1}=1$ .

We will prove that, for every $n$ , there is indeed exactly one determinantal function and we will actually provide a formula for it; it will be called the determinant. First we draw some conclusions from the definition.

If $D$ is a determinantal function on $\mathbb{R}^n$ , then it has the following properties.

$D(\vec{a}_1,\ldots, \vec{a} ,\ldots, \vec{a} ,\ldots ,\vec{a}_n)=0$ : if two vectors among $\vec{a}_1,\ldots ,\vec{a}_n$ are the same, then the determinant is equal to $0$ .
$D (\vec{a}_1,\ldots ,\vec{a}_n)=0$ if the system $\vec{a}_1,\ldots ,\vec{a}_n$ is linearly dependent.

1. By interchanging the two arguments that are equal to the vector $\vec{a}$ , the determinant transforms into its negative, but at the same time, the arguments do not change, so the determinant remains the same. This is only possible if the determinant is $0$ .

2. Assume for the sake of convenience that $\vec{a}_1=\alpha_2\vec{a}_2+\cdots +\alpha_n\vec{a}_n$ Then $\begin{array}{rl}D(\vec{a}_1,\vec{a}_2,\ldots ,\vec{a}_n)& =D(\sum_{k=2}^n\alpha_k\vec{a}_k,\vec{a}_2,\ldots ,\vec{a}_n)\\& =\sum_{k=2}^n\alpha_k D(\vec{a}_k ,\vec{a}_2,\ldots ,\vec{a}_n)\\&=0\end{array}$ because of the first part.

If $D$ is multilinear, then the antisymmetry may also be formulated as $D(\vec{a}_1, \ldots , \vec{a}_n) = 0$ when two arguments are equal (that is, if $\vec{a}_i=\vec{a}_j$ for mutually different $i$ and $j$ ).

In item 1 of the proof we already saw that the assumption that $D$ is antisymmetric implies that $D(\vec{a}_1, \ldots , \vec{a}_{n}) = 0$ when two arguments are equal. Below, we establish the converse implication that says that it follows from the assumption that $D(\vec{a}_1, \ldots , \vec{a}_{n}) = 0$ whenever $\vec{a}_i =\vec{a}_j$ , that $D$ is antisymmetric in the arguments $i$ and $j$ .

We focus on the first two arguments and write $F(\vec{a}_1,\vec{a}_2) = D(\vec{a}_1,\vec{a}_2,\vec{a}_3,\ldots,\vec{a}_n)$ . Thanks to the multilinearity we have

$F(\vec{a}_1+\vec{a}_2,\vec{a}_1+\vec{a}_2) = F(\vec{a}_1,\vec{a}_2)+F(\vec{a}_2,\vec{a}_1)+F(\vec{a}_1,\vec{a}_1)+F(\vec{a}_2,\vec{a}_2)$

If $D(\vec{a}_1, \ldots , \vec{a}_n) = 0$ whenever two arguments are the same, then $F(\vec{x}, \vec{x}) = 0$ for all vectors $\vec{x}$ . Therefore, the above equality becomes $0= F(\vec{a}_1,\vec{a}_2)+F(\vec{a}_2,\vec{a}_1)$ from which antisymmetry of $F$ follows, and so also antisymmetry of $D$ in the first two arguments.

Antisymmetry for every other pair of arguments can be shown in the same way.

For the following characterization of determinantal functions we use some facts about permutations.

Consider the function $\det$ on $\mathbb{R}^n$ defined by
$\det(\vec{a}_1,\ldots ,\vec{a}_n)=\sum_{\sigma}\text{sg}(\sigma)\cdot a_{1\sigma(1)}\cdots a_{n\sigma(n)}$ where the sum runs over all $n!$ permutations $\sigma$ of $\{1,\ldots ,n\}$ . Here, $\vec{a}_i = \rv{a_{i1},a_{i2},\ldots,a_{in}}$ , so $\det(\vec{a}_1,\ldots ,\vec{a}_n) =\det(A)$ , the determinant of $A$ , the $(n\times n)$ -matrix whose $i$ -th row is equal to $\vec{a}_i$ .

