Question

How can you conclude a function with given properties is uniquely determined?

Determinant(function on square matrix) is a function with 3 properties.
1. D(I) = 1
2. You get to multiply by -1 when you switch two rows.
3. D is linear on each row.

I can follow these 3 properties lead to the Big formular.
But does it guarantee there's no another function different from the big formular?

How do you know that there's no other function(apart from big formular) that also satisfies the 3 properties?

Answer 1

You can prove it by induction on n. The determinant is a function from the space of all n x n matrices to the real numbers. The basic step is immediate for n = 1. Assume it is true for n - 1. Then expand by co-factors from the first row of the n x n matrix. You use property (2) to determine the sign pattern +1, -1, +1, .... You use property (3) to be able to expand by one row only. It is the basic induction step. Since the proof is constructive, there can be no other such function.

Answer 2

Mustangtiger, if it is true for the basis step of n=1, why do you do n-1 and not n+1? Just small details, thanks.

Answer 3

mustangtiger, I don't quite get how it works. The essential step I don't get is, "Since the proof is constructive, there can be no other such function".
How do I know for sure other way of construction wouldn't lead to another function?

Answer 4

jlengyel: I just as well have assumed it true for n and then proved it for n + 1. I would expand by the 1st column, knowing the value of all the co-factors, which would be the determinants of n x n matrices

pcompassion: The determinant of an n x n matrix is the volume of a parallelepiped in R^n. If I tell you all the edges and the angles between the edges, then there is only one volume. You could tell me the area of a square is f(x) = x^2, or the area of a square is f(x) (1/8) * (2x)^2 * 2, but these are the same function. A function is a rule that maps from one set to another. The rules are the same--so is the function.

Answer 5

mustangtiger,

It may sound dumb but a parallel question is following.
if x-3 = 0 then x=3 ... what does this mean?
it says if there's an x such that x-3 = 0, it should be unique value 3.(uniqueness)
I can see 3 is indeed a solution to x-3=0 by substituting x by 3.(existence)

If I may apply the same reasoning to the question whether the determinant is unique,
I can say the following..

if a function satisfies 3 properties then the function is expressed in co-factor expansion.
If I can prove the above statement, I've already shown the uniqueness of the function.

Did we really have to apply the induction proof to show that determinant function is unique?

Answer 6

pcompassion. Consider to functions f(x) and g(x). They must be defined on the same domain X, and they must have the same range Y. Then f(.) = g(.) if and only f(x) = g(x) for every x in X. There is only one function f: R ==> R that has f(x) = 3 for every possible x.