basis-free definition of determinant

The definition of determinantMathworldPlanetmath as a multilinear mapping on rows can be modified to provide a basis-free definition of determinant. In order to make it clear that we are not using bases. we shall speak in terms of an endomorphism of a vector spaceMathworldPlanetmath over k rather than speaking of a matrix whose entries belong to k. We start by recalling some preliminary facts.

Suppose V is a finite-dimensional vector space of dimensionPlanetmathPlanetmath n over a field k. Recall that a multilinear map f:Vnk is alternating if f(x)=0 whenever there exist distinct indices i,j[n]={1,,n} such that xi=xj. Every alternating map f:Vnk is skew-symmetric, that is, for each permutation π𝔖n, we have that f(x)=sgn(π)f(xπ), where xπ denotes (xπ(i))i[n], the result of π permuting the entries of x.

Since the trivial map 0:Vnk is alternating and any linear combinationMathworldPlanetmath of alternating maps is alternating, it follows that alternating maps form a subspacePlanetmathPlanetmath of the space of multilinear maps. In the following propositionPlanetmathPlanetmathPlanetmath we show that this subspace is one-dimensional.


Suppose V is a finite-dimensional vector space of dimension n over a field k. Then the space of alternating maps from Vn to k is one-dimensional.


We use a basis here, but we will throw it away later. We need the basis here because each map we will consider has exactly as many elements as a basis of V. So let B={bi:i[n]} be a basis of V.

Suppose f and g are nontrivial alternating maps from Vn to k. We claim that f and g are linearly dependent. Let xVn. We may assume that the entries of x are basis vectors, that is, that X={xi:i[n]}{bi:i[n]}. If XB, then there exist distinct indices i,j[n] such that xi=xj. Since f and g are alternating, it follows that f(x)=g(x)=0, which implies that f(b)g(x)=g(b)f(x). On the other hand, if X=B, then there is a permutation π𝔖n such that x=bπ. Since f and g are skew-symmetric, it follows that


In either case we find that f(b)g(x)=g(b)f(x). Since f(b) and g(b) are fixed scalars, it follows that f and g are linearly dependent.

So far we have shown only that the dimension of the space of alternating maps is less than or equal to one. In order to show that the space is one-dimensional we simply need to find a nontrivial alternating form. To do this, let {bi*:i[n]} be the natural basis of V*, so that bi*(bj) is the Kronecker delta of i and j for any i,j[n]. Define a map f:Vnk by


One can check that f is multilinear and alternating. Moreover, f(b)=1, so it is nontrivial. Hence the space of alternating maps is one-dimensional. ∎

For an alternate view of the above results, we could look instead at linear maps from the exterior product nV into k. The proposition above can be viewed as saying that the dimension of nV is (nn)=1.

We define the determinant of an endomorphism in terms of the action of the endomorphism on alternating maps. Recall that if M:VV is an endomorphism, its pullback M* is the unique operator such that


Since the space of alternating maps is one-dimensional and endomorphisms of a one-dimensional space reduce to scalar multiplication, it follows that M*f is a scalar multiple of f. We call this scalar the determinant. It is well-defined because the scalar depends on M but not on f.


Suppose V is a finite-dimensional vector space of dimension n over a field k, and let M:VV be an endomorphism. Then the determinant of M is the unique scalar det(M) such that


for all alternating maps f:Vnk.

Title basis-free definition of determinant
Canonical name BasisfreeDefinitionOfDeterminant
Date of creation 2013-03-22 16:51:39
Last modified on 2013-03-22 16:51:39
Owner mps (409)
Last modified by mps (409)
Numerical id 9
Author mps (409)
Entry type Definition
Classification msc 15A15