The function $\det$ is a determinantal function.
The function $\det$ is the only determinantal function on $\mathbb{R}^n$ .
If $E$ is a function of $n$ arguments from $\mathbb{R}^n$ which is multilinear and antisymmetric, then $E(A) = E(I)\cdot \det(A)$ .

Instead of $\det\left(\matrix{a_{11}&\cdots&a_{1n}\\ \vdots &\ddots&\vdots\\ a_{n1}&\cdots&a_{nn}}\right)$ we also write $\left | \begin{array}{ccc}a_{11}&\cdots&a_{1n}\\ \vdots&\ddots&\vdots\\ a_{n1}&\cdots&a_{nn}\end {array} \right |$ . We will refer to the above expression for $\det$ as the sum formula for the determinant.

Take $n=2$ and consider the matrix $A = \matrix{a&b\\ c&d}$ Thus, the rows are $\vec{a}_1=\rv{a,b}$ and $\vec{a}_2=\rv{c,d}$ .

Now, there are only two permutations of $\{1,2\}$ , namely $\rv{1,2}$ and $\rv{2,1}$ . The first has already the correct order; the corresponding sign is thus $1$ , and the second one a transposition in the order becomes $(1,2)$ , so its sign is $-1$ . We thus find
$\det(A) = \left|\,\begin{array}{rr} a & b\\ c & d \end{array}\,\right|=a\,d - b\,c\$ in accordance with the definition given before.

Now $n=3$ . Consider the determinant of $(3\times3)$ -matrix $A$ with $(i,j)$ -element $a_{ij}$ :
$\det(A) = \left|\,\begin{array}{rrr} a_{11} & a_{12} & a_{13}\\ a_{21} & a_{22} & a_{23}\\ a_{31} & a_{32} & a_{33} \end{array}\,\right|$ There are six permutations of $\rv{1,2,3}$ ; those with a plus sign are $\rv{1,2,3}$ , $\rv{2,3,1}$ , and $\rv{3,1,2}$ , and those with a minus sign are $\rv{1,3,2}$ , $\rv{2,1,3}$ , and $\rv{3,2,1}$ . We find
$\left|\,\begin{array}{rrr} a_{11} & a_{12} & a_{13}\\ a_{21} & a_{22} & a_{23}\\ a_{31} & a_{32} & a_{33} \end{array}\,\right|=\begin{array}{l l} & a_{11} a_{22} a_{33} + a_{12} a_{23} a_{31} + a_{13} a_{21} a_{32}\\ &{}-a_{11} a_{23} a_{32} - a_{12} a_{21} a_{33} - a_{13} a_{22} a_{31} \end{array}$ This expression is known as the rule of Sarrus. It is easy to remember; place the first two columns behind the matrix
$\begin{array}{rrrrr} a_{11} & a_{12} & a_{13} & a_{11} & a_{12}\\ a_{21} & a_{22} & a_{23} & a_{21} & a_{22}\\ a_{31} & a_{32} & a_{33} & a_{31} & a_{32} \end{array}$ and now include the three terms on the main diagonal or parallel to it with a plus sign and the terms on the secondary diagonal (that is, $a_{13},a_{22},a_{31}$ ) or parallel to it with a minus sign.

In 2-Dimensional determinants we saw that $(2\times 2)$ -determinants have an interpretation as surface area. Something similar applies to $(n\times n)$ -determinants: these measure volumes of parallellopipida in $\mathbb{R}^n$ .

PermutationsExpanding the determinant as a sum of terms involving $D(\vec{e}_{j_1} ,\ldots ,\vec{e}_{j_n})$ , as is done in the proof, each list of indices $\rv{j_1,\ldots ,j_n}$ appears in this sum. However, many terms vanish. After all, if two of the indices are the same, then at the right-hand side there is a determinant with two equal vectors, and so it is $0$ . Consequently, if $D$ exists, then $D(\vec{a}_1,\ldots ,\vec{a}_n)=\sum_{\text{all}\ j_i\ \text{ distinct}}\ a_{1j_1}\cdots a_{nj_n} D(\vec{e}_{j_1} ,\ldots ,\vec{e}_{j_n})$ Which indices $\rv{j_1,\ldots ,j_n}$ appear in this sum, and how many terms are there in the sum? From the fact that all of the numbers $j_1,\ldots ,j_n$ have to be different, and that each lies between $1$ and $n$ , in the sequence $\rv{j_1,\ldots ,j_n}$ all of the numbers between $1$ and $n$ occur exactly once. Such a list is called a permutation of the numbers $1,\ldots ,n$ . All permutations of $\{1,2,3\}$ , for example, are

