I am looking for a non-permutational definition of determinant. The definition should have these properties:

1: Calculational power (easily applicable, it cold be used for practical calculations).

2: It should hold the usual properties (otherwise it wouldn't be a determinant, right!)

3: No permutation, No permutation, please no permutations.

I would also appreciate if you apply the definition and calculate the determinant of a $4×4$ matrix.

For all I care the definition could be from an obsolete parchment, but it needs to have those three properties.

Appreciate all the help!

  • 1
    Alternating scalar valued tensor with value 1 at the point $(e_1,...,e_n)$. It is **equivalent** to the permutation definition, so there is no escaping it. – copper.hat Jan 22 '19 at 21:38
  • @DietrichBurde I want one like Liebniz' notation, but without the permutation. – Bertrand Wittgenstein's Ghost Jan 22 '19 at 21:38
  • 1
    You cannot escape the permutations. – copper.hat Jan 22 '19 at 21:38
  • 1
    Then take the Leibniz formula as in my answer. The permutations are not directly visible, but still in the background, like copper.hat says. – Dietrich Burde Jan 22 '19 at 21:39
  • @copper.hat Nooo! may be its possible, let's see. – Bertrand Wittgenstein's Ghost Jan 22 '19 at 21:41
  • 1
    The determinant of a triangular matrix is the product of the diagonal coefficients. For the general matrix, perform Gauss-Jordan reduction ( https://en.m.wikipedia.org/wiki/Gaussian_elimination ) till you get a triangular matrix, the determinant of each step being easy to compute. You may also want to look at https://math.stackexchange.com/questions/668/whats-an-intuitive-way-to-think-about-the-determinant . – Aphelli Jan 22 '19 at 21:45
  • @Mindlack Hi, I know how to calculate determinants, but I was looking for a definition which has the positive aspect of being calculationally expressive. – Bertrand Wittgenstein's Ghost Jan 22 '19 at 21:57
  • 2
    @BertrandWittgenstein'sGhost: At some level, permutations are the essence of the determinant... Being multlinear means there are potentially $2^n$ computations, by being alternating it reduces to $n!$. A smaller number means you are dealing with a special case (Hessenberg, triangular, etc.) – copper.hat Jan 22 '19 at 22:04
  • 1
    ...there is no escape. – copper.hat Jan 22 '19 at 22:07
  • @copper.hat Thanks for that input, but it has to be proven one way or the other. We can't just assume permutation are necessary. If you thinks so, then prove it. Tbh, I am not sure if it is necessary or not. – Bertrand Wittgenstein's Ghost Jan 22 '19 at 22:41
  • 1
    Let $A=P^{-1} LU$ be the partial pivot LU factorisation of $A$, then define $\det A = {1 \over \det P} \det L \det U$. Since $L,U$ are triangular, their determinants are just the product of the diagonal elements, and $\det P$ is just $(-1)^k$, where $k$ is the number of row exchanges performed while pivoting. This satisfies 1,3, but showing 2 would be a lot of work. – copper.hat Jan 23 '19 at 00:00

1 Answers1


For the $3\times 3$-determinant we can use the Rule of Sarrus. Then the $4\times 4$ determinant reduces to the $3\times 3$ determinant, because of $${\begin{vmatrix}a&b&c&d\\e&f&g&h\\i&j&k&l\\m&n&o&p\end{vmatrix}}=a\,{\begin{vmatrix}f&g&h\\j&k&l\\n&o&p\end{vmatrix}}-b\,{\begin{vmatrix}e&g&h\\i&k&l\\m&o&p\end{vmatrix}}+c\,{\begin{vmatrix}e&f&h\\i&j&l\\m&n&p\end{vmatrix}}-d\,{\begin{vmatrix}e&f&g\\i&j&k\\m&n&o\end{vmatrix}}.$$

Dietrich Burde
  • 117,315
  • 7
  • 72
  • 137
  • Thanks for that, but it is not general. I need a general definition. The $4×4$ calculation was just so I can see how the def. Works when applied. – Bertrand Wittgenstein's Ghost Jan 22 '19 at 21:40
  • The general definition is, as already said, by the Weiserstrass axioms. From there $1\times 1$, $2\times 2$, $3\times 3$ are obvious. Then we also have $4\times 4$ from the above formula. – Dietrich Burde Jan 23 '19 at 09:05