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Given (finite dimensional?) vector spaces $V$ and $W$, we can define the transpose of a linear map $f:V \to W$ by the obvious map $W^* \to V^*$. Can we do a similar thing for the determinant? Can we define the determinant of an operator without making reference to a matrix, or bases? I feel we should be able to do this, given that the value of the determinant is independent of the basis chosen.

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Are you aware of the alternating product of vector spaces? – user27126 Jan 18 '13 at 4:41
up vote 3 down vote accepted

The determinant of a linear map $f:V\to W$ between finite dimensional vector spaces over a field $K$ can only be defined if $V=W \; !$
And be very careful: we really have to require that our vector spaces be equal and not merely isomorphic.

The intrinsic definition is then as follows :
If $V$ has dimension $n$ then the top exterior power $\wedge^n V$ has dimension one so that the induced linear mapping $\wedge^n f:\wedge^n V\to \wedge^n V$ is necessarily a homothety $\omega\mapsto d \cdot\omega$ i.e. the multiplication by some scalar $d\in K$ in the base field.
And that scalar is the required determinant: $\text {det} f\stackrel {\text {def}}{=}d$.

This approch (pioneered by Bourbaki around 1950, I think) is undoubtedly very elegant, but it requires a knowledge of multilinear algebra not taught in elementary undergraduate courses.

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What problem does it have if $V \cong W$? Surely we can still take the top exterior power on both sides? – user27126 Jan 18 '13 at 5:08
Dear @Sanchez: yes, you can take the top exterior product of both sides but the resulting linear map is not a homothety. The problem is that a linear map between two different vector spaces of dimension one is not a homothety and it is absolutely impossible to describe it by a single scalar unless you choose a base in each or the vector spaces. This is an elementary but subtle point – Georges Elencwajg Jan 18 '13 at 5:14
Ah, I see your point. Thanks! – user27126 Jan 18 '13 at 5:16
Yes, I see that the determinant can only be defined when $V=W$. I should have been more careful. – nigelvr Jan 18 '13 at 13:57
Dear nigelvr, don't be too harsh on yourself: this is a rather delicate point, perhaps not sufficiently emphasized in books. – Georges Elencwajg Jan 18 '13 at 14:50

Elaborating on Sanchez's answer, it's useful to consider the action of a linear operator on objects created by wedge products--which represent oriented planes, volumes, and so on, just as a single vector represents an oriented line.

We generally define the action of linear operators across the wedge product as follows:

$$\underline T(a \wedge b) \equiv \underline T(a) \wedge \underline T(b)$$

The operator on a wedge product is the wedge product of the operator on the individual vectors.

Now, consider the highest-dimensional object that can be formed by wedges. In an $N$-dimensional space, this is a wedge product of $N$ linearly independent vectors. This object itself forms a 1d vector space---all objects of this kind are scalar multiples of each other. For this reason, this object is often called the pseudoscalar. We'll call it $i_N$.

Now, what is $\underline T(i_N)$? First, linear operators that can be extended across wedges preserve the grade of their arguments--vectors go to vectors, planes to go planes, and the pseudoscalar can only go to (some multiple of) itself.

In other words,

$$\underline T(i_N) = \alpha i_N$$

We call the number $\alpha$ the determinant, and it describes how a unit volume in the space is dilated or shrunk by the linear operator. There is no need to resort to a matrix representation to do this, nor must we choose a specific basis to know this is so.

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