How to generalize the determinant as function

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

• Did you google for the word "determinant"? – Peter Franek Mar 3 '16 at 18:41
• @HagenvonEitzen Don't you need antisymmetry for that? – Peter Franek Mar 3 '16 at 18:47
• @RobArthan You're implying that the OP should just check that the determinant satifies the given properties. But that's not enough. It has to be shown that there is a unique multilinear form with those properties. – Fryie Mar 7 '16 at 1:51
• @Qualtiy: your question mentions nothing about "construction" and you said in an earlier comment that you knew about determinants. If you are still unable to prove that the determinant has the uniqueness properties that you require, then you have my sympathies. – Rob Arthan Mar 7 '16 at 2:01
• I just want to make sure: Do you know what is the determinant of a $n\times n$ matrix when $n>2$? – user99914 Mar 7 '16 at 5:25

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 $D_\beta$.

• NB: This only proves the uniqueness of the function, not its existence. – darij grinberg Mar 10 '16 at 20:00
• Why shouldn't it exist, as it takes well defined values when its arguments belong to basis $\beta$? From what I've written above it follows that if $\sigma$ is a permutation of $(w_1,\ldots w_n)$ then $D_\beta(\sigma)=\hbox{sgn}(\sigma)$. – Intelligenti pauca Mar 10 '16 at 23:42
• Great answer! This makes sense to me – Quality Mar 11 '16 at 2:40
• @Aretino: If You assume the existence of the sign function, then sure, it proves existence of the determinant. – darij grinberg Mar 11 '16 at 8:04

There is a complaint Answers that do not help but take away attention about the answers here; so I will make this CW and put up with negative comments for a time.

This result is done in exquisite detail in Greub, a book I no longer have.

It is likely done in Axler. The idea that Axler might include this is somewhat supported by his early article; his book is intended for undergraduates. I do not have a copy.

Each set of vectors $(v_1,...,v_n)$ corresponds to a matrix (the coordinates of $v_k$ are in the column $k$). Using properties 1 and 2, the matrix may be diagonalized without changing the determinant, remember? After that, using property 3, your function is the product of the diagonal terms. The cases with too many zeros give a null determinant, as you should verify to complete the proof.

Example. For $n=2$, $v_1=\sum_{i=1}^2 a_{i1}w_i$, $v_2=\sum_{i=1}^2 a_{i2}w_i$ and $(v_1,v_2)$ corresponds to the matrix $$A=\left( \begin{array}{cc} a_{11} & a_{12} \\ a_{21} & a_{22} \end{array} \right).$$ The properties 1 and 2 of your function imply that $$D\left( \begin{array}{cc} a_{11} & ca_{11}+a_{12} \\ a_{21} & ca_{21}+a_{22} \end{array} \right)=cD\left( \begin{array}{cc} a_{11} & a_{11} \\ a_{21} & a_{21} \end{array} \right) +D\left( \begin{array}{cc} a_{11} & a_{12} \\ a_{21} & a_{22} \end{array} \right)= D\left( \begin{array}{cc} a_{11} & a_{12} \\ a_{21} & a_{22} \end{array} \right)=D(A).$$ For $a_{11}\neq0$, choose $c$ such that $ca_{11}+a_{12}=0$ and obtain $$D\left( \begin{array}{cc} a_{11} & 0 \\ a_{21} & -\frac{a_{12}a_{21}}{a_{11}}+a_{22} \end{array} \right)=D\left( \begin{array}{cc} a_{11} & a_{12} \\ a_{21} & a_{22} \end{array} \right).$$ Now eliminate $a_{21}$ without changing $D(A)$ and obtain $$D\left( \begin{array}{cc} a_{11} & 0 \\ 0 & -\frac{a_{12}a_{21}}{a_{11}}+a_{22} \end{array} \right)=D(A).$$ Now use properties 2 and 3 and obtain $D(A)=-a_{12}a_{21}+a_{11}a_{22}$.

• Fixed,thanks. Property 1 is antisymmetry, 2 is linearity and 3 is normalization at the identity matrix (corresponding to (w1,w2)). – user299632 Mar 10 '16 at 11:12

The argument is that the properties can be satisfied, because Determinant relative to the basis $w_i$ satisfies them ; and if a function has those properties it is equal to the Determinant (in the same basis).
The proof of uniqueness needs the fact that the sign of a permutation is well defined, so that different ways of sorting a permutation by transpositions lead to the same $\pm$ sign.
The antisymmetry / skew-symmetry / alternating property (the terminology varies) is usually stated as one of the properties of $D$. This exercise tries to be cuter or more difficult by making the point that antisymmetry can be deduced from the vanishing when $v_i = v_j$. In practice, when the vanishing at equal arguments is known, skew-symmetry is also known without having to specially deduce it.