Proof of ##g(A_1, A_2, \cdots A_n) = c g (I_1, \cdots I_n)##.

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SUMMARY

The discussion centers on proving the equation $$g(A_1, \cdots A_n) = c g(I_1, \cdots I_n)$$, where g is an alternating function that exhibits anti-symmetry. The participants explore the implications of transforming a matrix A into an identity matrix I using the Gauss-Jordan elimination method, noting that this process involves n row interchanges and m scalar multiplications. The conclusion drawn is that $$g(A_1, \cdots, A_n) = (-1)^n c_1 \cdots c_m g(I_1, \cdots I_n)$$, emphasizing the relationship between the properties of g and the determinant.

PREREQUISITES
  • Understanding of alternating functions and anti-symmetry in mathematics.
  • Familiarity with the Gauss-Jordan elimination method for matrix transformations.
  • Knowledge of multilinear maps and their properties.
  • Basic concepts of determinants in linear algebra.
NEXT STEPS
  • Study the properties of alternating functions in linear algebra.
  • Learn about the Gauss-Jordan elimination method in detail.
  • Research multilinear maps and their applications in mathematics.
  • Explore the relationship between determinants and anti-symmetric functions.
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Mathematicians, students of linear algebra, and anyone interested in the properties of alternating functions and determinants will benefit from this discussion.

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Homework Statement
Let's say ##g## is function satisfying the following three axioms, where ##A_i## is any n-tuple vector and ##I_k## is the third unit vector,

1. ##g(A_1, \cdots tA_k, \cdots A_n) = t g(A_1, \cdots A_k, \cdots A_n)## for all ##t \in R## and any ##A_k##

2. ## g(A_1, \cdots A_k + C, \cdots A_n) = g(A_1\cdots A_k, \cdots A_n )+ g(A_1, \cdots, C ,\cdots A_n)## for any n-tuple vector C and any ##A_k##

3. ##g(A_1, \cdots A_n)=0 ## if for some i and j ##A_i =A_j##
Relevant Equations
In fact, those ##A_k## are rows of ##n \times n## matrix.
How can we prove that
$$
g(A_1, \cdots A_n)= c g(I_1 \cdots I_n)$$?

From the those three axioms we can prove a property of g that if any of two vectors in domain exchange their respective places the sign of output of g will be changed.

Now, do we have to argue that any matrix can be changed into identity matrix by Gauss-Jordan elimination method?
 
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What are your thoughts? That 3 (being alternating) means that ##g## is fully anti-symmetric is a good start. What can you say about the As in terms of the Is?

Edit: The problem is also not very well defined. What is ##c## allowed to depend on? That two numbers differ through the multiplication of a different number is not a very restrictive statement.
 
Considering each ##A_i## as rows of n-square matrix.

Let’s say moving from the matrix A to the identity matrix I, by Gauss-Jordan Method, it takes n number of interchangement of rows and m scalars were mutliplied: ##c_1, c_2, \cdots c_m## altogether in reaching from A to I, neglecting the adding of one row to the multiple of another, we can conclude:

1. Moving from A to I, ##g(A_1, \cdots, A_n)## changed signs n times, and

2. Moving from A to I, ##g(A_1, \cdots, A_n)## got multiplied by m different numbers, they are ##c_1 \cdots c_m##.

Therefore,
$$
g(A_1 \cdots, A_n) = (-1)^n c_1 \cdots c_m g (I_1, \cdots I_n)$$

I feel like I was not very rigorous.
 
I do not understand what you are trying to do here.
 
This reminds me of the properties defining the determinant, only multilinear map satisfying the conditions described in the OP. Iirc, it follows immediately by applying multilinearity .
 
I don't think the problem is very well stated. What is c? Is ## I_k ## really the 3rd unit vector for every k; that would make ## g(I_1, ..., I_n) = 0 ##. Also, please write ## , \ldots, ## instead of ## \cdots ## unless you really mean iterated multiplication.
 

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