$\rv{1,2,3},\, \rv{1,3,2},\, \rv{2,1,3},\, \rv{2,3,1},\, \rv{3,1,2},\, \rv{3,2,1}$ These permutations can be found as follows: we choose an element from $\rv{1,2,3}$ . This can be done in three ways. Thereafter, there are two choices left for the second element. The third element is then fixed. Thus, there are $3\times 2\times 1=3!=6$ possible permutations of $\{1,2,3\}$ .

Uniqueness: We are now able to write down the determinantal function for each $n$ . It is done in the same way as for $(2\times 2)$ -matrices: we will write each row as a linear combination of the standard basis vectors and use the multilinearity to write the determinant as a sum of many determinants of standard basis vectors. For this purpose consider the matrix
$A=\left(\,\begin{array}{cccc} a_{11} & a_{12} & \ldots & a_{1n}\\ a_{21} & a_{22} & \ldots & a_{2n}\\ \vdots & \vdots & \vdots & \vdots\\ a_{n1} & \ldots & \ldots & a_{nn} \end{array}\,\right)$ with rows $\vec{a}_1 ,\ldots ,\vec{a}_n$ . Then, for each $i$ ,
$\vec{a}_i=\sum_{j=1}^n\ a_{ij}\vec{e}_j$ and so $\begin{array}{ll} D\ (\vec{a}_1 ,\ldots ,\vec{a}_n)\ &=\ D\Big(\sum_{j_1=1}^na_{1j_1}\vec{e}_{j_1},\ldots ,\sum_{j_n=1}^n a_{nj_n}\vec{e}_{j_n}\Big)\\ &=\ \sum_{j_1=1}^n\ldots \sum_{j_n=1}^na_{1j_1}\ldots a_{nj_n}D(\vec{e}_{j_1},\ldots ,\vec{e}_{j_n}) \end{array}$ This is a sum of a lot of terms. There are $n$ summation indices, each with $n$ possible values, so the number of terms is $n^n$ . For a value as low as $n=8$ we have $16\,777\,216$ terms.

Since terms in which two indices $j_i$ and $j_k$ are equal constribute zero, we we can limit ourselves to permutations $\rv{j_1,\ldots ,j_n}$ of $\{1,2,\ldots,n\}$ . We count these permutations as follows. For the first element, there are $n$ possibilities, then for the second element there are $n-1$ possibilities, for the third element $n-2$ , $\ldots$ , for the penultimate $2$ , and finally, the last element is fixed. Thus, there are $n\cdot(n-1)\cdots 2\cdot 1=n!$ permutations of $\{1,\ldots ,n\}$ .

The sum for $D$ thus has $n!$ terms. That is a lot less than $n^n$ , but for $n=8$ there still are $8!=40\,320$ terms. If some terms $D(\vec{e}_{j_1},\ldots ,\vec{e}_{j_n})$ of the sum over all permutations $\rv{j_1,\ldots ,j_n}$ would be equal to $0$ , then the sum would consist of fewer terms. However, this is not the case. Because $\rv{j_1,\ldots, j_n}$ contains all numbers $1,\ldots ,n$ , we can obtain the order $1,\ldots ,n$ by repeated transpositions. Each transposition involves a factor $-1$ . If the number of permutations is even, then $D(\vec{e}_{j_1},\ldots ,\vec{e}_{j_n})$ equals $1$ and otherwise it is equal to $-1$ . Thus, we find:

If $D$ exists, then
$D(\vec{a}_1,\ldots ,\vec{a}_n)=\sum_{j_1,\ldots ,j_n}\pm a_{1j_1}\ldots a_{nj_n}$ where the sum runs over all $n!$ permutations of $\rv{1,\ldots ,n}$ and the sign must be $1$ if the permutation of the sequence $\rv{1,\ldots ,n}$ involves an even number of transpositions and $-1$ otherwise.

det is a determinantal function: To prove this we verify that $\det$ meets the three requirements for a determinantal function.

