Definition of Determinant

[Editor: The following conversation on determinants was prompted by questions on the general definition.]

By Kerwin Hui on Thursday, January 10, 2002 - 02:53 pm:

There are two equivalent ways to define a determinant - either a) by volume form:

$A = (a_{ij}) = (r_{1}, r_{2}, \dots, r_{n}) \in M_{n \times n} (F)$ ,

where $r_{i}$ are the rows of $A$ , $F$ is a field, then the determinant function is the unique $D : M_{n \times n} (F) \to F$ satisfying

$D$ is a linear function on the rows of $A$ , i.e.
$D (r_{1}, \dots, α r_{i} + r'_{i}, \dots, r_{n}) = α D (r_{1}, \dots, r_{i}, \dots, r_{n}) + D (r_{1}, \dots, r'_{i}, \dots, r_{n})$
If two adjacent rows of $A$ are equal, i.e. $r_{k} = r_{k + 1}$ for some $k$ , such that $k = 1$ , 2, ..., $n - 1$ , then $D (A) = 0$
$D (I_{n}) = 1$ , where $I_{n}$ is the $n \times n$ identity matrix.

or b) by the direct formula

det (\begin{matrix} a_{11} & \dots & a_{1 n} \\ : & ⋱ & : \\ a_{n 1} & \dots & a_{n n} \end{matrix}) = \sum_{σ \in S_{n}} ε (σ) Π_{i = 1}^{N} a_{i σ (i)}

We can prove that the two definitions are equivalent. Hence, your trick only works for

n = 3

Kerwin

By Yatir Halevi on Thursday, January 10, 2002 - 03:05 pm:

Kerwin, I didn't really understand some of the terminilogy you used, but I do know that I tried it out. and I got the exact same thing for n=4...

Could be that I made an error, but I don't know...

Yatir

By Kerwin Hui on Thursday, January 10, 2002 - 03:30 pm:

Yatir,

By the formula, we see that we can replace 'row' in definition a) with 'columns'. Then, since we have more than 1 column of 1s in the case n > = 4, the determinant you give is zero, whereas my function D does not necessarily give 0, e.g. take x₁ =y₁ =y₂ =x₄ =0, x₂ =x₃ =y₃ =y₄ =1.

Kerwin

By Kerwin Hui on Thursday, January 10, 2002 - 04:21 pm:

Yatir,

Actually I think I better explain a bit about the formula (I believe you understand the volume form definition, right?)

$S_{n}$ is the permutation group of $n$ elements, which we label, WLOG, as $T = {1, 2, \dots, n}$ . A permutation of the elements is a bijection $s : T \to T$ . We can write the permutation as (with a slight abuse of notation)

$σ = (\begin{matrix} 1 \dots n \\ σ (1) \dots σ (n) \end{matrix})$

There isa more compact way of writing the permutation (as a product of disjoint cycles), but for our purpose this will suffice. We define a function $ε : S_{n} \to {- 1, 1}$ by

$ε (σ) = \underset{i < j}{Π} \frac{σ (i) - σ (j)}{i - j}$

There is also a easier way to work out $ε (σ)$ . Writing the permutation in the above form, we now work out, for each $j$ , the number of $σ (i)$ with the property that $σ (i) > σ (j)$ and $i < j$ . Summing over the $j$ 's gives you a number $m$ . We have

$ε (σ) = (- 1)^{m}$ Further, the function $ε$ is a group homomorphism, since we have

$ε (s t) = ε (σ) ε (τ)$ , and

$ε (identity) = 1$

Kerwin

By Yatir Halevi on Thursday, January 10, 2002 - 05:13 pm:

I understand this, but:
take the following matrix:

matrix

and lets add two columns of 1s:
matrix+ones

Lets take the determinant of this square matrix (by adding left-to-right diagonals and subtracting the the right-to-left diagonals) and we get:
det

But now, you function gives the following result to our original matrix (before adding the 1s):

As you can clearly see those two are equal...
So, It could be that I am not getting what you mean...

Yatir

By Kerwin Hui on Thursday, January 10, 2002 - 05:22 pm:

Yatir,

There are 24 elements of S₄ , and this corresponds to 24 different terms in the determinant.

Take your determinant, for example,

s (1)	s (2)	s (3)	s (4)	e (s )	contribution
1	2	3	4	1	ad
1	2	4	3	-1	-ad
1	3	2	4	-1	-af
1	3	4	2	1	ah
1	4	2	3	1	af
1	4	3	2	-1	-ah
...

and you see the determinant is zero.

Kerwin

By Yatir Halevi on Thursday, January 10, 2002 - 05:39 pm:

I fail to understand what you mean...
Are you telling me that you can't take a determinant of a matrix by the diagonals way???

I'm sorry for not getting it...

Yatir

By Kerwin Hui on Thursday, January 10, 2002 - 05:43 pm:

Indeed, the 'wrapping' method only works for 3x3 matrices. (It fails for 2x2 or 1x1 in the obvious way.)

Kerwin

By Yatir Halevi on Thursday, January 10, 2002 - 07:22 pm:

WOW! I had no Idea! You just turned my world up-side down.
So for n> 3, I can divide the matrix into determinants of segments of the matrix, right?

Yatir

By David Loeffler on Saturday, January 12, 2002 - 01:19 pm:

Yes, that method (it's known as Laplace development) works for all matrices; there's a good summary of determinants here .

David