Please explain definition of determinant using permutations?

Question

Many people (in different texts) use the following famous definition of the determinant of a matrix $A$: \begin{align*} \det(A) = \sum_{\tau \in S_n}\operatorname{sgn}(\tau)\,a_{1,\tau(1)}a_{2,\tau(2)} \ldots a_{n,\tau(n)}, \end{align*} where the sum is over all permutations of $n$ elements over the symmetric group. None of them actually explains how one interprets this definition, so this makes me suspicious and think they don't know it either.

This is what I understand so far:

Definition: A permutation $\tau$ of $n$ elements is a bijective function having the set $ \left\{1, 2, ..., n\right\}$ both as its domain and codomain. The number of permutations of $n$ elements, and hence the cardinality of the set $S_n$ is $n!$

So for example, for every integer $i \in \left\{1, 2, ..., n\right\}$ there exists exactly one integer $j \in \left\{1, 2, ..., n\right\}$ for which $\tau(j) = i$.

Permutations can also be represented in matrices, for example if $\tau(1) = 3, \tau(2) = 1, \tau(3) =4, \tau(4) =5, \tau(5) =2$, then \begin{align*} \tau = \begin{pmatrix} 1 & 2 & 3 & 4 & 5 \\ 3 & 1 & 4 & 5 & 2 \end{pmatrix}. \end{align*} Definition: Let $\tau \in S_n$ be a permutation. Then an inversion pair $(i,j)$ of $\tau$ is a pair of positive integers $i, j \in \left\{1, 2, ..., n\right\}$ for which $i < j$ but $\tau(i) > \tau(j)$.

This determines how many elements are 'out of order'. For example if $\tau = \begin{pmatrix} 1 & 2 & 3 \\ 1 & 3 & 2 \end{pmatrix}$, then $\tau$ has one single inversion pair $(2,3)$, since $\tau(2) = 3 > \tau(3) = 2$.

Definition: A transposition, called $t_{ij}$, is the permutation that interchanges $i$ and $j$ while leaving all other integers fixed in place. The numbers of inversions in a transposition is always odd, because one can compute that the number of inversion pairs in $t_{ij}$ is exactly $2(j-1)-1$.

Definition: Let $\tau \in S_n$ be a permutation. Then the sign of $\tau$, denoted by sign$(\tau)$ is defined by \begin{align*} sign(\tau) = (-1)^{\text{# of inversion pairs in}\ \tau} \end{align*} This is $+1$ if the number of inversions is even, and $-1$ if the number is odd. Every transposition is an odd permutation.

This is all clear to me, but can someone explain to me, in an understandable fashion, how one interprets the definition of the determinant on the basis of all this information? That would be greatly appreciated (not only by me, but I think by many others aswell).

For example: what do I make of the $a_{1,\tau(1)}$ etc. in the definition of the determinant, all the way up to $n$? What do they represent?

Answer 1

I'm not sure I understand what you mean by how to "derive" this definition. Typically you simply define the determinant of a matrix this way and then show that it is useful because it has so many nice properties. Historically of course, this definition of the determinant was chosen because it had these nice properties, but these days you typically begin your discussion with the definition and then derive the properties. (The "nice properties" I am referring to here are things like the fact that a matrix is invertible iff its determinant is nonzero, etc.)

Historically, the determinant was first used in a system of linear equations as a measure of whether a unique solution to the system existed. If the determinant of the system was nonzero, then there was a unique solution. When we started doing linear algebra with matrices, this naturally became the determinant of a matrix. Suppose though, that didn't have this definition already but that we wanted to find some function on a matrix that told us whether the matrix was invertible. It turns out that pretty much the only such function is the determinant function as defined in your question.

In other words, there is a unique function, $| \cdot |:M_n(\mathbb{R}) \to \mathbb{R}$ that is linear in the rows of the matrix, zero when the matrix is not invertible, and such that $| I_n | = 1$ (where $I_n$ is the identity matrix). The last condition we require as a normalization. This unique function is called the determinant function, and you can prove that its form is the one given above in terms of permutations. It turns out that this function also has all sorts of other uses, such as giving the volume of the parallelepiped created by the column vectors, etc.

I must confess that I'm not exactly sure what you're looking for, so if this doesn't answer your question then feel free to comment and clarify.

Answer 2

The result is fairly easy to see if you're willing to accept the result that any alternating multilinear $n$-form on $\mathbb{R}^n$ is a scalar multiple of the determinant. A multilinear form $d(x_1,x_2,\ldots,x_n)$ is said to be alternating if it evaluates to 0 whenever two arguments are the same; the determinant is an alternating multilinear form where the arguments are the rows of the matrix. Let $d$ be our permutation form; I will show that it is alternating. Suppose without loss of generality that $x_1=x_2$. Then for every even permutation $\sigma$ there is a term $$x_{1,\sigma(1)} x_{2,\sigma(2)} \cdots x_{n,\sigma(n)}$$ There is a bijection $S_n\to S_n$ given by multiplication on the right by the transposition $(12)$ that induces a bijection between the even and odd permutations. Suppose $\sigma\cdot (12)=\tau$. The term $\tau$ contributes to the sum is $$-x_{1,\sigma(2)} x_{2,\sigma(1)}x_{3,\sigma(3)} \cdots x_{n,\sigma(n)}$$ Since $x_{1,\sigma(1)}= x_{2,\sigma(1)}$ and $x_{1,\sigma(2)}= x_{2,\sigma(2)}$, this exactly cancels the term coming from $\sigma$. Since every term is cancelled by another term, the form evaluates to 0, hence it is alternating and therefore a multiple of the determinant. When we evaluate it at the identity matrix we get 1, therefore it is equal to the determinant.