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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

2 Answers 2

First, some easy observations: T is injective since for any A, there is some B such that the matrices tA+B have different determinants for different scalars t (easy exercise). By multiplying T by T(1)^{-1}, it may be assumed that T(1)=1.

Now note that T preserves the rank of matrices. Indeed, T must preserve the rank n matrices, and then the rank n-1 matrices are just the nonsingular locus in the variety of matrices with determinant 0. This implies T preserves rank n-1 matrices. Rank n-2 matrices are then the nonsingular locus in rank < n-1 matrices so they are preserved, and so on.

Now rank k projections are exactly those rank k matrices which when subtracted from the identity give you something of rank n-k; this is easy to see from Jordan normal form. Thus T sends rank 1 projections to rank 1 projections. Two projections have disjoint ranges and commute iff their sum is also a projection. In particular, for Pi the projections onto a basis ei, T sends Pi to projections Qi onto some other basis fi. Now let U be the change of basis matrix from the ei to the fi. Conjugating T by U shows that we may assume T fixes each Pi. That is, picking the standard basis, T fixes all diagonal matrices.

Now matrices whose only nonzero entries are all either in the first row or first column are characterized by the fact that they are rank 1 and they remain rank 1 if their first diagonal entry changes but they become rank 2 if any other diagonal entry changes. Similar statements hold for other rows and columns. It follows that T(eij) is a multiple of either eij or eji for all j and i, for eij the matrix with ij entry 1 and all others 0. It is now easy to check that we must either always have T(eij)=eij or always have T(eij)=eji, i.e. T(A)=A or T(A)=A^t.

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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

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