# Linear transformation that preserves the determinant.

It seems "common knowledge" that the following holds:

Let T be a linear transformation on nxn matrices with complex coefficients that preserves the determinant. Then there exists matrices U and V whose product has determinant 1 such that one of the following holds:

a) For any matrix A we have T(A)=UAV
b) For any matrix A we have T(A)=UBV where B is the transpose of A

It seems quite reasonable, but as far as "common knowledge" goes, I have no clue right now on how to prove such a thing?

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What happened to the answer that was here a few minutes ago? –  Ohdarkdevil Oct 14 '09 at 22:59
It was deleted by the person who posted it. –  Anton Geraschenko Oct 14 '09 at 23:04

The conclusion you indicate is obtained as the main result in the following paper, but with an apparently stronger hypothesis: (EDIT: Not stronger at all, actually - just realized you're assuming the map is linear.)

Determinant preserving maps on matrix algebras

Gregor Dolinar and Peter Semrl

Linear Algebra and its Applications Volume 348, Issues 1-3, 15 June 2002, Pages 189-192

Let Mn be the algebra of all n×n complex matrices. If φ:Mn→Mn is a surjective mapping satisfying det(A+λB)=det(φ(A)+λφ(B)) then either φ is of the form φ(A)=MAN or φ is of the form φ(A)=MAtN where M,N are nonsingular matrices with det(MN)=1.

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It seems to me that the added hypothesis follows from the requirement that phi is linear. Am I missing something? –  David Speyer Oct 15 '09 at 0:11
Of course - just missed that. It may actually make the claim easier to prove, I'm not sure. –  Alon Amit Oct 15 '09 at 1:14