How is uniqueness about the determinant proved by this theorem?

[Hall]

TL;DR Summary

Uniqueness theorem.

Let me first list the four axioms that a determinant function follows:
1. $d (A_1, \cdots, t_kA_k, \cdots, A_n)=t_kd(A_1, \cdots A_k, \cdots, A_n)$ for any $A_k$ and $t_k$

2. $d(A_1, \cdots A_k + C , \cdots A_n)= d(A_1, \cdots A_k, \cdots A_n) + d(A_1, \cdots C, \cdots A_n)$ for any $A_k$ and $C$.

3. If any two row vectors are equal the determinant function would yield 0.

4 $d (I_1, \cdots, I_n) = 1$

Now, I will put the theorem (I'm putting it in spoiler otherwise the post would become too lengthy and we would need too much scrolling)

Spoiler

Let d be a determinant function satisfying all four axioms for a determinant function of order n, and let f be another function satisfying axiom 1,2 and 3. Then, for every choice of vectors $A_1, A_2, \cdots, A_n$ in n-space we have:
$$
f(A_1, \cdots A_n) = d(A_1, \cdots A_n) \dot f (I_1, \cdots I_n)$$

If f satisfies axiom 4 we have $f(A_1, \cdots, A_n)= d(A_1, \cdots, A_n)$.

Proof: Let
$$
g(A_1, \cdots, A_n) = f(A_1, \cdots A_n) - d(A_1, \cdots A_n) \dot f (I_1, \cdots, I_n)$$

As, f and d satisfy the axioms 1,2 and 3, so g would also satisfy those. And therefore we can have
$$
g(A_1, \cdots A_n) = c g(I_1, \cdots I_n)$$
Taking the matrix A =I
$$
g(I_1, \cdots I_n) = f(I_1, \cdots , I_n) - d(I_1, \cdots I_n) \dot f(I_1, \cdots, I_n)$$
$$1/c g(A_1, \cdots A_n ) = 0$$
$$
g(A_1, \cdots A_n)=0$$

Thus, proving the expression given in the statement.

But what does it prove? Of course, it proves what it wrote, but they christened it as "Uniqueness Theorem", does the name justify what the thing contains? Well, it may but I cannot see, help me seeing it, I implore.

 

[fresh_42]

Summary:: Uniqueness theorem.

Of course, it proves what it wrote, but they christened it as "Uniqueness Theorem", does the name justify what the thing contains? Well, it may but I cannot see, help me seeing it, I implore.

It says that any function $f$ with the properties 1-4 is automatically $f=\det$. There is no second function that does that, hence the determinant is unique among all functions that satisfy 1-4.

If we have only 1-3, then we get uniqueness up to a constant factor, namely $f(I_1,\ldots,I_n).$

It is a typical theorem that connects a specific function (here the determinant) with an at prior arbitrary function, that satisfies certain key properties. These properties are called functional equations. For the exponential function, it is $e(x+y)=e(x)\cdot e(y)$ and $e(0)=1$, for the Gamma function, it is $G(x+1)=x\cdot G(x)\, , \,G(1)=1$ and logarithmic convexity. They all have some central properties which make them unique up to a constant multiple, and a certain point to fix this factor.

The theorem says that the pictures on the left with their abstract definition pattern function plus functional equations look actually like the pictures on the right: properties 1-4 yield a unique dot (= function) which is the determinant, and properties 1-3 yield a straight $x\cdot \det$ with the determinant on it.

 

[Hall]

Somehow, I posted this in the Topology and Analysis sub-forum, can you please move it to the Linear Algebra sub-forum?

 

[Hall]

So, does it say that all the functions that satisfy axioms 1 to 3 will have this formula (no matter what)
$$
f (A_1, \cdots , A_n) = d (A_1, \cdots A_n) \cdot f(I_1, \cdots I_n)$$
?

 

[fresh_42]

So, does it say that all the functions that satisfy axioms 1 to 3 will have this formula (no matter what)
$$
f (A_1, \cdots , A_n) = d (A_1, \cdots A_n) \cdot f(I_1, \cdots I_n)$$
?

Yes. I wouldn't call it axioms though. Properties is in my opinion the better word here.

 

[Hall]

I just wanted to ask candidly, isn't it possible to prove directly that a function following all those four properties will be unique? Why they have gone for condition of following first three properties and then showing if it follows the fourth then it would be equal to the determinant function?

 

[PeroK]

I just wanted to ask candidly, isn't it possible to prove directly that a function following all those four properties will be unique? Why they have gone for condition of following first three properties and then showing if it follows the fourth then it would be equal to the determinant function?

Why is any proof the way it is? If you understand the given proof, then what is the issue? Learn from it and move on.

Or, you may learn a lot from trying to find your own proof.

 

[Hall]

Why is any proof the way it is? If you understand the given proof, then what is the issue? Learn from it and move on.

Or, you may learn a lot from trying to find your own proof.

I mean, it's a little different from all other uniqueness theorems where we assume another function following all those properties and then show that it's the same as the original one. In this way we have taken two steps, first we assumed a function which follows the first three axioms and then proved if it follows the fourth it would be same as the original one.

 

[fresh_42]

I just wanted to ask candidly, isn't it possible to prove directly that a function following all those four properties will be unique? Why they have gone for condition of following first three properties and then showing if it follows the fourth then it would be equal to the determinant function?

Because this is a piece of important information, too. Properties 1-3 are the algebraic properties of the determinant. They are what you check first. However, this solution is only almost unique. Up to a factor is pretty good.

It is a standard procedure in mathematics: Solve the problem in principle, e.g. solve a differential equation system and get a vector field, and specify the unique solutions among all of them, e.g. choose an initial value for the differential equation to find the unique path through this vector field.

The determinant is an oriented volume. Property 4 says that we can gauge this volume.

There are used different factors in the more general context of homological algebra, where some authors divide the boundary operator by $1/n!$ and some don't. It is important to know, that this doesn't change the function's qualities.

 

[Hall]

Properties 1-3 are the algebraic properties of the determinant.

The determinant is an oriented volume. Property 4 says that we can gauge this volume.

Those are really remarkable points!

Now, I see the whole point, the author wanted me to observe that there can be so many different functions following properties 1-3 and but as soon as we introduce the 4th property (which is not an algebraic one) there remains just one which follows it along with the previous three.

 

[Infrared]

Those are really remarkable points!

Now, I see the whole point, the author wanted me to observe that there can be so many different functions following properties 1-3 and but as soon as we introduce the 4th property (which is not an algebraic one) there remains just one which follows it along with the previous three.

I think the simpler way to read it is that if a function satisfies the first three requirements, it must be a multiple of the determinant function (where the proportionality constant is $f(I_1,\ldots,I_n).$)