basis-free definition of determinant
The definition of determinant 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 space over rather than speaking of a matrix whose entries belong to . We start by recalling some preliminary facts.
Suppose is a finite-dimensional vector space of dimension over a field . Recall that a multilinear map is alternating if whenever there exist distinct indices such that . Every alternating map is skew-symmetric, that is, for each permutation , we have that , where denotes , the result of permuting the entries of .
Since the trivial map is alternating and any linear combination of alternating maps is alternating, it follows that alternating maps form a subspace of the space of multilinear maps. In the following proposition we show that this subspace is one-dimensional.
Suppose is a finite-dimensional vector space of dimension over a field . Then the space of alternating maps from to 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 . So let be a basis of .
Suppose and are nontrivial alternating maps from to . We claim that and are linearly dependent. Let . We may assume that the entries of are basis vectors, that is, that . If , then there exist distinct indices such that . Since and are alternating, it follows that , which implies that . On the other hand, if , then there is a permutation such that . Since and are skew-symmetric, it follows that
In either case we find that . Since and are fixed scalars, it follows that and 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 be the natural basis of , so that is the Kronecker delta of and for any . Define a map by
One can check that is multilinear and alternating. Moreover, , so it is nontrivial. Hence the space of alternating maps is one-dimensional. ∎
Since the space of alternating maps is one-dimensional and endomorphisms of a one-dimensional space reduce to scalar multiplication, it follows that is a scalar multiple of . We call this scalar the determinant. It is well-defined because the scalar depends on but not on .
Suppose is a finite-dimensional vector space of dimension over a field , and let be an endomorphism. Then the determinant of is the unique scalar such that
for all alternating maps .
|Title||basis-free definition of determinant|
|Date of creation||2013-03-22 16:51:39|
|Last modified on||2013-03-22 16:51:39|
|Last modified by||mps (409)|