Multilinearity: Let $i$ be one of the numbers $1,\ldots,n$ . We want to show that $\det$ is linear in the $i$ -th argument $\vec{a}_i$ . Because $\det(\vec{a}_1,\ldots ,\vec{a}_n)$ is a sum of terms of the form $\text{sg}(\sigma) a_{1\sigma(1)}\cdots a_{n\sigma(n)}$ , it suffices to verify that each of these terms is linear in $\vec{a}_i$ . That is indeed the case, because only the factor $a_{i\sigma(i)}$ comes from $\vec{a}_i$ .

Antisymmetry: upon an interchange of two vectors in the arguments, the value of the determinant changes to its negative. If we interchange the arguments $i$ and $j$ , we get the same expression, with $a_{i\sigma(i)}$ and $a_{j\sigma(j)}$ in each term replaced by $a_{j\sigma(i)}$ and $a_{i\sigma(j)}$ , respectively. Because it only involves a transposition, we may suppose $i\lt j$ . So if we let $\tau$ be the composition of $\sigma$ and the transposition of $i$ and $j$ , we get

$\begin{array}{rcl}&&a_{1\sigma(1)}\cdots a_{i-1\sigma(i-1)}a_{j\sigma(i)}a_{i+1\sigma(i+1)}\cdots a_{j-1\sigma(j-1)}a_{i\sigma(j)}a_{j+1\sigma(j+1)}\cdots a_{n\sigma(n)}\\&&\phantom{xx}=a_{1\tau(1)}\cdots a_{i-1\tau(i-1)}a_{j\tau(j)}a_{i+1\tau(i+1)}\cdots a_{j-1\tau(j-1)}a_{i\tau(i)}a_{j+1\tau(j+1)}\cdots a_{n\tau(n)}\\ &&\phantom{xx}= a_{1\tau(1)}\cdots a_{n\tau(n)}\end{array}$ For clarity, we treat these equalities again, but with more words:

$\begin{array}{lcc}&{\left.a_{1\sigma(1)}\cdots a_{n\sigma(n)}\right.}&\qquad\left({\text{with }a_{i\sigma(i)} \text{ and }a_{j\sigma(j)}\text{ replaced by }a_{j\sigma(i)}\text{ and }a_{i\sigma(j)} }\right)\\ &\phantom{1234}={\left.a_{1\tau(1)}\cdots a_{n\tau(n)}\right.}&\qquad\left({\text{with }a_{i\tau(i)}\text{ and }a_{j\tau(j)}\text{ interchanged} }\right)\\ &\phantom{1234}=a_{1\tau(1)}\cdots a_{n\tau(n)}&\qquad\left(\text{in a product of scalars the sequence does not matter}\right)\end{array}$

From $\tau\equiv\sigma\,(i,j)$ it follows that $\sigma$ has one transposition less than $\tau$ such that $\text{sg}(\sigma)=-\text{sg}(\tau)$ . The sum over all permutations $\sigma$ equals the sum over all permutations $\tau$ . With all these results we find:

