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?

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    $\begingroup$ Your suspicion is a little presumptuous! $\endgroup$ – copper.hat Jan 6 '15 at 17:34
  • $\begingroup$ There's a nice definition when the field is $\mathbb R$, you can define it as the signed volume of the generalized parallelogram given by the vectors. It's not hard to prove it matches the definition by examining the linear transformations corresponding to scaling and adding a multiple of one row to some other row (if you already know how they act on the permutational determinant). $\endgroup$ – user2345215 Jan 6 '15 at 17:36
  • $\begingroup$ I know there are many definitions of the determinant, and I know it is the volume of a parallelepiped. But that's not what I was asking. $\endgroup$ – Kamil Jan 6 '15 at 17:41
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    $\begingroup$ The expression follows from three conditions (i) the determinant is a multilinear function of the columns (or rows), (ii) it is alternating (that is, it switches sign if you switch two columns (rows) and (iii) $\det I = 1$. (See math.stackexchange.com/a/888762/27978.) These are reasonable conditions for a signed volume. $\endgroup$ – copper.hat Jan 6 '15 at 17:45
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    $\begingroup$ It is reasonable to ask how a particular definition came about, in the case $\det$ has a very geometric interpretation and can be derived from fairly elementary requirements. $\endgroup$ – copper.hat Jan 6 '15 at 17:48

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.

  • $\begingroup$ Sorry, I don't know what you mean with $x_{1,\delta(1)}, ...$ etc. This is exactly what I was asking. Can u explain how you arrive at that definition? $\endgroup$ – Kamil Jan 6 '15 at 18:03
  • $\begingroup$ @Kamil suppose $\sigma(1)=2$, $\sigma(2)=3$, $\sigma(3)=1$, or more succinctly $\sigma=231$. Then $x_{1, \sigma(1)} x_{2, \sigma(2)} x_{3, \sigma(3)}=x_{1,2} x_{2,3} x_{3,1}$. Thus you are multiplying the second entry of the first row, the third entry of the second row, and the first entry of the third row. $\endgroup$ – Matt Samuel Jan 6 '15 at 18:08
  • $\begingroup$ I think I understand it better now. Let's say we would have the set $(1,2,3,4)$, and we wanted to know the permutations and their sign so that we could compute the determinant of a $4 \times 4$ matrix. Is there any way you can compute the permutations in a fast manner? $\endgroup$ – Kamil Jan 6 '15 at 18:39
  • $\begingroup$ @Kamil no. While theoretically interesting, this is an absurdly inefficient way to compute the determinant. The most efficient way I can think of that you can do by hand is to remember the permutations of length 3, pick the first number and get six permutations by putting the other 3 numbers in order of the permutations if length 3. Thus for 1 we would have $1234,1324,1243,1423,1342,1432$, then you can compute the number of inversions. This will not be a fun calculation, but it's doable. $\endgroup$ – Matt Samuel Jan 6 '15 at 18:47

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.

  • $\begingroup$ Maybe I expressed myself in a wrong fashion. I didn't mean how the mathematicians arrived at this definition (historically), but how I can work with it/interpret it in a practical way. For example if I have a $3 \times 3$-matrix, how does one compute the determinant using the given definition above? $\endgroup$ – Kamil Jan 6 '15 at 18:05
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    $\begingroup$ Practically, to find the determinant of an nxn matrix, you could write down all permutations in $S_n$ and then find their sign, multiply the elements of the matrix corresponding to that permutation together, etc. This is, however, an extremely inefficient way to find the determinant. A much faster way is to row-reduce the matrix to upper (or lower) triangular form and keep track of how many times you had to switch rows of the matrix. If there are $k$ such switches, then the determinant is the product of the diagonal elements of this reduced matrix, times $(-1)^k$. $\endgroup$ – JotThisDown Jan 6 '15 at 19:33

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