Htydjtk

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!

The only way I can think of is using Weierstrass' axioms. He assumed that there is a function $$det: mathrm{M}(ntimes n;; F) rightarrow F, ; A mapsto det A $$ from the set of all $n times n$ Matrices with entries from a field $F$ to the field $F$ such that the following three Axioms hold:

$(mathrm{D}1)$ $det$ is linear in each row $a_i$:

If $a_i = a_i' +a_i''$, then $$ det begin{bmatrix}vdots\ a_i \ vdots end{bmatrix} =detbegin{bmatrix}vdots\ a_i' \ vdots end{bmatrix}+det begin{bmatrix}vdots\ a_i'' \ vdots end{bmatrix} $$

If $a_i = lambda a_i'$, then $$ det begin{bmatrix}vdots\ a_i \ vdots end{bmatrix} =lambda cdotdetbegin{bmatrix}vdots\ a_i' \ vdots end{bmatrix} $$

where the vertical dots indicate that the rows $a_1, a_2, dots, a_{i-1}, a_{i+1},dots, a_n$ remain unchanged.

$(mathrm{D}2)$ $det$ is alternating:

If $A$ has two rows $a_i, a_j$ with $i neq j$ and $a_i = a_j$, then $$det A = 0$$

$(mathrm{D}3)$ $det$ is normalized:

$$det I_n = 1 $$

Using only these three properties you can find all other properties of the determinant.

However one problem remains. Is there a function that fulfills all of these three axioms? And if there is, how unique is it? We can show that it is in fact unique, i.e. there can't be two different functions that both fulfill the axioms. Can we show that there exists such a function? Most definitely! How? By using the Leibniz formula with permutations. Luckily, this is rather technical and in practice you will rarely have to use the Leibniz formula when you just use the axioms stated above and its immediate corollaries.

Here are a few important and handy corollaries:

$$det (lambda cdot A) = lambda^n cdot det A tag{D4}$$

If you switch two rows of a matrix $A$ to get a matrix $B$, then $$ det A = -det B tag{D5}$$

If you add $lambda a_i$ to a row $a_j$ (where $a_i, a_j$ are rows of $A$ with $j neq i$) to get a matrix $B$, then $$det A = det B tag{D6}$$

And most importantly:

$$det begin{bmatrix}lambda_1 &dots&\ &ddots&vdots\0&&lambda_nend{bmatrix} = lambda_1 cdot dots cdot lambda_n tag{D7}$$

Because of $mathrm{D}5$ and $mathrm{D}6$, you can always get to $mathrm{D}7$. So you in practice you can avoid permutations by forming for example a $4times4$-matrix into row echelon form, and then multiply all diagonal entries to get the determinant. If you have switch two rows an even-numbered amount of times, then the sign of the determinant is positive. If you have switch two rows an odd-numbered amount of times, then the sign of the determiant is negative. Sometimes it is also nice to use the LaPlace-Expansion.

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