$\begin{array}{rcll}\det\overbrace{(\vec{a}_1,\ldots ,\vec{a}_n)}^{\vec{a}_i\quad\leftrightarrow\quad\vec{a}_j}&=&\sum_{\sigma}\text{sg}(\sigma)\cdot \overbrace{a_{1\sigma(1)}\cdots a_{n\sigma(n)}}^{\begin{array}{rcl}a_{i\sigma(i)}&\leftrightarrow&a_{j\sigma(i)}\\a_{j\sigma(j)}&\leftrightarrow&a_{i\sigma(j)}\end{array}}&\color{blue}{\text{definition }\det}\\&=&\sum_{\sigma}\text{sg}(\sigma)\cdot a_{1\tau(1)}\cdots a_{n\tau(n)}&\color{blue}{\text{result from above}}\\&=&\sum_{\sigma}-\text{sg}(\tau)\cdot a_{1\tau(1)}\cdots a_{n\tau(n)}&\color{blue}{\text{sg}(\sigma)=-\text{sg}(\tau)}\\&=&\sum_{\tau}-\text{sg}(\tau)\cdot a_{1\tau(1)}\cdots a_{n\tau(n)}&\color{blue}{\sum_{\sigma}=\sum_{\tau}}\\&=&-\sum_{\tau}\text{sg}(\tau)\cdot a_{1\tau(1)}\cdots a_{n\tau(n)}&\color{blue}{\text{minus sign moved forward}}\\&=&-\det(\vec{a}_1,\ldots ,\vec{a}_n)&\color{blue}{\text{definition }\det}\end{array}$

Normalization: $D(\vec{e}_1,\vec{e}_2,\ldots ,\vec{e}_n)=1$ .

A standard basis vector $\vec{e}_i$ of $\mathbb{R}^n$ consists of $n-1$ zeroes and one $1$ at position $i$ : $\vec{e}_i=\rv{e_{i1},e_{i2},\ldots,e_{in}}=[0,0,\ldots,0,\underbrace{1}_{\text{position }i},0,\ldots,0,0]$ This means: $e_{ij}=\left\{\begin{array}{rl}1&\text{for }i=j\\0&\text{for }i\neq j\end{array}\right.$ Below, in the sum over all permutations $\sigma$ the only term unequal to zero is therefore $e_{11}\cdots e_{nn}$ belonging to $\sigma=\rv{1,2,\ldots,n}$ : $\begin{array}{rcl}\det(\vec{e}_1,\vec{e}_2,\ldots ,\vec{e}_n) &=&\sum_{\sigma}\text{sg}(\sigma)\cdot e_{1\sigma(1)}\cdots e_{n\sigma(n)}\\ &=&\text{sg}(\rv{1,\ldots,n})\cdot e_{11}\cdots e_{nn}\\&=& 1\cdot 1\cdots 1\\&=& 1\end{array}$

Last statement: If $E(I) = 0$ , then each term in the expansion of $E(A)$ is zero because it contains a factor $E(I)$ . Then $E(A) = 0 = E(I)\cdot \det(A)$ , as required. Suppose, therefore, $E(I)\ne0$ . Then $\frac{E(A)}{E(I)}$ is defined; it is a function that satisfies all three requirements for a determinantal function so because of the uniqueness we have $\frac{E(A)}{E(I)} = \det(A)$ The statement now follows after multiplying both members by $E(I)$ .

The sum formula has $n!$ terms, where $0! = 1$ and $n! = n\cdot (n-1)!$ for $n\gt0$ . This is way too much to be of use even for relatively small $n$ . However, the formula implies results that we will use to calculate determinants.

For what value of $k$ is the determinant of the matrix $A$ below equal to zero?
$A = \matrix{1 & 2 & -5\\ 3 & 7 &-2\\ 6& k& -4}$
Give your answer in the form of an integer or a simplified fraction.

$k =$ $14$

We calculate the determinant of $A$ :
$\begin{array}{rcl}\det(A)&=& a_{11}\cdot a_{22}\cdot a_{33} -a_{11}\cdot a_{23}\cdot a_{32} \\ &&{}-a_{12}\cdot a_{21}\cdot a_{33}+a_{12}\cdot a_{23}\cdot a_{31} \\ &&{}+a_{13}\cdot a_{21}\cdot a_{32}-a_{13}\cdot a_{22}\cdot a_{31} \\ &=& 1\cdot 7\cdot (-4) -1\cdot (-2)\cdot k \\ &&{}-2\cdot 3\cdot (-4)+2\cdot (-2)\cdot 6 \\ &&{}-5\cdot 3\cdot k+5\cdot 7\cdot 6 \\ &=&182-13 k \end{array}$ Thus, the determinant is equal to zero if and only if $k = 14$ .

New example