Hi I was asked to show that for any vector space $V$ over a field $\mathbb{F}$ of arbitrary dimension $n$ that if we fix some basis $\beta=\{w_1,\ldots,w_n\}$ that there is a unique function
$D_\beta : V \times \cdots \times V \to \mathbb{F}$ that will satisfy the three properties,
$D_\beta(v_1,\ldots,v_n)=0$ if some $v_i=v_j, i \neq j$
$D_\beta$ is linear in each factor and
$D_\beta(w_1,\ldots,w_n)=1$
Moreover, I want to be able to prove some things using this.
For example that Applying D to some vectors v will be the same as taking the determinant of the coordinate representation of those v, that applying D and getting something non zero to a set of vectors implys those vectors are not a basis, etc.
I know that in the case of two dimensional, we can do this by using the idea that the determinant gives the area of a parallelepiped, etc. And if we define the function to be that that satisfies the above properties then it can be easily shown to be unique.
But in cases when it is not dim 2, since I dont have a general formula for the n vectors , how can this be done? Is it safe to assume that such a function exists and then just show it is unique, or must it be shown that even such a function exists at all?
And say we could just define the function that satisfies those properties. Then how would we show it is unique, etc?
Is it possible to maybe do this problem using matrices? Ie, we can define the determinant on the nxn matrices as our inputs, and somehow use that to prove things?
I have not learned about things called alternating maps.
I am still trying to do this , using advice in answers, is this how it should be approached?
Define a mapping $f: L(V^{n}, \mathbb{F}) \to \mathbb{F}$ by $f \to f(e_1,…,e_n)$, or should it be as $f \to f(x_1,…x_n)$?
Then $f$ is injective as if $f_{1}(e_1,…,e_n)=f_{2}(e_1,…,e_n)$ then would this imply they are equal as how a function acts on a basis completely determines it? I a just really confused. and then If I could show that it is subjective and thus an isomorphism I could say choose that unique f which is the one that gives $f(x_1,\ldots,x_n)=0$ and claim this is the function $D_\beta$ I wanted. But I am having lots of trouble putting it all together.
I still dont understand. It seems like the top answer is getting many vote, but I dont understand it. This is only a first course in linear algebra by the way, so I do not know about many advanced results.
Maybe the answer could rely on sgn function and permutations. Then I could have a general formula
I dont know how in some of the answers we can just simply say suppose D is alternating etc, when this is one of the things I want to prove. That is what is also confusing me. I dont know what we can even start with. Or if we must start from scratch completely.
Looking for advice.
If this cannot be done so easily, I would also be happy to see that if we assume such a function does exist, then at least prove it is unique. Without just saying the determinant is that function and the determinant is unique.
Thank you
Best Answer
Notice that $$ D_\beta(\ldots, \underbrace{w_i+w_j}_{k\text{-th place}},\ldots, \underbrace{w_i+w_j}_{m\text{-th place}},\ldots)=0 $$ because there are two equal vectors. On the other hand, by linearity: $$ D_\beta(\ldots, {w_i+w_j},\ldots, {w_i+w_j},\ldots)= D_\beta(\ldots, w_i,\ldots, w_j,\ldots)+ D_\beta(\ldots, w_j,\ldots, w_i,\ldots), $$ because $D_\beta(\ldots, w_i,\ldots, w_i,\ldots)=D_\beta(\ldots, w_j,\ldots, w_j,\ldots)=0$. You get then: $$ D_\beta(\ldots, w_i,\ldots, w_j,\ldots)= -D_\beta(\ldots, w_j,\ldots, w_i,\ldots), $$ that is your function, applied to basis vectors, changes of sign whenever you exchange two of its arguments. Together with the request $D_\beta(w_1, \ldots, w_n)=1$ this completely determines the values of $D_\beta$ when its arguments are basis vectors. It follows by linearity that $D_\beta$ is uniquely determined.
To show that applying $D_\beta$ to some vectors will be the same as taking the determinant of the coordinate representation of those vectors, one would need a definition of determinant. My favourite definition of determinant is indeed the same as the definition of $D_\beta$.